Re: [R] bootstrapping quantile regression
sry, I forgot to replace rlm() - but actually I tried both and the question
applies to both approaches..
Am 31.10.2012 00:19 schrieb Kay Cichini kay.cich...@gmail.com:
HI everyone,
I try to get some bootstrap CIs for coefficients obtained by quantile
regression. I have influencial values and thus switched to quantreg..
The data is clustered and within clusters the variance of my DV = 0..
Is this sensible for the below data? And what about the warnings?
Thanks in advance for any guidance,
Kay
dput(d)
structure(list(Porenfläche = c(4990L, 7002L, 7558L, 7352L, 7943L,
7979L, 9333L, 8209L, 8393L, 6425L, 9364L, 8624L, 10651L, 8868L,
9417L, 8874L, 10962L, 10743L, 11878L, 9867L, 7838L, 11876L, 12212L,
8233L, 6360L, 4193L, 7416L, 5246L, 6509L, 4895L, 6775L, 7894L,
5980L, 5318L, 7392L, 7894L, 3469L, 1468L, 3524L, 5267L, 5048L,
1016L, 5605L, 8793L, 3475L, 1651L, 5514L, 9718L), P.Perimeter = c(2791.9,
3892.6, 3930.66, 3869.32, 3948.54, 4010.15, 4345.75, 4344.75,
3682.04, 3098.65, 4480.05, 3986.24, 4036.54, 3518.04, 3999.37,
3629.07, 4608.66, 4787.62, 4864.22, 4479.41, 3428.74, 4353.14,
4697.65, 3518.44, 1977.39, 1379.35, 1916.24, 1585.42, 1851.21,
1239.66, 1728.14, 1461.06, 1426.76, 990.388, 1350.76, 1461.06,
1376.7, 476.322, 1189.46, 1644.96, 941.543, 308.642, 1145.69,
2280.49, 1174.11, 597.808, 1455.88, 1485.58), P.Form = c(0.0903296,
0.148622, 0.183312, 0.117063, 0.122417, 0.167045, 0.189651, 0.164127,
0.203654, 0.162394, 0.150944, 0.148141, 0.228595, 0.231623, 0.172567,
0.153481, 0.204314, 0.262727, 0.200071, 0.14481, 0.113852, 0.291029,
0.240077, 0.161865, 0.280887, 0.179455, 0.191802, 0.133083, 0.225214,
0.341273, 0.311646, 0.276016, 0.197653, 0.326635, 0.154192, 0.276016,
0.176969, 0.438712, 0.163586, 0.253832, 0.328641, 0.230081, 0.464125,
0.420477, 0.200744, 0.262651, 0.182453, 0.200447), Durchlässigkeit =
c(6.3,
6.3, 6.3, 6.3, 17.1, 17.1, 17.1, 17.1, 119, 119, 119, 119, 82.4,
82.4, 82.4, 82.4, 58.6, 58.6, 58.6, 58.6, 142, 142, 142, 142,
740, 740, 740, 740, 890, 890, 890, 890, 950, 950, 950, 950, 100,
100, 100, 100, 1300, 1300, 1300, 1300, 580, 580, 580, 580), Gebiete =
structure(c(1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 6L, 6L, 6L, 6L, 4L, 4L, 4L, 4L, 3L,
3L, 3L, 3L, 7L, 7L, 7L, 7L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L,
11L, 11L, 11L, 11L, 5L, 5L, 5L, 5L, 12L, 12L, 12L, 12L, 8L, 8L,
8L, 8L), .Label = c(6.3, 17.1, 58.6, 82.4, 100, 119,
142, 580, 740, 890, 950, 1300), class = factor)), .Names =
c(Porenfläche,
P.Perimeter, P.Form, Durchlässigkeit, Gebiete), row.names =
c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46,
47, 48), class = data.frame)
## do quantile regression and bootstrap the coefficients, allowing for
clustered data
## by putting Gebiet as strata argument (?),
## dv variation within clusters/Gebiet = 0!
bs - function(formula, data, indices) {
d - data[indices, ] # allows boot to select sample
fit - rlm(formula, data = d)
return(coef(fit))
}
results - boot(data = d, statistic = bs, strata = d$Gebiete,
R = 199, formula = Durchlässigkeit ~ P.Perimeter + P.Form)
# get 99% confidence intervals
boot.ci(results, type=bca, index=1, conf = .99) # intercept
boot.ci(results, type=bca, index=2, conf = .99) # P.Perimeter
boot.ci(results, type=bca, index=3, conf = .99) # P.Form
--
Kay Cichini, MSc Biol
Grubenweg 22, 6071 Aldrans
E-Mail: kay.cich...@gmail.com
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Re: [R] POSIXct date missing time component
I did not read that close enough; thanks again. arun kirshna wrote Hi, ?as.POSIXct() format: character string giving a date-time format as used by ‘strptime’. ?strptime() format: A character string. The default for the ‘format’ methods is ‘%Y-%m-%d %H:%M:%S’ if any component has a time component which is not midnight, and ‘%Y-%m-%d’ otherwise. If ‘options(digits.secs)’ is set, up to the specified number of digits will be printed for seconds. So, may be it is better to add a sec and later subtract it. dates.mine2-dates.mine+1 dates.mine2[3] #[1] 2009-05-22 00:00:01 GMT dates.mine3-dates.mine2-1 dates.mine3[3] #[1] 2009-05-22 GMT A.K. - Original Message - From: chuck.01 lt; CharlieTheBrown77@ gt; To: r-help@ Cc: Sent: Tuesday, October 30, 2012 5:21 PM Subject: Re: [R] POSIXct date missing time component Um, OK. My dates have times, they are 00:00:00 (i.e. midnight) I'll just add a fraction of a second my dates and go with it. Thanks for the reply. arun kirshna wrote HI, Please check this link (http://rss.acs.unt.edu/Rdoc/library/base/html/as.POSIXlt.html). Dates without times are treated as being at midnight UTC. May be you can try this (if it doesn't create additional problems): dates.mine2-dates.mine+1 dates.mine2 #[1] 2009-05-21 23:30:01 GMT 2009-05-21 23:45:01 GMT #[3] 2009-05-22 00:00:01 GMT 2009-05-22 00:15:01 GMT dates.mine2[3] #[1] 2009-05-22 00:00:01 GMT str(dates.mine2) #POSIXct[1:4], format: 2009-05-21 23:30:01 2009-05-21 23:45:01 ... A.K. -- View this message in context: http://r.789695.n4.nabble.com/POSIXct-date-missing-time-component-tp4647932p4647939.html Sent from the R help mailing list archive at Nabble.com. __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/POSIXct-date-missing-time-component-tp4647932p4647963.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R crashing after successfully running compiled code
I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32 and 2.13.0x64. I am trying to run compiled C code using the .C convention. The code compiles without problems, dynamically loads within the R workspace with no problems, and even runs and gives correct results with no problems. However, R will randomly crash within a few minutes of successfully using the compiled function. For example, if I run my compiled function using: dyn.load(mycfun.dll) answer-.C(mycfun, parameters...), I get a completely sensible result that gets stored to answer. However, if I try to do too many things to answer, the R exits without warning. I've tried dyn.unload in hopes that R would become stable afterwards, but in this case using the function crashes R without fail. Usually, I can either plot, or view, or save answer to a file - but never take more than a single action before R exits. This does not appear to depend on how long R has been open. Initially, I thought it was a bug in the inline function, but I'm finding the same problem now that I'm using the dynamically loaded file directly. I'm used to R being insanely stable, and am somewhat mystified by this whole problem. My next move is to learn the .Call convention, as I suspect that my problem is related to my C function using memory that R doesn't know is used. But - before I invest a while lot more time on this, I'd like to know whether anybody things this is likely to solve the problem. If not, I may just want to run my code entirely in C, and forget the R problem. -- Adam Clark University of Minnesota, EEB 100 Ecology Building 1987 Upper Buford Circle St. Paul, MN 55108 (857)-544-6782 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] IBrokers readBin is of argument NULL
hi
I ran the below code, its throwing some error.
contract - twsEquity(FAS)
reqMktData(tws, contract)
Error in if (curMsg == .twsIncomingMSG$TICK_PRICE) { :
argument is of length zero
So i ran this following code to check the location of the problem
socketSelect(list(con), write=FALSE, timeout=15)
[1] TRUE
readBin(con,character(),1) # which is the curMsg
character(0)
The above readBin command is always giving the same result as above. I think
its not able to read any msg from TWS API or TWS API is not sending any msg
(may be i should change settings). Can some one please help with this
problem ?
Thanks
Prabakar
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Re: [R] R crashing after successfully running compiled code
On Oct 31, 2012, at 3:13 AM, Adam Clark atcl...@umn.edu wrote: I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32 and 2.13.0x64. I am trying to run compiled C code using the .C convention. The code compiles without problems, dynamically loads within the R workspace with no problems, and even runs and gives correct results with no problems. However, R will randomly crash within a few minutes of successfully using the compiled function. For example, if I run my compiled function using: dyn.load(mycfun.dll) answer-.C(mycfun, parameters...), I get a completely sensible result that gets stored to answer. However, if I try to do too many things to answer, the R exits without warning. I've tried dyn.unload in hopes that R would become stable afterwards, but in this case using the function crashes R without fail. Usually, I can either plot, or view, or save answer to a file - but never take more than a single action before R exits. This does not appear to depend on how long R has been open. Initially, I thought it was a bug in the inline function, but I'm finding the same problem now that I'm using the dynamically loaded file directly. I'm used to R being insanely stable, and am somewhat mystified by this whole problem. My next move is to learn the .Call convention, as I suspect that my problem is related to my C function using memory that R doesn't know is used. But - before I invest a while lot more time on this, I'd like to know whether anybody things this is likely to solve the problem. If not, I may just want to run my code entirely in C, and forget the R problem. Hi Adam, Can you make a minimal reproducible example of your C sources available? I'm relatively certain that the problem is in the memory management therein, but I obviously can't say more without seeing the code. Michael -- Adam Clark University of Minnesota, EEB 100 Ecology Building 1987 Upper Buford Circle St. Paul, MN 55108 (857)-544-6782 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] POSIXct date missing time component
Rather than altering your data, it is better to use the same date time format at all times, and specify it explicitly when converting to or from character. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. chuck.01 charliethebrow...@gmail.com wrote: I did not read that close enough; thanks again. arun kirshna wrote Hi, ?as.POSIXct() format: character string giving a date-time format as used by ‘strptime’. ?strptime() format: A character string. The default for the ‘format’ methods is ‘%Y-%m-%d %H:%M:%S’ if any component has a time component which is not midnight, and ‘%Y-%m-%d’ otherwise. If ‘options(digits.secs)’ is set, up to the specified number of digits will be printed for seconds. So, may be it is better to add a sec and later subtract it. dates.mine2-dates.mine+1 dates.mine2[3] #[1] 2009-05-22 00:00:01 GMT dates.mine3-dates.mine2-1 dates.mine3[3] #[1] 2009-05-22 GMT A.K. - Original Message - From: chuck.01 lt; CharlieTheBrown77@ gt; To: r-help@ Cc: Sent: Tuesday, October 30, 2012 5:21 PM Subject: Re: [R] POSIXct date missing time component Um, OK. My dates have times, they are 00:00:00 (i.e. midnight) I'll just add a fraction of a second my dates and go with it. Thanks for the reply. arun kirshna wrote HI, Please check this link (http://rss.acs.unt.edu/Rdoc/library/base/html/as.POSIXlt.html). Dates without times are treated as being at midnight UTC. May be you can try this (if it doesn't create additional problems): dates.mine2-dates.mine+1 dates.mine2 #[1] 2009-05-21 23:30:01 GMT 2009-05-21 23:45:01 GMT #[3] 2009-05-22 00:00:01 GMT 2009-05-22 00:15:01 GMT dates.mine2[3] #[1] 2009-05-22 00:00:01 GMT str(dates.mine2) #POSIXct[1:4], format: 2009-05-21 23:30:01 2009-05-21 23:45:01 ... A.K. -- View this message in context: http://r.789695.n4.nabble.com/POSIXct-date-missing-time-component-tp4647932p4647939.html Sent from the R help mailing list archive at Nabble.com. __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/POSIXct-date-missing-time-component-tp4647932p4647963.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] standard error for quantile
On 10/31/2012 12:46 AM, PIKAL Petr wrote: Dear all I have a question about quantiles standard error, partly practical partly theoretical. I know that x-rlnorm(10, log(200), log(2)) quantile(x, c(.10,.5,.99)) computes quantiles but I would like to know if there is any function to find standard error (or any dispersion measure) of these estimated values. And here is a theoretical one. I feel that when I compute median from given set of values it will have lower standard error then 0.1 quantile computed from the same set of values. Is it true? If yes can you point me to some reasoning? Hi Petr, Using a resampling method, it depends upon the distribution of the values. If you have a love-hate distribution (bimodal and heavily weighted toward extreme values), the median standard error can be larger. Try this: x-sample(-5:5,1000,TRUE, prob=c(0.2,0.1,0.05,0.04,0.03,0.02,0.03,0.04,0.05,0.1,0.2)) x-ifelse(x0,x+runif(1000),x-runif(1000)) hist(x) mcse.q(x, 0.1) $est [1] -3.481419 $se [1] 0.06887319 mcse.q(x, 0.5) $est [1] 1.088475 $se [1] 0.3440115 mcse.q(x, 0.1) $est [1] -3.481419 $se [1] 0.06887319 Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] aggregate.formula: formula from string
Dear all, I want to use aggregate.formula to conveniently summarize a data.frame. I have quiet some variables in the data.frame and thus I don't want to write all these names by hand, but instead create them on the fly. This approach has the advantage that if there will be even more columns in the data.frame I don't have to change the code. I've hence tried to construct a formula object and to pass that to aggregate: d - expand.grid(a = factor(1:3), b = factor(LETTERS[1:2])) d - rbind(d,d,d) d$y - rnorm(18) d$z - rnorm(18) mF - as.formula(paste(cbind(, paste(names(d)[-(1:2)], collapse = ,), ) ~ a + b, sep = )) But if I try to pass that formula to aggregate aggregate(mF, d, mean) I get the following error: Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable But if I pass the formula directly: aggregate(cbind(y, z) ~ a + b, d, mean) Everything is working as expected. So I was wondering what went wrong? I know I could use a formula like . ~ a + b instead and this would work fine, but I'm just interested in why the outlined approach does not work as expected, and where my mistake lies? (that means in particular I am not asking for a solution of how to get the thing done - there are plenty of alternatives - but instead to understand why this very approach does not work) Thanks for your help! Kind Regards, Thorn Thaler Mathematician Applied Mathematics Nestec Ltd, Nestlé Research Center PO Box 44 CH-1000 Lausanne 26 Phone: +41 21 785 8220 Fax: +41 21 785 9486 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] standard error for quantile
Thanks Jim. After reinstall of new R version all mentioned packages work. I tested various functions which revealed that on my lognorm data there is no big difference in error of median or 10% quantile. I also found some function for quantile se computing in Hmisc package. Petr -Original Message- From: Jim Lemon [mailto:j...@bitwrit.com.au] Sent: Wednesday, October 31, 2012 9:56 AM To: PIKAL Petr Cc: r-help@r-project.org Subject: Re: [R] standard error for quantile On 10/31/2012 12:46 AM, PIKAL Petr wrote: Dear all I have a question about quantiles standard error, partly practical partly theoretical. I know that x-rlnorm(10, log(200), log(2)) quantile(x, c(.10,.5,.99)) computes quantiles but I would like to know if there is any function to find standard error (or any dispersion measure) of these estimated values. And here is a theoretical one. I feel that when I compute median from given set of values it will have lower standard error then 0.1 quantile computed from the same set of values. Is it true? If yes can you point me to some reasoning? Hi Petr, Using a resampling method, it depends upon the distribution of the values. If you have a love-hate distribution (bimodal and heavily weighted toward extreme values), the median standard error can be larger. Try this: x-sample(-5:5,1000,TRUE, prob=c(0.2,0.1,0.05,0.04,0.03,0.02,0.03,0.04,0.05,0.1,0.2)) x-ifelse(x0,x+runif(1000),x-runif(1000)) hist(x) mcse.q(x, 0.1) $est [1] -3.481419 $se [1] 0.06887319 mcse.q(x, 0.5) $est [1] 1.088475 $se [1] 0.3440115 mcse.q(x, 0.1) $est [1] -3.481419 $se [1] 0.06887319 Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] standard error for quantile
Hi Bert -Original Message- From: Bert Gunter [mailto:gunter.ber...@gene.com] Sent: Tuesday, October 30, 2012 3:37 PM To: PIKAL Petr Cc: r-help@r-project.org Subject: Re: [R] standard error for quantile Petr: 1. Not an R question. Partly. I asked also for pointing me to some R functions which can compute such standard error, which is by my humble opinion valid R question. 2. You want the distribution of order statistics. Search on that. It's basically binomial/beta. Hm. I looked at some web info, which is quite good for trained statistician but at the edge of my understanding as chemist (and sometimes beyound:-). Thanks anyway. Regards Petr -- Bert On Tue, Oct 30, 2012 at 6:46 AM, PIKAL Petr petr.pi...@precheza.cz wrote: Dear all I have a question about quantiles standard error, partly practical partly theoretical. I know that x-rlnorm(10, log(200), log(2)) quantile(x, c(.10,.5,.99)) computes quantiles but I would like to know if there is any function to find standard error (or any dispersion measure) of these estimated values. And here is a theoretical one. I feel that when I compute median from given set of values it will have lower standard error then 0.1 quantile computed from the same set of values. Is it true? If yes can you point me to some reasoning? Thanks for all answers. Regards Petr PS. I found mcmcse package which shall compute the standard error but which I could not make to work probably because I do not have recent R-devel version installed Error in eval(expr, envir, enclos) : could not find function .getNamespace Error : unable to load R code in package 'mcmcse' Error: package/namespace load failed for 'mcmcse' Maybe I will also something find in quantreg package, but I did not went through it yet. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb- biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Hausman test error solve
Liebe Gloria,
the error says it all: you have a singular covariance matrix (the
vcov(q) term in the Hausman test); this is not invertible and therefore
the test fails.
As for the reasons, this might be either a bug or perhaps an
ill-conditioned problem: it is impossible for me to tell without a
reproducible example. I can only suggest an alternative strategy (which
you must tailor to your example).
Another (computationally more robust) way to perform a Hausman test is
through an auxiliary regression (see Wooldridge, Econometric analysis
of cross-section and panel data, 10.7.3)
Hausman test by auxiliary regression
library(plm)
## standard Munnell example
data(Produc)
fm-log(gsp)~log(pcap)+log(pc)+log(emp)+unemp
re-plm(fm, data=Produc, model=random)
fe-plm(fm, data=Produc, model=within)
## extract transformed data
y.re-pmodel.response(re)
X.re-model.matrix(re)
X.fe-model.matrix(fe)
## check coefficients
coef(re)
lm(y.re~X.re-1)
coef(fe)
lm(pmodel.response(fe)~X.fe-1)
## make aux dataset
auxdata-data.frame(cbind(y.re,X.re,X.fe))
colnames(auxdata)-c(y, paste(x, 1:9, sep=))
## auxiliary model
auxmod-lm(y~x1+x2+x3+x4+x5+x6+x7+x8+x9-1, auxdata)
## regression-based Hausman test is exclusion test for 6-9
library(lmtest)
waldtest(auxmod, 6:9)
I will eventually add a production version of this to 'plm' as
phtest2().
Best wishes nach Koeln
Giovanni
Giovanni Millo, PhD
Research Dept.,
Assicurazioni Generali SpA
Via Machiavelli 4,
34132 Trieste (Italy)
tel. +39 040 671184
fax +39 040 671160
original message --
Date: Mon, 29 Oct 2012 13:14:09 -0700 (PDT)
From: gloria gbus...@smail.uni-koeln.de
To: r-help@r-project.org
Subject: [R] Hausman test error solve
Message-ID: 1351541649650-4647793.p...@n4.nabble.com
Content-Type: text/plain; charset=UTF-8
Hello,
I am trying to conduct a Hausman test to choose between FE estimators
and RE
estimators.
When I try to run:
library(plm)
fixed - plm(ROS ~ DiffClosenessC +ZZiele + AggSK + nRedundantStrecken +
Degree + KantenGew + BetweennessC + SitzKappazitaet,
data=Panel,index=c(id,time),model=within)
summary(fixed)
fixef(fixed)
random -plm(ROS ~ DiffClosenessC +ZZiele + AggSK + nRedundantStrecken +
Degree + KantenGew + BetweennessC + SitzKappazitaet,
data=Panel,index=c(id,time), model=random)
summary(random)
phtest(fixed, random)
I get an error from phtest(fixed, random)
Fehler in solve.default(dvcov) :
System ist f?r den Rechner singul?r: reziproke Konditionszahl = ...
Error in solve.default(dvcov) :
system is computationally singular: reciprocal condition number =...
Can someone Help me with this problem or give me a hint where to look?
Thanks
gloria
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Re: [R] boxplots of various levels
Have a look at this:
http://stats.stackexchange.com/questions/15486/plotting-a-boxplot-against-multiple-factors-in-r-with-ggplot2
Hope it helps.
José
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of dysonsphere
Sent: 30 October 2012 18:22
To: r-help@r-project.org
Subject: [R] boxplots of various levels
noob here
trying to make boxplots of some data
i would like to separate the boxplots according to conditons of various
levels
for example:
i have
group:1 and 2, each group performed tests consisting of
condition A,B,C,D
side: left and right
time: 1 to 10
I would like separate boxplots of the results (x) of the tests (numeric) for
each group under each condition on each side over time.
so far i have set it up like this:
boxplot(test$x~test$time)
this gives me the plot for all vaues of x in each time bin. basicaly i
would need a command that tells R to include only the data that agrees with
the group, condition, and side I set.
something like
boxplot(test$x~test$time) where test$group=1,test$condition=A,test$side=left
can this be done?
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Re: [R] standard error for quantile
Hi Ted -Original Message- From: ted@deb [mailto:ted@deb] On Behalf Of Ted Harding Sent: Tuesday, October 30, 2012 6:41 PM To: r-help@r-project.org snip The general asymptotic result for the pth quantile (0p1) X.p of a sample of size n is that it is asymptotically Normally distributed with mean the pth quantile Q.p of the parent distribution and var(X.p) = p*(1-p)/(n*f(Q.p)^2) where f(x) is the probability density function of the parent distribution. So if I understand correctly p*(1-p) is biggest when p=0.5 and decreases with smaller or bigger p. The var(X.p) then depends on ratio to parent distribution at this p probability. For lognorm distribution and 200 values the resulting var is (0.5*(1-.5))/(200*qlnorm(.5, log(200), log(2))^2) [1] 3.125e-08 (0.1*(1-.1))/(200*qlnorm(.1, log(200), log(2))^2) [1] 6.648497e-08 so 0.1 var is slightly bigger than 0.5 var. For different distributions this can be reversed as Jim pointed out. Did I manage to understand? Thank you very much. Regards Petr This is not necessarily very helpful for small sample sizes (depending on the parent distribution). However, it is possible to obtain a general result giving an exact confidence interval for Q.p given the entire ordered sample, though there is only a restricted set of confidence levels to which it applies. If you'd like more detail about the above, I could write up derivations and make the write-up available. Hoping this helps, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 30-Oct-2012 Time: 17:40:55 This message was sent by XFMail - __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] standard error for quantile
Hi Roger Thanks for your answer. I tried it with partial success. I got values for tau but I did not manage to evaluate errors or variances of these values. from help page this shall compute median rq(rnorm(50) ~ 1, ci=FALSE) and I assumed some kind of confidence interval is computed when ci=TRUE, but no avail. In the mean time I have got several other answers which helped me to understand the topic. Thanks again. Regards Petr -Original Message- From: Roger Koenker [mailto:rkoen...@illinois.edu] Sent: Tuesday, October 30, 2012 3:42 PM To: PIKAL Petr Cc: r-help@r-project.org help Subject: Re: [R] standard error for quantile Petr, You can do: require(quantreg) summary(rq(x ~ 1, tau = c(.10,.50,.99)) url:www.econ.uiuc.edu/~rogerRoger Koenker emailrkoen...@uiuc.eduDepartment of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Urbana, IL 61801 On Oct 30, 2012, at 9:37 AM, Bert Gunter wrote: Petr: 1. Not an R question. 2. You want the distribution of order statistics. Search on that. It's basically binomial/beta. -- Bert On Tue, Oct 30, 2012 at 6:46 AM, PIKAL Petr petr.pi...@precheza.cz wrote: Dear all I have a question about quantiles standard error, partly practical partly theoretical. I know that x-rlnorm(10, log(200), log(2)) quantile(x, c(.10,.5,.99)) computes quantiles but I would like to know if there is any function to find standard error (or any dispersion measure) of these estimated values. And here is a theoretical one. I feel that when I compute median from given set of values it will have lower standard error then 0.1 quantile computed from the same set of values. Is it true? If yes can you point me to some reasoning? Thanks for all answers. Regards Petr PS. I found mcmcse package which shall compute the standard error but which I could not make to work probably because I do not have recent R-devel version installed Error in eval(expr, envir, enclos) : could not find function .getNamespace Error : unable to load R code in package 'mcmcse' Error: package/namespace load failed for 'mcmcse' Maybe I will also something find in quantreg package, but I did not went through it yet. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional- groups/pdb -biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] NA-friendly operator
@sarah, it's true , but I need a boolean vector. @berend good idea @David, yes %in% work fine but I need = and = @bill yes, it could be a solution, but i prefer a simple operator.. so i will build one. thank you all 2012/10/30 Sarah Goslee sarah.gos...@gmail.com Here's one option: vec-c(3,4,5,NA,1,NA,9,NA,1) subset(vec, vec 2) [1] 3 4 5 9 subset(vec, vec == 1) [1] 1 1 Sarah On Tue, Oct 30, 2012 at 5:08 PM, vincent guyader vincent.guya...@gmail.com wrote: Hi everyone, i'm looking for a NA-friendly operator I explain : vec-c(3,4,5,NA,1,NA,9,NA,1) vec[vec == 1] # NA 1 NA NA 1 I dont want the NA's : vec[vec == 1 ! is.na(vec)]# 1 1 is the same as vec[vec %in% 1] # 1 1 %in% is NA-friendly :) But if i want 2 without the NA's : vec[vec2] #3 4 5 NA NA 9 NA if i dont want the NA i have to do : vec[vec2 !is.na(vec)] #3 4 5 9 is there an opérator to directly do that? any idea? thx a lot. -- Sarah Goslee http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] invalid labels; length 2 should be 1 or 0
I had the same problem, but found this note in the MICE package documentation (at http://cran.r-project.org/web/packages/mice/mice.pdf): Added June 25, 2012: The currently implemented algorithm does not handle predictors that are specified as fixed effects (type=1). When using mice.impute.2l.norm(), the current advice is to specify all predictors as random effects (type=2). Warning: The assumption of heterogeneous variances requires that in every class at least one observation has a response in y. -- View this message in context: http://r.789695.n4.nabble.com/invalid-labels-length-2-should-be-1-or-0-tp4643712p4647974.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] RCurl - curlPerform - Time out?!?
Hi,
I am working with the RCurl package and I am using the curlPerform
function for an soap-query.
The problem is that the code is usually working well, but sometimes the
connection gets lost.
So I wrote a while-loop to repeat the query if anything might happened
so that the same query runs again, but if the query-faults it takes a
very long time for the repetition.
My question is if there is any possibility to force a time out for the
curlPerform function or something like that?
Thanks!
run = 1
i=0
while(run==1)
{
i=i+1
try(
run - curlPerform(url =
http://search.webofknowledge.com/esti/wokmws/ws/WokSearchLite.cgi;,
httpheader=c(Accept-Encoding=gzip,deflate,Content-Type=text/xml;charset=UTF-8,'SOAPAction'='',
Cookie=paste('SID=',s_session,'',sep=),Content-Length=paste(nchar(s_body)),Host=search.webofknowledge.com,Connection=Keep-Alive,User-Agent=Apache-HttpClient/4.1.1
(java 1.5)),
postfields=s_body,
writefunction = h$update
,verbose = TRUE)
,TRUE)
print(i)
}
[[alternative HTML version deleted]]
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Re: [R] NA-friendly operator
On Wed, Oct 31, 2012 at 10:54 AM, vincent guyader vincent.guya...@gmail.com wrote: yes, it could be a solution, but i prefer a simple operator.. so i will build one. See this thread for both advice and pitfalls: https://stat.ethz.ch/pipermail/r-help/2012-June/316240.html Michael Weylandt __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R crashing after successfully running compiled code
On 12-10-30 11:13 PM, Adam Clark wrote: I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32 and 2.13.0x64. I am trying to run compiled C code using the .C convention. The code compiles without problems, dynamically loads within the R workspace with no problems, and even runs and gives correct results with no problems. However, R will randomly crash within a few minutes of successfully using the compiled function. For example, if I run my compiled function using: dyn.load(mycfun.dll) answer-.C(mycfun, parameters...), I get a completely sensible result that gets stored to answer. However, if I try to do too many things to answer, the R exits without warning. I've tried dyn.unload in hopes that R would become stable afterwards, but in this case using the function crashes R without fail. Usually, I can either plot, or view, or save answer to a file - but never take more than a single action before R exits. This does not appear to depend on how long R has been open. Initially, I thought it was a bug in the inline function, but I'm finding the same problem now that I'm using the dynamically loaded file directly. I'm used to R being insanely stable, and am somewhat mystified by this whole problem. My next move is to learn the .Call convention, as I suspect that my problem is related to my C function using memory that R doesn't know is used. But - before I invest a while lot more time on this, I'd like to know whether anybody things this is likely to solve the problem. If not, I may just want to run my code entirely in C, and forget the R problem. I think your C code has a bug in it. The bug might go away when you rewrite the function to work within the .Call convention, but it is probably easier to find the bug and fix it with the current code. Things to look for: Are you fully allocating all arrays in R before passing them to C? The C code receives a pointer and will happily write to it, whether that makes sense or not. Are you careful with your limits on vectors? In R, a vector is indexed from 1 to n, but the same vector in C is indexed from 0 to n-1. If the C code writes to entry n, that will eventually cause problems. Are you allocating memory in your C code? There are several ways to do that, depending on how you want it managed. If you do it one way and expect it to be managed in a different way, you'll get problems. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R crashing after successfully running compiled code
On Wed, 31 Oct 2012, Duncan Murdoch wrote: On 12-10-30 11:13 PM, Adam Clark wrote: I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32 and 2.13.0x64. I am trying to run compiled C code using the .C convention. The code compiles without problems, dynamically loads within the R workspace with no problems, and even runs and gives correct results with no problems. However, R will randomly crash within a few minutes of successfully using the compiled function. For example, if I run my compiled function using: dyn.load(mycfun.dll) answer-.C(mycfun, parameters...), I get a completely sensible result that gets stored to answer. However, if I try to do too many things to answer, the R exits without warning. I've tried dyn.unload in hopes that R would become stable afterwards, but in this case using the function crashes R without fail. Usually, I can either plot, or view, or save answer to a file - but never take more than a single action before R exits. This does not appear to depend on how long R has been open. Initially, I thought it was a bug in the inline function, but I'm finding the same problem now that I'm using the dynamically loaded file directly. I'm used to R being insanely stable, and am somewhat mystified by this whole problem. My next move is to learn the .Call convention, as I suspect that my problem is related to my C function using memory that R doesn't know is used. But - before I invest a while lot more time on this, I'd like to know whether anybody things this is likely to solve the problem. If not, I may just want to run my code entirely in C, and forget the R problem. I think your C code has a bug in it. The bug might go away when you rewrite the function to work within the .Call convention, but it is probably easier to find the bug and fix it with the current code. Things to look for: Are you fully allocating all arrays in R before passing them to C? The C code receives a pointer and will happily write to it, whether that makes sense or not. Are you careful with your limits on vectors? In R, a vector is indexed from 1 to n, but the same vector in C is indexed from 0 to n-1. If the C code writes to entry n, that will eventually cause problems. Using R-devel and the following new feature • There is a new option, options(CBoundsCheck=), which controls how .C() and .Fortran() pass arguments to compiled code. If true (which can be enabled by setting the environment variable R_C_BOUNDS_CHECK to yes), raw, integer, double and complex arguments are always copied, and checked for writing off either end of the array on return from the compiled code (when a second copy is made). This also checks individual elements of character vectors passed to .C(). This is not intended for routine use, but can be very helpful in finding segfaults in package code. makes checking these two points a lot easier. Are you allocating memory in your C code? There are several ways to do that, depending on how you want it managed. If you do it one way and expect it to be managed in a different way, you'll get problems. If you can run your code under valgrind (see 'Writing R Extensions') you will usually get pointed to the exact cause. But that's for Linux, and with some care MacOS X. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] boxplots of various levels
I am not sure I understand exactly what you want but does this do anything like what you want? library(ggplot2) mydata - data.frame(result = rnorm(100), group = rep(c(1, 2), each = 50), side = sample(c(L, R), 100, replace = TRUE) , dtime = rep(1:10, each=10)) p - ggplot(mydata , aes( group, result, fill = side ))+ geom_boxplot() + facet_wrap(~dtime) p John Kane Kingston ON Canada -Original Message- From: dysonspher...@gmail.com Sent: Tue, 30 Oct 2012 11:21:49 -0700 (PDT) To: r-help@r-project.org Subject: [R] boxplots of various levels noob here trying to make boxplots of some data i would like to separate the boxplots according to conditons of various levels for example: i have group:1 and 2, each group performed tests consisting of condition A,B,C,D side: left and right time: 1 to 10 I would like separate boxplots of the results (x) of the tests (numeric) for each group under each condition on each side over time. so far i have set it up like this: boxplot(test$x~test$time) this gives me the plot for all vaues of x in each time bin. basicaly i would need a command that tells R to include only the data that agrees with the group, condition, and side I set. something like boxplot(test$x~test$time) where test$group=1,test$condition=A,test$side=left can this be done? -- View this message in context: http://r.789695.n4.nabble.com/boxplots-of-various-levels-tp4647917.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Send any screenshot to your friends in seconds... Works in all emails, instant messengers, blogs, forums and social networks. TRY IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if2 for FREE __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate.formula: formula from string
Hi, Try this: res- aggregate(eval(mF),d,mean) res # a b NA NA #1 1 A -1.48354978 -0.37141485 #2 2 A -0.08862713 0.35359250 #3 3 A 1.17519518 -0.47595290 #4 1 B 0.10214686 -0.70005131 #5 2 B 0.41185154 0.03707291 #6 3 B 0.20507062 -0.67946389 res1-aggregate(cbind(y, z) ~ a + b, d, mean) colnames(res)[3:4]-colnames(res1)[3:4] identical(res,res1) #[1] TRUE A.K. - Original Message - From: Thaler,Thorn,LAUSANNE,Applied Mathematics thorn.tha...@rdls.nestle.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Wednesday, October 31, 2012 5:46 AM Subject: [R] aggregate.formula: formula from string Dear all, I want to use aggregate.formula to conveniently summarize a data.frame. I have quiet some variables in the data.frame and thus I don't want to write all these names by hand, but instead create them on the fly. This approach has the advantage that if there will be even more columns in the data.frame I don't have to change the code. I've hence tried to construct a formula object and to pass that to aggregate: d - expand.grid(a = factor(1:3), b = factor(LETTERS[1:2])) d - rbind(d,d,d) d$y - rnorm(18) d$z - rnorm(18) mF - as.formula(paste(cbind(, paste(names(d)[-(1:2)], collapse = ,), ) ~ a + b, sep = )) But if I try to pass that formula to aggregate aggregate(mF, d, mean) I get the following error: Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable But if I pass the formula directly: aggregate(cbind(y, z) ~ a + b, d, mean) Everything is working as expected. So I was wondering what went wrong? I know I could use a formula like . ~ a + b instead and this would work fine, but I'm just interested in why the outlined approach does not work as expected, and where my mistake lies? (that means in particular I am not asking for a solution of how to get the thing done - there are plenty of alternatives - but instead to understand why this very approach does not work) Thanks for your help! Kind Regards, Thorn Thaler Mathematician Applied Mathematics Nestec Ltd, Nestlé Research Center PO Box 44 CH-1000 Lausanne 26 Phone: +41 21 785 8220 Fax: +41 21 785 9486 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] box() doesnt work
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example John Kane Kingston ON Canada -Original Message- From: cat...@bas.ac.uk Sent: Tue, 30 Oct 2012 10:27:29 -0700 (PDT) To: r-help@r-project.org Subject: [R] box() doesnt work Hi, when plotting a graphic i find that the surrounding box disappears if I adjust the margins with par(mar=..). Ive tried reassigning it with box() but it doesnt seem to make any difference. Does anyone know a way to overcome this? Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/box-doesnt-work-tp4647909.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate.formula: formula from string
Dear Arun, Thanks for your suggestion, that does the trick. Just because I'm curious, where does the problem come from? I figured that wrapping the formula object in brackets would work as well: aggregate((mF), d, mean) So I guess it has something to do with the scope of mF, or what is the root cause? Thanks for your help! KR, -Thorn -Original Message- From: arun [mailto:smartpink...@yahoo.com] Sent: Mittwoch, 31. Oktober 2012 13:15 To: Thaler,Thorn,LAUSANNE,Applied Mathematics Cc: R help Subject: Re: [R] aggregate.formula: formula from string Hi, Try this: res- aggregate(eval(mF),d,mean) res # a b NA NA #1 1 A -1.48354978 -0.37141485 #2 2 A -0.08862713 0.35359250 #3 3 A 1.17519518 -0.47595290 #4 1 B 0.10214686 -0.70005131 #5 2 B 0.41185154 0.03707291 #6 3 B 0.20507062 -0.67946389 res1-aggregate(cbind(y, z) ~ a + b, d, mean) colnames(res)[3:4]-colnames(res1)[3:4] identical(res,res1) #[1] TRUE A.K. - Original Message - From: Thaler,Thorn,LAUSANNE,Applied Mathematics thorn.tha...@rdls.nestle.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Wednesday, October 31, 2012 5:46 AM Subject: [R] aggregate.formula: formula from string Dear all, I want to use aggregate.formula to conveniently summarize a data.frame. I have quiet some variables in the data.frame and thus I don't want to write all these names by hand, but instead create them on the fly. This approach has the advantage that if there will be even more columns in the data.frame I don't have to change the code. I've hence tried to construct a formula object and to pass that to aggregate: d - expand.grid(a = factor(1:3), b = factor(LETTERS[1:2])) d - rbind(d,d,d) d$y - rnorm(18) d$z - rnorm(18) mF - as.formula(paste(cbind(, paste(names(d)[-(1:2)], collapse = ,), ) ~ a + b, sep = )) But if I try to pass that formula to aggregate aggregate(mF, d, mean) I get the following error: Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable But if I pass the formula directly: aggregate(cbind(y, z) ~ a + b, d, mean) Everything is working as expected. So I was wondering what went wrong? I know I could use a formula like . ~ a + b instead and this would work fine, but I'm just interested in why the outlined approach does not work as expected, and where my mistake lies? (that means in particular I am not asking for a solution of how to get the thing done - there are plenty of alternatives - but instead to understand why this very approach does not work) Thanks for your help! Kind Regards, Thorn Thaler Mathematician Applied Mathematics Nestec Ltd, Nestlé Research Center PO Box 44 CH-1000 Lausanne 26 Phone: +41 21 785 8220 Fax: +41 21 785 9486 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate.formula: formula from string
Hi Thorn, May be it is a bug in aggregate.formula(). Not sure about it. The solutions that I gave and yours were returning the colnames as NA. I guess this should be much better in terms of getting the result in one step. do.call(aggregate,list(mF,d,mean)) # a b y z #1 1 A -1.48354978 -0.37141485 #2 2 A -0.08862713 0.35359250 #3 3 A 1.17519518 -0.47595290 #4 1 B 0.10214686 -0.70005131 #5 2 B 0.41185154 0.03707291 #6 3 B 0.20507062 -0.67946389 A.K. - Original Message - From: Thaler,Thorn,LAUSANNE,Applied Mathematics thorn.tha...@rdls.nestle.com To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Wednesday, October 31, 2012 8:30 AM Subject: RE: [R] aggregate.formula: formula from string Dear Arun, Thanks for your suggestion, that does the trick. Just because I'm curious, where does the problem come from? I figured that wrapping the formula object in brackets would work as well: aggregate((mF), d, mean) So I guess it has something to do with the scope of mF, or what is the root cause? Thanks for your help! KR, -Thorn -Original Message- From: arun [mailto:smartpink...@yahoo.com] Sent: Mittwoch, 31. Oktober 2012 13:15 To: Thaler,Thorn,LAUSANNE,Applied Mathematics Cc: R help Subject: Re: [R] aggregate.formula: formula from string Hi, Try this: res- aggregate(eval(mF),d,mean) res # a b NA NA #1 1 A -1.48354978 -0.37141485 #2 2 A -0.08862713 0.35359250 #3 3 A 1.17519518 -0.47595290 #4 1 B 0.10214686 -0.70005131 #5 2 B 0.41185154 0.03707291 #6 3 B 0.20507062 -0.67946389 res1-aggregate(cbind(y, z) ~ a + b, d, mean) colnames(res)[3:4]-colnames(res1)[3:4] identical(res,res1) #[1] TRUE A.K. - Original Message - From: Thaler,Thorn,LAUSANNE,Applied Mathematics thorn.tha...@rdls.nestle.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Wednesday, October 31, 2012 5:46 AM Subject: [R] aggregate.formula: formula from string Dear all, I want to use aggregate.formula to conveniently summarize a data.frame. I have quiet some variables in the data.frame and thus I don't want to write all these names by hand, but instead create them on the fly. This approach has the advantage that if there will be even more columns in the data.frame I don't have to change the code. I've hence tried to construct a formula object and to pass that to aggregate: d - expand.grid(a = factor(1:3), b = factor(LETTERS[1:2])) d - rbind(d,d,d) d$y - rnorm(18) d$z - rnorm(18) mF - as.formula(paste(cbind(, paste(names(d)[-(1:2)], collapse = ,), ) ~ a + b, sep = )) But if I try to pass that formula to aggregate aggregate(mF, d, mean) I get the following error: Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable But if I pass the formula directly: aggregate(cbind(y, z) ~ a + b, d, mean) Everything is working as expected. So I was wondering what went wrong? I know I could use a formula like . ~ a + b instead and this would work fine, but I'm just interested in why the outlined approach does not work as expected, and where my mistake lies? (that means in particular I am not asking for a solution of how to get the thing done - there are plenty of alternatives - but instead to understand why this very approach does not work) Thanks for your help! Kind Regards, Thorn Thaler Mathematician Applied Mathematics Nestec Ltd, Nestlé Research Center PO Box 44 CH-1000 Lausanne 26 Phone: +41 21 785 8220 Fax: +41 21 785 9486 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bootstrapping quantile regression
There is no automatic clustering option for QR bootstrapping.
You will have to roll your own.
url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Urbana, IL 61801
On Oct 31, 2012, at 1:38 AM, Kay Cichini wrote:
sry, I forgot to replace rlm() - but actually I tried both and the question
applies to both approaches..
Am 31.10.2012 00:19 schrieb Kay Cichini kay.cich...@gmail.com:
HI everyone,
I try to get some bootstrap CIs for coefficients obtained by quantile
regression. I have influencial values and thus switched to quantreg..
The data is clustered and within clusters the variance of my DV = 0..
Is this sensible for the below data? And what about the warnings?
Thanks in advance for any guidance,
Kay
dput(d)
structure(list(Porenfläche = c(4990L, 7002L, 7558L, 7352L, 7943L,
7979L, 9333L, 8209L, 8393L, 6425L, 9364L, 8624L, 10651L, 8868L,
9417L, 8874L, 10962L, 10743L, 11878L, 9867L, 7838L, 11876L, 12212L,
8233L, 6360L, 4193L, 7416L, 5246L, 6509L, 4895L, 6775L, 7894L,
5980L, 5318L, 7392L, 7894L, 3469L, 1468L, 3524L, 5267L, 5048L,
1016L, 5605L, 8793L, 3475L, 1651L, 5514L, 9718L), P.Perimeter = c(2791.9,
3892.6, 3930.66, 3869.32, 3948.54, 4010.15, 4345.75, 4344.75,
3682.04, 3098.65, 4480.05, 3986.24, 4036.54, 3518.04, 3999.37,
3629.07, 4608.66, 4787.62, 4864.22, 4479.41, 3428.74, 4353.14,
4697.65, 3518.44, 1977.39, 1379.35, 1916.24, 1585.42, 1851.21,
1239.66, 1728.14, 1461.06, 1426.76, 990.388, 1350.76, 1461.06,
1376.7, 476.322, 1189.46, 1644.96, 941.543, 308.642, 1145.69,
2280.49, 1174.11, 597.808, 1455.88, 1485.58), P.Form = c(0.0903296,
0.148622, 0.183312, 0.117063, 0.122417, 0.167045, 0.189651, 0.164127,
0.203654, 0.162394, 0.150944, 0.148141, 0.228595, 0.231623, 0.172567,
0.153481, 0.204314, 0.262727, 0.200071, 0.14481, 0.113852, 0.291029,
0.240077, 0.161865, 0.280887, 0.179455, 0.191802, 0.133083, 0.225214,
0.341273, 0.311646, 0.276016, 0.197653, 0.326635, 0.154192, 0.276016,
0.176969, 0.438712, 0.163586, 0.253832, 0.328641, 0.230081, 0.464125,
0.420477, 0.200744, 0.262651, 0.182453, 0.200447), Durchlässigkeit =
c(6.3,
6.3, 6.3, 6.3, 17.1, 17.1, 17.1, 17.1, 119, 119, 119, 119, 82.4,
82.4, 82.4, 82.4, 58.6, 58.6, 58.6, 58.6, 142, 142, 142, 142,
740, 740, 740, 740, 890, 890, 890, 890, 950, 950, 950, 950, 100,
100, 100, 100, 1300, 1300, 1300, 1300, 580, 580, 580, 580), Gebiete =
structure(c(1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 6L, 6L, 6L, 6L, 4L, 4L, 4L, 4L, 3L,
3L, 3L, 3L, 7L, 7L, 7L, 7L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L,
11L, 11L, 11L, 11L, 5L, 5L, 5L, 5L, 12L, 12L, 12L, 12L, 8L, 8L,
8L, 8L), .Label = c(6.3, 17.1, 58.6, 82.4, 100, 119,
142, 580, 740, 890, 950, 1300), class = factor)), .Names =
c(Porenfläche,
P.Perimeter, P.Form, Durchlässigkeit, Gebiete), row.names =
c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46,
47, 48), class = data.frame)
## do quantile regression and bootstrap the coefficients, allowing for
clustered data
## by putting Gebiet as strata argument (?),
## dv variation within clusters/Gebiet = 0!
bs - function(formula, data, indices) {
d - data[indices, ] # allows boot to select sample
fit - rlm(formula, data = d)
return(coef(fit))
}
results - boot(data = d, statistic = bs, strata = d$Gebiete,
R = 199, formula = Durchlässigkeit ~ P.Perimeter + P.Form)
# get 99% confidence intervals
boot.ci(results, type=bca, index=1, conf = .99) # intercept
boot.ci(results, type=bca, index=2, conf = .99) # P.Perimeter
boot.ci(results, type=bca, index=3, conf = .99) # P.Form
--
Kay Cichini, MSc Biol
Grubenweg 22, 6071 Aldrans
E-Mail: kay.cich...@gmail.com
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] standard error for quantile
The rank test inversion option that you are trying to use won't work with only one coefficient, and therefore with univariate quantiles, if you use summary(rq(rnorm(50) ~ 1, tau = .9), cov = TRUE) you will have better luck. url:www.econ.uiuc.edu/~rogerRoger Koenker emailrkoen...@uiuc.eduDepartment of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Urbana, IL 61801 On Oct 30, 2012, at 9:42 AM, Koenker, Roger W wrote: Petr, You can do: require(quantreg) summary(rq(x ~ 1, tau = c(.10,.50,.99)) url:www.econ.uiuc.edu/~rogerRoger Koenker emailrkoen...@uiuc.eduDepartment of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Urbana, IL 61801 On Oct 30, 2012, at 9:37 AM, Bert Gunter wrote: Petr: 1. Not an R question. 2. You want the distribution of order statistics. Search on that. It's basically binomial/beta. -- Bert On Tue, Oct 30, 2012 at 6:46 AM, PIKAL Petr petr.pi...@precheza.cz wrote: Dear all I have a question about quantiles standard error, partly practical partly theoretical. I know that x-rlnorm(10, log(200), log(2)) quantile(x, c(.10,.5,.99)) computes quantiles but I would like to know if there is any function to find standard error (or any dispersion measure) of these estimated values. And here is a theoretical one. I feel that when I compute median from given set of values it will have lower standard error then 0.1 quantile computed from the same set of values. Is it true? If yes can you point me to some reasoning? Thanks for all answers. Regards Petr PS. I found mcmcse package which shall compute the standard error but which I could not make to work probably because I do not have recent R-devel version installed Error in eval(expr, envir, enclos) : could not find function .getNamespace Error : unable to load R code in package 'mcmcse' Error: package/namespace load failed for 'mcmcse' Maybe I will also something find in quantreg package, but I did not went through it yet. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] issues with krige function
My apologies to all, the package is gstat that I am using not geoR. thanks! A On Tue, Oct 30, 2012 at 1:35 PM, Adrienne Wootten amwoo...@ncsu.edu wrote: Greetings all, Ran into a strange problem with the krige function from geoR. The problem that I am having is that while the krige function seems to work well, the resulting predicted values are all NAs. Given the size of the datasets I am working with can't attach it, but I can provide snippets of the datasets. casedata station year month day obs mpebias type lat lon 140 319147 2011 8 28 0.00 0.0 0. COOP 35.48667 -82.96833 141 319354 2011 8 28 0.02 0.001305799 -0.01869420 COOP 34.25722 -78.68722 142 319357 2011 8 28 0.00 0.045194085 0.04519409 COOP 34.40944 -78.79139 143 319440 2011 8 28 6.92 5.530313974 -1.38968603 COOP 35.85278 -77.03056 144 319461 2011 8 28 0.98 0.949224103 -0.03077590 COOP 34.14250 -77.87861 145 319467 2011 8 28 0.30 1.492852601 1.19285260 COOP 34.32083 -77.92056 146 319476 2011 8 28 2.95 3.284362345 0.33436235 COOP 35.69389 -77.94556 147 319555 2011 8 28 0.00 0.0 0. COOP 36.13083 -81.22750 148 319667 2011 8 28 0.00 0.014472943 0.01447294 COOP 35.85583 -80.36083 149 319675 2011 8 28 0.00 0.0 0. COOP 36.13056 -80.54806 v.fit = vgm(phi=1.2914,model=Mat,range=14.1611,nugget=0.23,kappa=83.1047) krige.pred.out = krige(formula=bias~1,locations=~lon+lat,data=casedata,newdata=gridmeta,model=v.fit) [using ordinary kriging] krige.pred.out lon lat var1.pred var1.var 1 -86.28771 33.91335 NaN NaN 2 -86.27388 33.94722 NaN NaN 3 -86.26003 33.98111 NaN NaN 4 -86.24615 34.01500 NaN NaN 5 -86.23225 34.04890 NaN NaN 6 -86.21835 34.08280 NaN NaN 7 -86.20441 34.11672 NaN NaN 8 -86.19045 34.15064 NaN NaN 9 -86.17648 34.18457 NaN NaN 10 -86.16249 34.21851 NaN NaN summary(krige.pred.out) lon lat var1.predvar1.var Min. :-86.29 Min. :30.51 Min. : NA Min. : NA 1st Qu.:-82.38 1st Qu.:33.73 1st Qu.: NA 1st Qu.: NA Median :-79.88 Median :35.22 Median : NA Median : NA Mean :-79.91 Mean :35.26 Mean :NaN Mean :NaN 3rd Qu.:-77.48 3rd Qu.:36.76 3rd Qu.: NA 3rd Qu.: NA Max. :-73.16 Max. :40.16 Max. : NA Max. : NA NA's :44408 NA's :44408 Any ideas are greatly appreciated, I'm not getting any warnings or errors from krige, so this is perplexing to me. -- Adrienne Wootten Graduate Research Assistant State Climate Office of North Carolina Department of Marine, Earth and Atmospheric Sciences North Carolina State University -- Adrienne Wootten Graduate Research Assistant State Climate Office of North Carolina Department of Marine, Earth and Atmospheric Sciences North Carolina State University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bootstrapping quantile regression
A piece of this is solved by the rms package's Rq and bootcov functions.
-Frank
Roger Koenker-3 wrote
There is no automatic clustering option for QR bootstrapping.
You will have to roll your own.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email
rkoenker@
Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Urbana, IL 61801
On Oct 31, 2012, at 1:38 AM, Kay Cichini wrote:
sry, I forgot to replace rlm() - but actually I tried both and the
question
applies to both approaches..
Am 31.10.2012 00:19 schrieb Kay Cichini lt;
kay.cichini@
gt;:
HI everyone,
I try to get some bootstrap CIs for coefficients obtained by quantile
regression. I have influencial values and thus switched to quantreg..
The data is clustered and within clusters the variance of my DV = 0..
Is this sensible for the below data? And what about the warnings?
Thanks in advance for any guidance,
Kay
dput(d)
structure(list(Porenfläche = c(4990L, 7002L, 7558L, 7352L, 7943L,
7979L, 9333L, 8209L, 8393L, 6425L, 9364L, 8624L, 10651L, 8868L,
9417L, 8874L, 10962L, 10743L, 11878L, 9867L, 7838L, 11876L, 12212L,
8233L, 6360L, 4193L, 7416L, 5246L, 6509L, 4895L, 6775L, 7894L,
5980L, 5318L, 7392L, 7894L, 3469L, 1468L, 3524L, 5267L, 5048L,
1016L, 5605L, 8793L, 3475L, 1651L, 5514L, 9718L), P.Perimeter =
c(2791.9,
3892.6, 3930.66, 3869.32, 3948.54, 4010.15, 4345.75, 4344.75,
3682.04, 3098.65, 4480.05, 3986.24, 4036.54, 3518.04, 3999.37,
3629.07, 4608.66, 4787.62, 4864.22, 4479.41, 3428.74, 4353.14,
4697.65, 3518.44, 1977.39, 1379.35, 1916.24, 1585.42, 1851.21,
1239.66, 1728.14, 1461.06, 1426.76, 990.388, 1350.76, 1461.06,
1376.7, 476.322, 1189.46, 1644.96, 941.543, 308.642, 1145.69,
2280.49, 1174.11, 597.808, 1455.88, 1485.58), P.Form = c(0.0903296,
0.148622, 0.183312, 0.117063, 0.122417, 0.167045, 0.189651, 0.164127,
0.203654, 0.162394, 0.150944, 0.148141, 0.228595, 0.231623, 0.172567,
0.153481, 0.204314, 0.262727, 0.200071, 0.14481, 0.113852, 0.291029,
0.240077, 0.161865, 0.280887, 0.179455, 0.191802, 0.133083, 0.225214,
0.341273, 0.311646, 0.276016, 0.197653, 0.326635, 0.154192, 0.276016,
0.176969, 0.438712, 0.163586, 0.253832, 0.328641, 0.230081, 0.464125,
0.420477, 0.200744, 0.262651, 0.182453, 0.200447), Durchlässigkeit =
c(6.3,
6.3, 6.3, 6.3, 17.1, 17.1, 17.1, 17.1, 119, 119, 119, 119, 82.4,
82.4, 82.4, 82.4, 58.6, 58.6, 58.6, 58.6, 142, 142, 142, 142,
740, 740, 740, 740, 890, 890, 890, 890, 950, 950, 950, 950, 100,
100, 100, 100, 1300, 1300, 1300, 1300, 580, 580, 580, 580), Gebiete =
structure(c(1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 6L, 6L, 6L, 6L, 4L, 4L, 4L, 4L, 3L,
3L, 3L, 3L, 7L, 7L, 7L, 7L, 9L, 9L, 9L, 9L, 10L, 10L, 10L, 10L,
11L, 11L, 11L, 11L, 5L, 5L, 5L, 5L, 12L, 12L, 12L, 12L, 8L, 8L,
8L, 8L), .Label = c(6.3, 17.1, 58.6, 82.4, 100, 119,
142, 580, 740, 890, 950, 1300), class = factor)), .Names =
c(Porenfläche,
P.Perimeter, P.Form, Durchlässigkeit, Gebiete), row.names =
c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46,
47, 48), class = data.frame)
## do quantile regression and bootstrap the coefficients, allowing for
clustered data
## by putting Gebiet as strata argument (?),
## dv variation within clusters/Gebiet = 0!
bs - function(formula, data, indices) {
d - data[indices, ] # allows boot to select sample
fit - rlm(formula, data = d)
return(coef(fit))
}
results - boot(data = d, statistic = bs, strata = d$Gebiete,
R = 199, formula = Durchlässigkeit ~ P.Perimeter +
P.Form)
# get 99% confidence intervals
boot.ci(results, type=bca, index=1, conf = .99) # intercept
boot.ci(results, type=bca, index=2, conf = .99) # P.Perimeter
boot.ci(results, type=bca, index=3, conf = .99) # P.Form
--
Kay Cichini, MSc Biol
Grubenweg 22, 6071 Aldrans
E-Mail:
kay.cichini@
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[R] Mixed modelling course with R in Portugal
We would like to announce the following statistics course: Introduction to linear mixed effects modelling course with R. 11 - 15 February 2013. Lisbon, Portugal. For details, see: http://www.highstat.com/statscourse.htm Course flyer: http://www.highstat.com/Courses/Flyer2013FebSIM_Lisbon.pdf Kind regards, Alain Zuur -- Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 4. Zero Inflated Models and Generalized Linear Mixed Models with R. (2012) Zuur, Saveliev, Ieno. http://www.highstat.com/book4.htm Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: highs...@highstat.com URL: www.highstat.com URL: www.brodgar.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] issues with krige function
Dear Adrienne, What is the output of summary(casestudy) and summary(gridmeta)? What happens if you set nmax to 10? krige(formula=bias~1,locations=~lon+lat,data=casedata,newdata=gridmeta ,model=v.fit, nmax = 10) ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Adrienne Wootten Verzonden: dinsdag 30 oktober 2012 18:36 Aan: r-help@r-project.org Onderwerp: [R] issues with krige function Greetings all, Ran into a strange problem with the krige function from geoR. The problem that I am having is that while the krige function seems to work well, the resulting predicted values are all NAs. Given the size of the datasets I am working with can't attach it, but I can provide snippets of the datasets. casedata station year month day obs mpebias type lat lon 140 319147 2011 8 28 0.00 0.0 0. COOP 35.48667 -82.96833 141 319354 2011 8 28 0.02 0.001305799 -0.01869420 COOP 34.25722 -78.68722 142 319357 2011 8 28 0.00 0.045194085 0.04519409 COOP 34.40944 -78.79139 143 319440 2011 8 28 6.92 5.530313974 -1.38968603 COOP 35.85278 -77.03056 144 319461 2011 8 28 0.98 0.949224103 -0.03077590 COOP 34.14250 -77.87861 145 319467 2011 8 28 0.30 1.492852601 1.19285260 COOP 34.32083 -77.92056 146 319476 2011 8 28 2.95 3.284362345 0.33436235 COOP 35.69389 -77.94556 147 319555 2011 8 28 0.00 0.0 0. COOP 36.13083 -81.22750 148 319667 2011 8 28 0.00 0.014472943 0.01447294 COOP 35.85583 -80.36083 149 319675 2011 8 28 0.00 0.0 0. COOP 36.13056 -80.54806 v.fit = vgm(phi=1.2914,model=Mat,range=14.1611,nugget=0.23,kappa=83.1047) krige.pred.out = krige(formula=bias~1,locations=~lon+lat,data=casedata,newdata=gridmeta ,model=v.fit) [using ordinary kriging] krige.pred.out lon lat var1.pred var1.var 1 -86.28771 33.91335 NaN NaN 2 -86.27388 33.94722 NaN NaN 3 -86.26003 33.98111 NaN NaN 4 -86.24615 34.01500 NaN NaN 5 -86.23225 34.04890 NaN NaN 6 -86.21835 34.08280 NaN NaN 7 -86.20441 34.11672 NaN NaN 8 -86.19045 34.15064 NaN NaN 9 -86.17648 34.18457 NaN NaN 10 -86.16249 34.21851 NaN NaN summary(krige.pred.out) lon lat var1.predvar1.var Min. :-86.29 Min. :30.51 Min. : NA Min. : NA 1st Qu.:-82.38 1st Qu.:33.73 1st Qu.: NA 1st Qu.: NA Median :-79.88 Median :35.22 Median : NA Median : NA Mean :-79.91 Mean :35.26 Mean :NaN Mean :NaN 3rd Qu.:-77.48 3rd Qu.:36.76 3rd Qu.: NA 3rd Qu.: NA Max. :-73.16 Max. :40.16 Max. : NA Max. : NA NA's :44408 NA's :44408 Any ideas are greatly appreciated, I'm not getting any warnings or errors from krige, so this is perplexing to me. -- Adrienne Wootten Graduate Research Assistant State Climate Office of North Carolina Department of Marine, Earth and Atmospheric Sciences North Carolina State University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R crashing after successfully running compiled code
Thanks for the advice.
I'll go ahead and dig through my C code. It's helpful to know that my C
code can cause R to crash AFTER successfully implementing the code.
I have made sure to account for C's vector indexing, and I think I'm
allocating my C memory, and passing information back and forth between C
and R, as I should. I'm including my input/output stuff below.
I tried including options(CBoundsCheck=TRUE) in the script both before and
after loading the C function, which doesn't seem to do much. To get it to
work, do I actually need to go into the R configuration file and edit the
default?
It would be exceedingly helpful if anybody could give me tips on where I'm
misusing pointers in the example below. That said, I certainly don't expect
the R community to debug my C code for me. If I come up with a solution,
I'll email it out over the list.
In R, I run the script:
dyn.load(mycfun.dll)
set.seed(1031)
A-1:100
B-runif(100)
myfunC-function(A, B, M, N) {
result-as.double(rep(0, (length(A)- N -(M+1
plengtht-as.integer(length(A))
Aest-as.numeric(rep(0, (length(A)- N -(M+1
distances-as.numeric(rep(0, length(A)))
neighbors-as.integer(rep(0, (M+1)))
u-as.numeric(rep(0, (M+1)))
w-as.numeric(rep(0, (M+1)))
return(.C(mycfun, as.double(A), as.double(B), as.integer(M),
as.integer(N),
result=as.double(result), as.integer(plengtht), as.double(Aest),
as.double(distances),
as.integer(neighbors), as.double(u), as.double(w))$result)
}
fun_result-myfunC(A,B,3,1)
This corresponds to the C code (input output only):
#include R.h
#include Rmath.h
void mycfun(double *A, double *B, int *pM, int *pN, double *result, int
*plengtht,
double *Aest, double *distances, int *neighbors, double *u, double *w) {
int t, i, j, n, from, to, nneigh;
double distsv, sumu, sumaest, corout;
int M = *pM;
int N = *pN;
int lengtht= *plengtht;
n=0;
# running various loops over variables #
result[n]=corout;
n=n+1;
}
# END #
I also have two sub-functions that manipulate neighbors and distances -
I can send the i/o for those as well, but they seem much more
straightforward, since I don't need to pass all my arguments as pointers. I
pass the pointers to internal variables at the beginning because I couldn't
index any C arrays using *pM or *pN.
Many thanks,
Adam
On Wed, Oct 31, 2012 at 7:12 AM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote:
On Wed, 31 Oct 2012, Duncan Murdoch wrote:
On 12-10-30 11:13 PM, Adam Clark wrote:
I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32
and
2.13.0x64.
I am trying to run compiled C code using the .C convention. The
code compiles without problems, dynamically loads within the R
workspace with no problems, and even runs and gives correct results with
no problems.
However, R will randomly crash within a few minutes of successfully using
the compiled function.
For example, if I run my compiled function using:
dyn.load(mycfun.dll)
answer-.C(mycfun, parameters...), I get a completely sensible
result that gets stored to answer.
However, if I try to do too many things to answer, the R exits
without warning.
I've tried dyn.unload in hopes that R would become stable afterwards, but
in this case using the function crashes R without fail.
Usually, I can either plot, or view, or save answer to a file - but
never
take more than a single action before R exits. This does not appear to
depend on how long R has been open. Initially, I thought it was a bug in
the inline function, but I'm finding the same problem now that I'm
using
the dynamically loaded file directly. I'm used to R being insanely
stable,
and am somewhat mystified by this whole problem.
My next move is to learn the .Call convention, as I suspect that
my problem is related to my C function using memory that R doesn't
know is used. But - before I invest a while lot more time on this, I'd
like to know whether anybody things this is likely to solve the problem.
If not, I may just want to run my code entirely in C, and forget the
R problem.
I think your C code has a bug in it. The bug might go away when you
rewrite the function to work within the .Call convention, but it is
probably easier to find the bug and fix it with the current code.
Things to look for:
Are you fully allocating all arrays in R before passing them to C? The C
code receives a pointer and will happily write to it, whether that makes
sense or not.
Are you careful with your limits on vectors? In R, a vector is indexed
from 1 to n, but the same vector in C is indexed from 0 to n-1. If the C
code writes to entry n, that will eventually cause problems.
Using R-devel and the following new feature
There is a new option, options(CBoundsCheck=), which controls how
.C() and .Fortran() pass arguments to compiled code. If true
(which can be enabled by setting the environment variable
R_C_BOUNDS_CHECK to yes), raw, integer, double and complex
arguments are
[R] Snallball, rJava, and R 2.15.1
I just bought a new machine and installed the latest release of R 2.15.1 two days ago. Loaded libraries yesterday (all reported successful loads). While running scripts, rJava and Snowball would not load. Here is Snowball successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/Snowball_0.0-8.zip' Content type 'application/zip' length 386840 bytes (377 Kb) opened URL downloaded 377 Kb package ‘Snowball’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages Here is rJava' successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/rJava_0.9-3.zip' Content type 'application/zip' length 746108 bytes (728 Kb) opened URL downloaded 728 Kb package ‘rJava’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages The directories for rJava and Snowball are viewable in the Libraries directory. Tried to run my scripts but rJava and Snowball failed to load giving the following error when the libraries try to load: library(rJava) Error : .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry Error: package/namespace load failed for ‘rJava’ library(Snowball) Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: NULL error: .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry In addition: Warning message: package ‘Snowball’ was built under R version 2.15.2 Error: package/namespace load failed for ‘Snowball’ I then found a different copy of rJava on R-Force last night. Installed it and it seemed to fix the rJava problem. However, Snowball continues to be a problem: Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: get(Info[i, 1], envir = env) error: internal error -3 in R_decompress1 In addition: Warning messages: 1: package ‘lsa’ was built under R version 2.15.2 2: package ‘Snowball’ was built under R version 2.15.2 Error: package ‘Snowball’ could not be loaded To illustrate this problem, I reinstalled the rJava package from Cran this morning. After writing this up, went back to R-force to reinstall a working copy of rJava, but it seems they have now pulled the rJava library so I am now stuck with two broken libraries. Thought the problem might be with R 2.15.1. Installed 2.15.0 but same results. One other point: at one time last night there was a window popping up with a warning message - something to do with Rf_copyListMatrix. That window is not coming up this morning. Let me know if there is anything else you need to figure this out. Thanks, Triss -- View this message in context: http://r.789695.n4.nabble.com/Snallball-rJava-and-R-2-15-1-tp4648007.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Cannot rescale a constant/zero column error.
I am trying to run the R Script below, I have actually simplified it to just this part that is causing issues. When I run this script I continue to get an error that says cannot rescale a constant/zero column to a unit variance. I cannot figure out what is going on here. I have stripped down my data file so it is more manageable so I can try to figure this out. The data.txt file that is being read looks like this: I have made this file very basic on purpose to see if I could get this to work, but it is not working. Of course once I get this to actually work I will expand the data file to match the data I am actually using. If I change the attribute in the prcomp function to scale=FALSE of course I can run my script. But if it is scaling...which is causing the issues, it errors. Any help would be GREATLY appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Cannot-rescale-a-constant-zero-column-error-tp4647996.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lavaan model
Dear R-users, Does somebody know what does the Estimate reported by the Lavaan model tell us? I assume this tells the relative strength of the dyadic relations. Thank you for your help! Regards, Sylvain -- View this message in context: http://r.789695.n4.nabble.com/Lavaan-model-tp4648004.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Mean Help
I have a dataframe. Let's suppose that i have two columns. The first one contains height, the second one contains eye color that can be Green, Blue or Brown. I want to calculate the aritmetic mean of the height only for those people who have Blue eyes. How can I do it? Thank you for your availability. -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] From summary function to formula
Hi there! I need to generate a final general linear model from a set of coefficients of the variables within the summary function. When I have the summary-function, HOW do I create a final model (response variable = bla bla bla)? LordSword -- View this message in context: http://r.789695.n4.nabble.com/From-summary-function-to-formula-tp4648008.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Forward and backward algorithm in R?
How can I wrtie and calculate alpha and beta in the forward backward algorithm in R ? -- View this message in context: http://r.789695.n4.nabble.com/Forward-and-backward-algorithm-in-R-tp4648012.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] gamlss mu.start vector ?
Dear All, I'd like to set up a loop whereby successive parameter values are used as start values in gamlss (yes I know this isn't usually necessary ! - unfortunately for my truncated data it is), to return the estimated parameters etc. giving the lowest AIC value. I notice that mu.start can take a vector of values - but I can't find any information as to what the function actually does with a vector of eg. mu.start values - does it cycle through the values and then give the result with the lowest AIC value ?? If it doesn't I'd appreciate any coding which has already been implemented (presumably using for loops) to achieve the above. Many thanks. -- View this message in context: http://r.789695.n4.nabble.com/gamlss-mu-start-vector-tp4648010.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Snallball, rJava, and R 2.15.1
Hello, It is said In addition: Warning messages: 1: package ‘lsa’ was built under R version 2.15.2 2: package ‘Snowball’ was built under R version 2.15.2 Install R 2.15.2 first. Regards, Pascal Le 31/10/12 23:29, Triss.Ashton a écrit : I just bought a new machine and installed the latest release of R 2.15.1 two days ago. Loaded libraries yesterday (all reported successful loads). While running scripts, rJava and Snowball would not load. Here is Snowball successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/Snowball_0.0-8.zip' Content type 'application/zip' length 386840 bytes (377 Kb) opened URL downloaded 377 Kb package ‘Snowball’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages Here is rJava' successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/rJava_0.9-3.zip' Content type 'application/zip' length 746108 bytes (728 Kb) opened URL downloaded 728 Kb package ‘rJava’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages The directories for rJava and Snowball are viewable in the Libraries directory. Tried to run my scripts but rJava and Snowball failed to load giving the following error when the libraries try to load: library(rJava) Error : .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry Error: package/namespace load failed for ‘rJava’ library(Snowball) Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: NULL error: .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry In addition: Warning message: package ‘Snowball’ was built under R version 2.15.2 Error: package/namespace load failed for ‘Snowball’ I then found a different copy of rJava on R-Force last night. Installed it and it seemed to fix the rJava problem. However, Snowball continues to be a problem: Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: get(Info[i, 1], envir = env) error: internal error -3 in R_decompress1 In addition: Warning messages: 1: package ‘lsa’ was built under R version 2.15.2 2: package ‘Snowball’ was built under R version 2.15.2 Error: package ‘Snowball’ could not be loaded To illustrate this problem, I reinstalled the rJava package from Cran this morning. After writing this up, went back to R-force to reinstall a working copy of rJava, but it seems they have now pulled the rJava library so I am now stuck with two broken libraries. Thought the problem might be with R 2.15.1. Installed 2.15.0 but same results. One other point: at one time last night there was a window popping up with a warning message - something to do with Rf_copyListMatrix. That window is not coming up this morning. Let me know if there is anything else you need to figure this out. Thanks, Triss -- View this message in context: http://r.789695.n4.nabble.com/Snallball-rJava-and-R-2-15-1-tp4648007.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Forward and backward algorithm in R?
On Wed, Oct 31, 2012 at 2:50 PM, quantum quan...@live.dk wrote: How can I wrtie and calculate alpha and beta in the forward backward algorithm in R ? You might need to be more specific Here's a smattering of links you should probably look into as well: http://r-manuals.flakery.org/R-intro.html http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example www.catb.org/esr/faqs/smart-questions.html http://mattgemmell.com/2008/12/08/what-have-you-tried/ Finally, I note you're posting from Nabble. Please include context in your reply -- I don't believe Nabble does this automatically, so you'll need to manually include it. Most of the regular respondents on this list don't use Nabble -- it is a _mailing list_ after all -- so we don't get the forum view you do, only emails of the individual posts. Combine that with the high volume of posts, and it's quite difficult to trace a discussion if we all don't make sure to include context. Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mean Help
Many forms, but I'll recommend this one this time around: aggregate(height ~ eyes, DATA, mean) See ?aggregate for an explanation of what aggregate() does and ?formula for an explanation of the tilde syntax. Note that this assumes your column names are height and eyes. Adjust as needed. Michael On Wed, Oct 31, 2012 at 1:18 PM, Hard Core gi...@hotmail.it wrote: I have a dataframe. Let's suppose that i have two columns. The first one contains height, the second one contains eye color that can be Green, Blue or Brown. I want to calculate the aritmetic mean of the height only for those people who have Blue eyes. How can I do it? Thank you for your availability. -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Snallball, rJava, and R 2.15.1
You forgot to install a JRE (Java runtime). And please use R 2.15.2, which is current, if you are installing from scratch. Some packages built with 2.15.2 (e.g. Matrix) do not work with 2.15.[01]. rJava is on CRAN/rforge.net, not 'R-force' (sic). PLEASE do read the posting guide http://www.R-project.org/posting-guide.html which includes updating your R before posting. On 31/10/2012 14:29, Triss.Ashton wrote: I just bought a new machine and installed the latest release of R 2.15.1 two days ago. Loaded libraries yesterday (all reported successful loads). While running scripts, rJava and Snowball would not load. Here is Snowball successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/Snowball_0.0-8.zip' Content type 'application/zip' length 386840 bytes (377 Kb) opened URL downloaded 377 Kb package ‘Snowball’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages Here is rJava' successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/rJava_0.9-3.zip' Content type 'application/zip' length 746108 bytes (728 Kb) opened URL downloaded 728 Kb package ‘rJava’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages The directories for rJava and Snowball are viewable in the Libraries directory. Tried to run my scripts but rJava and Snowball failed to load giving the following error when the libraries try to load: library(rJava) Error : .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry Error: package/namespace load failed for ‘rJava’ library(Snowball) Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: NULL error: .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry In addition: Warning message: package ‘Snowball’ was built under R version 2.15.2 Error: package/namespace load failed for ‘Snowball’ I then found a different copy of rJava on R-Force last night. Installed it and it seemed to fix the rJava problem. However, Snowball continues to be a problem: Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: get(Info[i, 1], envir = env) error: internal error -3 in R_decompress1 In addition: Warning messages: 1: package ‘lsa’ was built under R version 2.15.2 2: package ‘Snowball’ was built under R version 2.15.2 Error: package ‘Snowball’ could not be loaded To illustrate this problem, I reinstalled the rJava package from Cran this morning. After writing this up, went back to R-force to reinstall a working copy of rJava, but it seems they have now pulled the rJava library so I am now stuck with two broken libraries. Thought the problem might be with R 2.15.1. Installed 2.15.0 but same results. One other point: at one time last night there was a window popping up with a warning message - something to do with Rf_copyListMatrix. That window is not coming up this morning. Let me know if there is anything else you need to figure this out. Thanks, Triss -- View this message in context: http://r.789695.n4.nabble.com/Snallball-rJava-and-R-2-15-1-tp4648007.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] HELP!! how to remove 10% of data randomly in R
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Eugenie Sent: Wednesday, October 31, 2012 5:42 AM To: r-help@r-project.org Subject: Re: [R] HELP!! how to remove 10% of data randomly in R tDate tTimeO3 No2 Temp Sun Wspeed Wdirect Hum Indicator 119980101 2400 0.065 0.036 31.4 7659.9 351 NA 1 219980102 2400 0.053 0.025 31.8 6247.7 351 NA 1 319980103 2400 0.027 0.033 31.5 8528.8 331 NA 2 419980104 2400 0.034 0.023 30.7 6797.0 338 NA 2 519980105 2400 0.019 0.016 28.1 3769.6 354 NA 1 619980106 2400 0.021 0.018 29.9 6039.3 356 NA 1 719980107 2400 0.026 0.047 31.2 857 10.7 336 NA 1 819980108 2400 0.024 0.014 31.1 6357.8 330 NA 1 919980109 2400 0.058 0.033 32.5 742 10.7 334 NA 1 10 19980110 2400 0.026 0.032 33.9 923 10.6 347 NA 2 11 19980111 2400 0.064 0.034 32.5 7516.3 355 NA 2 12 19980112 2400 0.066 0.034 33.3 6978.5 319 NA 1 13 19980113 2400 0.026 0.030 33.4 992 12.5 341 NA 1 14 19980114 2400 0.101 0.028 33.8 7058.7 349 NA 1 15 19980115 2400 0.069 0.030 33.3 718 11.4 348 NA 1 16 19980116 2400 0.054 0.026 33.4 639 10.9 354 NA 1 17 19980117 2400 0.090 0.039 33.1 653 13.2 342 NA 2 18 19980118 2400 0.048 0.017 33.2 825 10.8 323 NA 2 19 19980119 2400 0.038 0.027 33.7 984 10.3 353 NA 1 20 19980120 2400 0.026 0.032 34.2 994 15.0 357 NA 1 21 19980121 2400 0.065 0.044 33.8 999 17.5 343 NA 1 22 19980122 2400 0.046 0.024 33.5 931 10.1 332 NA 1 23 19980123 2400 0.050 0.041 33.9 881 11.3 353 NA 1 24 19980124 2400 0.036 0.027 33.8 8779.1 328 NA 2 25 19980125 2400 0.043 0.021 33.2 777 10.5 340 NA 2 26 19980126 2400 0.029 0.016 33.1 999 14.1 341 NA 1 27 19980127 2400 0.033 0.030 33.9 943 12.9 344 NA 1 28 19980128 2400 0.040 0.022 33.7 805 12.6 354 NA 1 29 19980129 2400 0.029 0.015 30.2 5127.4 356 NA 1 30 19980130 2400 0.027 0.013 31.7 656 13.9 349 NA 1 if given data like this,how to remove the data in O3,NO2,sun,temp,wspeed randomly??(missing values in these rows columns) If your data frame is called dat, then something like this may do what you want, but since you haven't given an example of what you want the output to look like I am only guessing. dat[sample(1:30,5),3:8] - NA hope this is helpful, Dan Daniel J. Nordlund Olympia, WA 98504-5204 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mean Help
please provide a bit of the dataframe in the future using the dput() function x-data.frame(height=1:10, color=sample(c(blue,green,brown),10,replace=T)) x height color 1 1 blue 2 2 green 3 3 blue 4 4 brown 5 5 blue 6 6 brown 7 7 green 8 8 brown 9 9 green 10 10 brown blue.eyes - x[x$color==blue, ] blue.eyes height color 1 1 blue 3 3 blue 5 5 blue should create a new dataframe with only those that have blue eyes, then mean(blue.eyes$height) [1] 3 should provide you with the mean of those. On 31.10.2012, at 14:18, Hard Core wrote: I have a dataframe. Let's suppose that i have two columns. The first one contains height, the second one contains eye color that can be Green, Blue or Brown. I want to calculate the aritmetic mean of the height only for those people who have Blue eyes. How can I do it? Thank you for your availability. -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R crashing after successfully running compiled code
Aha - got it.
My problem was that I had a pointer (*Aest) that had less memory allocated
to it than I ended up storing in it (e.g. I had *Aest = 1:100, but stored
into it values at positions 5:105). With that fixed, all works flawlessly.
Thanks a lot for the help. What I hadn't realized was that .C allowed you
to exceed memory allocations - I'd have assumed that it would just crash
while I was running the function as soon as it ran out of space. Instead, I
guess it must have been writing beyond the space allocated for *Aest, and
not running into any trouble until R tried to store something else in the
same spot later (e.g. while plotting a figure).
It was immensely helpful to hear from you all that the function could still
have a bug, even if it ran successfully. As I understand it, the way
.Call passes variables, this sort of a mistake would not be possible. Is
this true? I'm tempted to learn the new syntax, but I also like how .C
allows me to keep things looking more or less like normal C code.
Adam
PS - As you mention, Prof. Ripley, I am insane for trying to do this
without a good debugger. Also, as you point out, valgrind doesn't run on
Cygwin. If you know of any useful PC debuggers, I'd be most grateful - but
if all else fails for debugging, I can just run Ubuntu in an Oracle
VirtualBox.
On Wed, Oct 31, 2012 at 9:38 AM, Adam Clark atcl...@umn.edu wrote:
Thanks for the advice.
I'll go ahead and dig through my C code. It's helpful to know that my C
code can cause R to crash AFTER successfully implementing the code.
I have made sure to account for C's vector indexing, and I think I'm
allocating my C memory, and passing information back and forth between C
and R, as I should. I'm including my input/output stuff below.
I tried including options(CBoundsCheck=TRUE) in the script both before
and after loading the C function, which doesn't seem to do much. To get it
to work, do I actually need to go into the R configuration file and edit
the default?
It would be exceedingly helpful if anybody could give me tips on where I'm
misusing pointers in the example below. That said, I certainly don't expect
the R community to debug my C code for me. If I come up with a solution,
I'll email it out over the list.
In R, I run the script:
dyn.load(mycfun.dll)
set.seed(1031)
A-1:100
B-runif(100)
myfunC-function(A, B, M, N) {
result-as.double(rep(0, (length(A)- N -(M+1
plengtht-as.integer(length(A))
Aest-as.numeric(rep(0, (length(A)- N -(M+1
distances-as.numeric(rep(0, length(A)))
neighbors-as.integer(rep(0, (M+1)))
u-as.numeric(rep(0, (M+1)))
w-as.numeric(rep(0, (M+1)))
return(.C(mycfun, as.double(A), as.double(B), as.integer(M),
as.integer(N),
result=as.double(result), as.integer(plengtht), as.double(Aest),
as.double(distances),
as.integer(neighbors), as.double(u), as.double(w))$result)
}
fun_result-myfunC(A,B,3,1)
This corresponds to the C code (input output only):
#include R.h
#include Rmath.h
void mycfun(double *A, double *B, int *pM, int *pN, double *result, int
*plengtht,
double *Aest, double *distances, int *neighbors, double *u, double *w) {
int t, i, j, n, from, to, nneigh;
double distsv, sumu, sumaest, corout;
int M = *pM;
int N = *pN;
int lengtht= *plengtht;
n=0;
# running various loops over variables #
result[n]=corout;
n=n+1;
}
# END #
I also have two sub-functions that manipulate neighbors and distances
- I can send the i/o for those as well, but they seem much more
straightforward, since I don't need to pass all my arguments as pointers. I
pass the pointers to internal variables at the beginning because I couldn't
index any C arrays using *pM or *pN.
Many thanks,
Adam
On Wed, Oct 31, 2012 at 7:12 AM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:
On Wed, 31 Oct 2012, Duncan Murdoch wrote:
On 12-10-30 11:13 PM, Adam Clark wrote:
I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32
and
2.13.0x64.
I am trying to run compiled C code using the .C convention. The
code compiles without problems, dynamically loads within the R
workspace with no problems, and even runs and gives correct results with
no problems.
However, R will randomly crash within a few minutes of successfully
using
the compiled function.
For example, if I run my compiled function using:
dyn.load(mycfun.dll)
answer-.C(mycfun, parameters...), I get a completely sensible
result that gets stored to answer.
However, if I try to do too many things to answer, the R exits
without warning.
I've tried dyn.unload in hopes that R would become stable afterwards,
but
in this case using the function crashes R without fail.
Usually, I can either plot, or view, or save answer to a file - but
never
take more than a single action before R exits. This does not appear to
depend on how long R has been open. Initially, I thought it was a bug in
the inline function, but I'm finding the same problem now
Re: [R] issues with krige function
Thierry, To answer your first question summary(casedata) station year monthday obs mpe Length:236 Min. :2011 Min. :8 Min. :28 Min. :0. Min. :0.0 Class :character 1st Qu.:2011 1st Qu.:8 1st Qu.:28 1st Qu.:0. 1st Qu.:0.0 Mode :character Median :2011 Median :8 Median :28 Median :0. Median :0.0 Mean :2011 Mean :8 Mean :28 Mean :0.6119 Mean :0.60163 3rd Qu.:2011 3rd Qu.:8 3rd Qu.:28 3rd Qu.:0.0225 3rd Qu.:0.03953 Max. :2011 Max. :8 Max. :28 Max. :9.3600 Max. :9.05766 bias typelat lon city Min. :-4.75583 Length:236 Min. :33.53 Min. :-84.84 Length:236 1st Qu.: 0.0 Class :character 1st Qu.:34.37 1st Qu.:-82.71 Class :character Median : 0.0 Mode :character Median :35.06 Median :-81.18 Mode :character Mean :-0.01028 Mean :35.18 Mean :-81.00 3rd Qu.: 0.03rd Qu.:35.99 3rd Qu.:-79.37 Max. : 2.82117 Max. :36.98 Max. :-76.33 statepreciptime Length:236 Length:236 Class :character Class :character Mode :character Mode :character summary(gridmeta) lon latxcell ycell row col Min. :-86.29 Min. :30.51 Min. :1895112 Min. :-6545105 Min. : 1.00 Min. : 1.0 1st Qu.:-82.38 1st Qu.:33.73 1st Qu.:2184464 1st Qu.:-6330770 1st Qu.: 61.75 1st Qu.: 46.0 Median :-79.88 Median :35.22 Median :2473816 Median :-6114053 Median :122.50 Median : 91.5 Mean :-79.91 Mean :35.26 Mean :2473816 Mean :-6114053 Mean :122.50 Mean : 91.5 3rd Qu.:-77.48 3rd Qu.:36.76 3rd Qu.:2763169 3rd Qu.:-5897337 3rd Qu.:183.25 3rd Qu.:137.0 Max. :-73.16 Max. :40.16 Max. :3052521 Max. :-5683002 Max. :244.00 Max. :182.0 To answer your second question, I'm not quite sure why, but apparently constraining nmax like that was exactly what it needed, because now it is producing values. v.fit = vgm(psill=1.2914,model=Mat,range=14.1611,nugget=0.23,kappa=83.1047) krige.pred.out = krige(formula=bias~1,locations=~lon+lat,data=casedata,newdata=gridmeta,model=v.fit) summary(krige.pred.out) lon lat var1.pred var1.var Min. :-86.29 Min. :30.51 Min. :-1.083 Min. :0.253 1st Qu.:-82.38 1st Qu.:33.73 1st Qu.:-0.044 1st Qu.:0.253 Median :-79.88 Median :35.22 Median : 0.000 Median :0.253 Mean :-79.91 Mean :35.26 Mean :-0.051 Mean :0.253 3rd Qu.:-77.48 3rd Qu.:36.76 3rd Qu.: 0.000 3rd Qu.:0.253 Max. :-73.16 Max. :40.16 Max. : 1.252 Max. :0.254 NA's :14925NA's :31483 Thanks for the help. Adrienne On Wed, Oct 31, 2012 at 10:28 AM, ONKELINX, Thierry thierry.onkel...@inbo.be wrote: Dear Adrienne, What is the output of summary(casestudy) and summary(gridmeta)? What happens if you set nmax to 10? krige(formula=bias~1,locations=~lon+lat,data=casedata,newdata=gridmeta ,model=v.fit, nmax = 10) ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Adrienne Wootten Verzonden: dinsdag 30 oktober 2012 18:36 Aan: r-help@r-project.org Onderwerp: [R] issues with krige function Greetings all, Ran into a strange problem with the krige function from geoR. The problem that I am having is that while the krige function seems to work well, the resulting predicted values are all NAs. Given the size of the datasets I am working with can't attach it, but I can provide snippets of the datasets. casedata station year month day obs mpebias type lat lon 140 319147 2011 8 28 0.00 0.0 0. COOP 35.48667 -82.96833 141 319354 2011 8 28 0.02 0.001305799 -0.01869420 COOP 34.25722 -78.68722 142 319357 2011 8 28 0.00 0.045194085 0.04519409 COOP 34.40944 -78.79139 143 319440 2011 8 28 6.92 5.530313974 -1.38968603 COOP 35.85278
Re: [R] HELP!! how to remove 10% of data randomly in R
On Oct 31, 2012, at 5:42 AM, Eugenie wrote:
tDate tTimeO3 No2 Temp Sun Wspeed Wdirect Hum Indicator
119980101 2400 0.065 0.036 31.4 7659.9 351 NA 1
219980102 2400 0.053 0.025 31.8 6247.7 351 NA 1
319980103 2400 0.027 0.033 31.5 8528.8 331 NA 2
419980104 2400 0.034 0.023 30.7 6797.0 338 NA 2
519980105 2400 0.019 0.016 28.1 3769.6 354 NA 1
619980106 2400 0.021 0.018 29.9 6039.3 356 NA 1
719980107 2400 0.026 0.047 31.2 857 10.7 336 NA 1
819980108 2400 0.024 0.014 31.1 6357.8 330 NA 1
919980109 2400 0.058 0.033 32.5 742 10.7 334 NA 1
10 19980110 2400 0.026 0.032 33.9 923 10.6 347 NA 2
11 19980111 2400 0.064 0.034 32.5 7516.3 355 NA 2
12 19980112 2400 0.066 0.034 33.3 6978.5 319 NA 1
13 19980113 2400 0.026 0.030 33.4 992 12.5 341 NA 1
14 19980114 2400 0.101 0.028 33.8 7058.7 349 NA 1
15 19980115 2400 0.069 0.030 33.3 718 11.4 348 NA 1
16 19980116 2400 0.054 0.026 33.4 639 10.9 354 NA 1
17 19980117 2400 0.090 0.039 33.1 653 13.2 342 NA 2
18 19980118 2400 0.048 0.017 33.2 825 10.8 323 NA 2
19 19980119 2400 0.038 0.027 33.7 984 10.3 353 NA 1
20 19980120 2400 0.026 0.032 34.2 994 15.0 357 NA 1
21 19980121 2400 0.065 0.044 33.8 999 17.5 343 NA 1
22 19980122 2400 0.046 0.024 33.5 931 10.1 332 NA 1
23 19980123 2400 0.050 0.041 33.9 881 11.3 353 NA 1
24 19980124 2400 0.036 0.027 33.8 8779.1 328 NA 2
25 19980125 2400 0.043 0.021 33.2 777 10.5 340 NA 2
26 19980126 2400 0.029 0.016 33.1 999 14.1 341 NA 1
27 19980127 2400 0.033 0.030 33.9 943 12.9 344 NA 1
28 19980128 2400 0.040 0.022 33.7 805 12.6 354 NA 1
29 19980129 2400 0.029 0.015 30.2 5127.4 356 NA 1
30 19980130 2400 0.027 0.013 31.7 656 13.9 349 NA 1
if given data like this,how to remove the data in O3,NO2,sun,temp,wspeed
randomly??(missing values in these rows columns)
Not clear whether those entries are to be NA or that you wanted a reduced size
dataframe. Perhaps:
is.na(dfrm[ sample(1:NROW(dfrm) , c('O3','NO2','sun','temp','wspeed')]) - TRUE
Note that the spellings of your column names and specified targets are not the
same, and so there is a further problem with you problem specification.
--
David Winsemius, MD
Alameda, CA, USA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cannot rescale a constant/zero column error.
On Oct 31, 2012, at 5:47 AM, fillay89 wrote: I am trying to run the R Script below, There is no script (or data description) appearing in this posting to the Rhelp Mailing List. Nabble is not Rhelp, despite Nabble's effort to get you to think it is an archive. Please review the Posting Guide that I know you were offered when you posted from Nabble. Yes, we could view this in Nabble, but most of us choose not to do so. This is a technical mailing list, not a website or chat-room. -- David. I have actually simplified it to just this part that is causing issues. When I run this script I continue to get an error that says cannot rescale a constant/zero column to a unit variance. I cannot figure out what is going on here. I have stripped down my data file so it is more manageable so I can try to figure this out. The data.txt file that is being read looks like this: I have made this file very basic on purpose to see if I could get this to work, but it is not working. Of course once I get this to actually work I will expand the data file to match the data I am actually using. If I change the attribute in the prcomp function to scale=FALSE of course I can run my script. But if it is scaling...which is causing the issues, it errors. Any help would be GREATLY appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Cannot-rescale-a-constant-zero-column-error-tp4647996.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R crashing after successfully running compiled code
As long as you use C (or C++ or Fortran ...), using memory that
you don't own is possible. This is one reason people use languages
like R.
(If you program microprocessors or write operating system code
then C's willingness to let you read or write any at address is essential.)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Adam Clark
Sent: Wednesday, October 31, 2012 8:47 AM
To: Adam Clark
Cc: r-help@r-project.org; Prof Brian Ripley
Subject: Re: [R] R crashing after successfully running compiled code
Aha - got it.
My problem was that I had a pointer (*Aest) that had less memory allocated
to it than I ended up storing in it (e.g. I had *Aest = 1:100, but stored
into it values at positions 5:105). With that fixed, all works flawlessly.
Thanks a lot for the help. What I hadn't realized was that .C allowed you
to exceed memory allocations - I'd have assumed that it would just crash
while I was running the function as soon as it ran out of space. Instead, I
guess it must have been writing beyond the space allocated for *Aest, and
not running into any trouble until R tried to store something else in the
same spot later (e.g. while plotting a figure).
It was immensely helpful to hear from you all that the function could still
have a bug, even if it ran successfully. As I understand it, the way
.Call passes variables, this sort of a mistake would not be possible. Is
this true? I'm tempted to learn the new syntax, but I also like how .C
allows me to keep things looking more or less like normal C code.
Adam
PS - As you mention, Prof. Ripley, I am insane for trying to do this
without a good debugger. Also, as you point out, valgrind doesn't run on
Cygwin. If you know of any useful PC debuggers, I'd be most grateful - but
if all else fails for debugging, I can just run Ubuntu in an Oracle
VirtualBox.
On Wed, Oct 31, 2012 at 9:38 AM, Adam Clark atcl...@umn.edu wrote:
Thanks for the advice.
I'll go ahead and dig through my C code. It's helpful to know that my C
code can cause R to crash AFTER successfully implementing the code.
I have made sure to account for C's vector indexing, and I think I'm
allocating my C memory, and passing information back and forth between C
and R, as I should. I'm including my input/output stuff below.
I tried including options(CBoundsCheck=TRUE) in the script both before
and after loading the C function, which doesn't seem to do much. To get it
to work, do I actually need to go into the R configuration file and edit
the default?
It would be exceedingly helpful if anybody could give me tips on where I'm
misusing pointers in the example below. That said, I certainly don't expect
the R community to debug my C code for me. If I come up with a solution,
I'll email it out over the list.
In R, I run the script:
dyn.load(mycfun.dll)
set.seed(1031)
A-1:100
B-runif(100)
myfunC-function(A, B, M, N) {
result-as.double(rep(0, (length(A)- N -(M+1
plengtht-as.integer(length(A))
Aest-as.numeric(rep(0, (length(A)- N -(M+1
distances-as.numeric(rep(0, length(A)))
neighbors-as.integer(rep(0, (M+1)))
u-as.numeric(rep(0, (M+1)))
w-as.numeric(rep(0, (M+1)))
return(.C(mycfun, as.double(A), as.double(B), as.integer(M),
as.integer(N),
result=as.double(result), as.integer(plengtht), as.double(Aest),
as.double(distances),
as.integer(neighbors), as.double(u), as.double(w))$result)
}
fun_result-myfunC(A,B,3,1)
This corresponds to the C code (input output only):
#include R.h
#include Rmath.h
void mycfun(double *A, double *B, int *pM, int *pN, double *result, int
*plengtht,
double *Aest, double *distances, int *neighbors, double *u, double *w) {
int t, i, j, n, from, to, nneigh;
double distsv, sumu, sumaest, corout;
int M = *pM;
int N = *pN;
int lengtht= *plengtht;
n=0;
# running various loops over variables #
result[n]=corout;
n=n+1;
}
# END #
I also have two sub-functions that manipulate neighbors and distances
- I can send the i/o for those as well, but they seem much more
straightforward, since I don't need to pass all my arguments as pointers. I
pass the pointers to internal variables at the beginning because I couldn't
index any C arrays using *pM or *pN.
Many thanks,
Adam
On Wed, Oct 31, 2012 at 7:12 AM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:
On Wed, 31 Oct 2012, Duncan Murdoch wrote:
On 12-10-30 11:13 PM, Adam Clark wrote:
I'm running R 2.15.1x64, though the same problem persists for 2.13.0x32
and
2.13.0x64.
I am trying to run compiled C code using the .C convention. The
code compiles without problems, dynamically loads within the R
workspace with no problems, and even runs and gives correct results with
no problems.
Re: [R] Snallball, rJava, and R 2.15.1
No difference.
R version 2.15.2 (2012-10-26) -- Trick or Treat
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-w64-mingw32/x64 (64-bit)
chooseCRANmirror()
utils:::menuInstallPkgs()
trying URL 'http://cran.stat.ucla.edu/bin/windows/contrib/2.15/rJava_0.9-3.zip'
Content type 'application/zip' length 746108 bytes (728 Kb)
opened URL
downloaded 728 Kb
package ‘rJava’ successfully unpacked and MD5 sums checked
The downloaded binary packages are in
C:\Users\Triss\AppData\Local\Temp\RtmpEJUIPF\downloaded_packages
local({pkg - select.list(sort(.packages(all.available = TRUE)),graphics=TRUE)
+ if(nchar(pkg)) library(pkg, character.only=TRUE)})
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: fun(libname, pkgname)
error: JAVA_HOME cannot be determined from the Registry
Error: package/namespace load failed for ‘rJava’
utils:::menuInstallPkgs()
also installing the dependencies ‘RWekajars’, ‘RWeka’
trying URL
'http://cran.stat.ucla.edu/bin/windows/contrib/2.15/RWekajars_3.7.7-1.zip'
Content type 'application/zip' length 5521316 bytes (5.3 Mb)
opened URL
downloaded 5.3 Mb
trying URL 'http://cran.stat.ucla.edu/bin/windows/contrib/2.15/RWeka_0.4-12.zip'
Content type 'application/zip' length 527842 bytes (515 Kb)
opened URL
downloaded 515 Kb
trying URL
'http://cran.stat.ucla.edu/bin/windows/contrib/2.15/Snowball_0.0-8.zip'
Content type 'application/zip' length 386840 bytes (377 Kb)
opened URL
downloaded 377 Kb
package ‘RWekajars’ successfully unpacked and MD5 sums checked
package ‘RWeka’ successfully unpacked and MD5 sums checked
package ‘Snowball’ successfully unpacked and MD5 sums checked
The downloaded binary packages are in
C:\Users\Triss\AppData\Local\Temp\RtmpEJUIPF\downloaded_packages
local({pkg - select.list(sort(.packages(all.available = TRUE)),graphics=TRUE)
+ if(nchar(pkg)) library(pkg, character.only=TRUE)})
Error : .onLoad failed in loadNamespace() for 'Snowball', details:
call: NULL
error: .onLoad failed in loadNamespace() for 'rJava', details:
call: fun(libname, pkgname)
error: JAVA_HOME cannot be determined from the Registry
Error: package/namespace load failed for ‘Snowball’
From: Pascal Oettli [kri...@ymail.com]
Sent: Wednesday, October 31, 2012 10:15 AM
To: Ashton, Triss
Cc: r-help@r-project.org
Subject: Re: [R] Snallball, rJava, and R 2.15.1
Hello,
It is said
In addition: Warning messages:
1: package ‘lsa’ was built under R version 2.15.2
2: package ‘Snowball’ was built under R version 2.15.2
Install R 2.15.2 first.
Regards,
Pascal
Le 31/10/12 23:29, Triss.Ashton a écrit :
I just bought a new machine and installed the latest release of R 2.15.1 two
days ago. Loaded libraries yesterday (all reported successful loads).
While running scripts, rJava and Snowball would not load. Here is Snowball
successful install:
utils:::menuInstallPkgs()
trying URL
'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/Snowball_0.0-8.zip'
Content type 'application/zip' length 386840 bytes (377 Kb)
opened URL
downloaded 377 Kb
package ‘Snowball’ successfully unpacked and MD5 sums checked
The downloaded binary packages are in
C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages
Here is rJava' successful install:
utils:::menuInstallPkgs()
trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/rJava_0.9-3.zip'
Content type 'application/zip' length 746108 bytes (728 Kb)
opened URL
downloaded 728 Kb
package ‘rJava’ successfully unpacked and MD5 sums checked
The downloaded binary packages are in
C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages
The directories for rJava and Snowball are viewable in the Libraries
directory. Tried to run my scripts but rJava and Snowball failed to load
giving the following error when the libraries try to load:
library(rJava)
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: fun(libname, pkgname)
error: JAVA_HOME cannot be determined from the Registry
Error: package/namespace load failed for ‘rJava’
library(Snowball)
Error : .onLoad failed in loadNamespace() for 'Snowball', details:
call: NULL
error: .onLoad failed in loadNamespace() for 'rJava', details:
call: fun(libname, pkgname)
error: JAVA_HOME cannot be determined from the Registry
In addition: Warning message:
package ‘Snowball’ was built under R version 2.15.2
Error: package/namespace load failed for ‘Snowball’
I then found a different copy of rJava on R-Force last night. Installed it
and it seemed to fix the rJava problem. However, Snowball continues to be a
problem:
Error : .onLoad failed in loadNamespace() for 'Snowball', details:
call: get(Info[i, 1], envir = env)
error: internal error -3 in R_decompress1
In addition: Warning messages:
1: package ‘lsa’ was built under R version 2.15.2
2: package ‘Snowball’ was built
Re: [R] From summary function to formula
?predict.xxx where xxx is either lm or glm, as I don't know what you mean by general linear model. I also don't know what you mean by summary function but have assumed it's fitted model. You need to read ?lm (or ?glm) carefully, and also follow some of the links therein. -- Bert On Wed, Oct 31, 2012 at 7:32 AM, LordSword kini...@student.sdu.dk wrote: Hi there! I need to generate a final general linear model from a set of coefficients of the variables within the summary function. When I have the summary-function, HOW do I create a final model (response variable = bla bla bla)? LordSword -- View this message in context: http://r.789695.n4.nabble.com/From-summary-function-to-formula-tp4648008.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Forward and backward algorithm in R?
R-site search with 'forward backward' returns references to packages for fitting hidden Markov models. hth, Ingmar https://www.google.com/search?q=forward+backwarddomains=r-project.orgsitesearch=r-project.orgbtnG=Google+Search On Wed, Oct 31, 2012 at 3:50 PM, quantum quan...@live.dk wrote: How can I wrtie and calculate alpha and beta in the forward backward algorithm in R ? -- View this message in context: http://r.789695.n4.nabble.com/Forward-and-backward-algorithm-in-R-tp4648012.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R crashing after successfully running compiled code
Adam Clark atcl...@umn.edu wrote: I'll go ahead and dig through my C code. My problem was that I had a pointer (*Aest) that had less memory allocated to it than I ended up storing in it ... On Oct 31, 2012, at 9:04 AM, William Dunlap wrote: As long as you use C (or C++ or Fortran ...), using memory that you don't own is possible. This is one reason people use languages like R. This seems suitable for the fortunes collection. Any seconds to this nomination? -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cannot rescale a constant/zero column error.
Sorry about that, this is my first time posting. Below is the complete post. I am trying to run the R Script below, I have actually simplified it to just this part that is causing issues. When I run this script I continue to get an error that says cannot rescale a constant/zero column to a unit variance. I cannot figure out what is going on here. I have stripped down my data file so it is more manageable so I can try to figure this out. The data.txt file that is being read looks like this: UserID,Q1,Q2 2342,3,2 I have made this file very basic on purpose to see if I could get this to work, but it is not working. Of course once I get this to actually work I will expand the data file to match the data I am actually using. library(stats) mydata - read.table(data.txt, header=TRUE, sep = ,) newdata -na.omit(mydata) pcdata - newdata[c(-1)] pcmatrix - data.frame(pcdata) # Add the difference so the totals in the scale = 100 pcmatrix[,1] - pcmatrix[,1] + 96 pcmatrix[,2] - pcmatrix[,2] + 96 x.pca - prcomp(pcmatrix, retx=TRUE, center=TRUE, scale=TRUE, cor=TRUE) weight - cbind(round(abs(x.pca$rotation[,1]),3)) write.table(weight,file=weights.txt) If I change the attribute in the prcomp function to scale=FALSE of course I can run my script. But if it is scaling...which is causing the issues, it errors. Any help would be GREATLY appreciated. On Oct 31, 2012, at 10:54 AM, David Winsemius wrote: On Oct 31, 2012, at 5:47 AM, fillay89 wrote: I am trying to run the R Script below, There is no script (or data description) appearing in this posting to the Rhelp Mailing List. Nabble is not Rhelp, despite Nabble's effort to get you to think it is an archive. Please review the Posting Guide that I know you were offered when you posted from Nabble. Yes, we could view this in Nabble, but most of us choose not to do so. This is a technical mailing list, not a website or chat-room. -- David. I have actually simplified it to just this part that is causing issues. When I run this script I continue to get an error that says cannot rescale a constant/zero column to a unit variance. I cannot figure out what is going on here. I have stripped down my data file so it is more manageable so I can try to figure this out. The data.txt file that is being read looks like this: I have made this file very basic on purpose to see if I could get this to work, but it is not working. Of course once I get this to actually work I will expand the data file to match the data I am actually using. If I change the attribute in the prcomp function to scale=FALSE of course I can run my script. But if it is scaling...which is causing the issues, it errors. Any help would be GREATLY appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Cannot-rescale-a-constant-zero-column-error-tp4647996.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Filling a covariance matrix
Hi Joshua,
The code you put together is very helpful. I have run into a small issue,
however, and I am surprised you weren't getting similar error message when
you tried running the nested for loops? As an example, I've pared down the
for loop a bit to highlight the error message I'm getting (note also that I
had to modify xs[[i]]$Q to xs[[i]]$*x*Q):
for(i in seq_along(xs)) {
for(j in seq_along(xs)) {
xcov[paste0(Q,i),paste0(Q,j)] - xcov[paste0(Q,j),paste0(Q,i)]
- cov(xs[[i]]$xQ, xs[[j]]$xQ, use=complete.obs)
}
}
#Error in cov(xs[[i]]$xQ, xs[[j]]$xQ, use = complete.obs) :
# incompatible dimensions
#Some investigation...
i
# 1
j
# 95
length(xs[[i]]$xQ)
# 96
length(xs[[j]]$xQ)
# 92
xs[[j]]$xQ[1:10]
#xQ
#2004-04-04 00:00:00 674
#2004-04-04 00:15:00 669
#2004-04-04 00:30:00 664
#2004-04-04 00:45:00 664
#2004-04-04 01:00:00 669
#2004-04-04 01:15:00 659
#2004-04-04 01:30:00 674
#2004-04-04 01:45:00 669
#2004-04-04 03:00:00 664
#2004-04-04 03:15:00 674
xs[[i]]$xQ[1:10]
# xQ
#2004-01-01 00:00:00 0.43
#2004-01-01 00:15:00 0.43
#2004-01-01 00:30:00 0.43
#2004-01-01 00:45:00 0.43
#2004-01-01 01:00:00 0.57
#2004-01-01 01:15:00 0.57
#2004-01-01 01:30:00 0.57
#2004-01-01 01:45:00 0.43
#2004-01-01 02:00:00 0.43
#2004-01-01 02:15:00 0.57
I suppose the reason use=complete.obs, use=pair, or
use=pairwise.complete.obs won't work is because the date stamp is
different despite the fact there are similar time stamps during the
respective days. Thus, I'm wondering if there is a way to direct the cov
function to make the calculation using only the time stamp (and not the date
stamp) to determine pairs?
Thanks, Eric
--
View this message in context:
http://r.789695.n4.nabble.com/Filling-a-covariance-matrix-tp4647170p4648015.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mean Help
Thank you ... can you explain what you've done? -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000p4648013.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mean Help
Thank you but with that formula i don't understand how R can select only the one with blue eyes -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000p4648030.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bootstrapping quantile regression
A possiblie solution might be to use the survey package. You could specify that the data is clustered using the svydesign function, and then speciy the replicate weights using the as.svrepdesign function. And then, it would be possible to use the withReplicates function to bootstrap the clusters A copy of Thomas Lumley's book - Complex Surveys - would probably help with this Hope this helps -- View this message in context: http://r.789695.n4.nabble.com/bootstrapping-quantile-regression-tp4647948p4648032.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mean Help
On Wed, Oct 31, 2012 at 4:31 PM, Hard Core gi...@hotmail.it wrote: Thank you but with that formula i don't understand how R can select only the one with blue eyes Assuming you're referring to my proposal of aggregate(height ~ eyes, DATA, mean) It's not blue eyes only: that will summarize height for each eye color by taking the mean. Then just select blue eyes and you're good to go. Finally, I note you're posting from Nabble. Please include context in your reply -- I don't believe Nabble does this automatically, so you'll need to manually include it. Most of the regular respondents on this list don't use Nabble -- it is a _mailing list_ after all -- so we don't get the forum view you do, only emails of the individual posts. Combine that with the high volume of posts, and it's quite difficult to trace a discussion if we all don't make sure to include context. I'm only following up because I have a hunch this involves my earlier reply, but it's really not clear. Cheers, Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] grep txt file names from html
Sorry, I know I should read a little 1st about this, but I am actually just helping somebody really quick and need help too. I want to grep all of the names of the .txt files mentioned on this html web page: http://www.epa.gov/emap/remap/html/three/data/index.html Thanks ahead of time. -- View this message in context: http://r.789695.n4.nabble.com/grep-txt-file-names-from-html-tp4648037.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Aggregate Table Data into Cell Frequencies
R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the temp (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency (Freq | counts) of each time each delayValue occurs in the cells. I've tried this command without luck, it seems to be dropping a bunch of values. tmp - reshape(sampleData, direction = wide,timevar = delayValue, new.row.names = unique(sampleData$id) ) structure(list(delayValue = c(0, NA, 7, 8, 9, 10, NA, NA, 4, 5, 6, 7, 8, NA, NA, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, NA, NA, 5, 6, 7, 8, 9, NA, 4, 5, 6, 7, 8, 9, 10, NA, 9, 10, NA, NA, NA, 4, 5, 6, NA, 5, 6, 7, NA, 7, 8, 9, 10, 11, NA, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, NA, 6, NA, 4, 7, NA, 4, NA, 2, 3, 4, NA, 5, NA, 5, 6, NA, 6, NA, 6, NA, 6, NA, 2, 3, NA, 5, NA, 3, 4, NA, 4, 5, NA, 4, NA, 3, NA, NA, 2, 3, 4, 5, 6, 7, 8, 9, NA, 9, NA, NA, 2, NA, 3, 4, 5, 6, NA, 4, 5, 6, NA, 5, NA, 2, 3, NA, 4, 5, 6, NA, 3, NA, NA, 2, NA, 2, NA, 6, NA, 2, NA, 4, NA, 2, 3, NA, 2, NA, 2, 3, 4, NA, 4, 5, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, 22, 23, 24, 25, 26, NA, NA, NA, NA, NA, 4, NA, NA, NA, NA, 5, 6, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, 14, NA, 10, NA, 11, 12, 13, 14, 15, 16, NA, 10, 11, 12, NA, 13, 14, 15, 16, 17, NA, 2, NA, 2, 3, 4, NA, 2, NA, 3, NA, 2, 3, 4, 5, 6, 5, 6, 7, NA, 6, 8, NA, 7, 8, NA, 6, 7, 8, 9, NA, NA, 6, 7, NA, NA, 5, 6, NA, 4, 5, 6, 7, NA, 6, NA, 7, 8, NA, 6, 7, 8, NA, 5, NA, 2, 3, 4, 5, NA, 7, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, 8, NA, 7, 8, NA, 5, 6, NA, 4, NA, 7, NA, 4, NA, 7, NA, 7, 8, NA, 8, NA, 7, 8, NA, 7, NA, 7, NA, 7, NA, NA, 16, 17, 18, NA, 19, 20, 21, NA, 21, 22, 23, 24, 25, 25, 26, 27, 28, NA, NA, NA, NA, NA, NA, NA, 4, 6, NA, NA, 4, 5, 6, 7, NA, 7, 8, 9, NA, 10, 11, 12, NA, 12, 13, 14, NA, 13, 14, NA, 12, 13, 14, NA, 15, 16, NA, NA, 2, 3, NA, 2, 3, NA, 4, 5, 6, NA, 5, 6, 7, 8, 9, 10, 11, 12, NA, 9, 10, 11, NA, 12, 13, NA, 9, 10, 11, NA, 12, 13, NA, 13, 14, 15, 16, 17, NA, 6, 7, NA, NA, 6, 7, 8, NA, 9, 10, 11, NA, NA, 5, 7, NA, 5, NA, NA, 2, 3, 4, 5, NA, NA, 8, 9, NA, 8, 9, NA, NA, NA, NA, NA, NA, NA, NA, NA, 6, NA, NA, 6, 7, 8, NA, 8, 9, NA, 9, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 7, 2, 3, NA, 2, NA, 2, 3, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, 3, 4, NA, 3, 4, NA, 5, NA, 2, 3, 4, NA, 3, 5, NA, 3, 4, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, 4, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, NA, 21, 22, 23, 24, NA, 23, 25, NA, 23, 24, 25, 26, NA, NA, 22, 23, 24, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, NA, 14, 15, NA, 13, 14, 15, NA, 14, 15, 16, NA, 15, 16, 17, 18, NA, 16, 17, NA, 7, 20, 21, 22, 23, NA, 22, 23, 24, NA, 25, 26, 27, NA, 27, 28, 29, 30, NA, 31, NA, NA, NA, NA, NA, NA, 7, 8, NA, NA, 8, NA, 8, 9, NA, 10, 11, 12, 13, NA, 13, 14, 15, NA, 15, 16, 17, NA, 15, 16, NA, 17, 18, 19, NA, NA, 12, 13, 14, NA, 13, NA, 9, 10, 11, NA, 12, 13, 14, 15, NA, NA, 9, 10, NA, NA, NA, NA, NA, 6, NA, NA, 4, 5, 6, NA, 3, 4, 5, NA, 6, 7, 8, NA, 9, NA, 3, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, 13, NA, 7, 13, NA, 11, 12, NA, 13, 14, 15, 16, 17, 14, NA, 13, 15, NA, 14, 15, 16, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, NA, 7, NA, 8, NA, 8, 9, 10, NA, 10, 11, 12, 13, NA, 13, 14, 15, 16, NA, 14, NA, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 4, 5, NA, 5, 6, NA, 4, 5, NA, 4, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, NA, 4, 5, 6, NA, 6, 7, 8, NA, 8, NA, 8, NA, 3, NA, 2, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 6, NA, 2, NA, 3, 4, 5, 6, 7, NA, 4, 5, NA, NA, NA, 2, 3, NA, 4, NA, 3, 4, NA, 7, 2, 3, 4, 5, 6, 4, 5, NA, 2, 3, NA, 3, 4, NA, 5, 6, 7, NA, 8, 9, 10, NA, 11, NA, 10, 12, NA, 9, 10, NA, 10, 11, NA, 6, 7, NA, 8, 9, 10, NA, 4, 5, 6, NA, 6, NA, 3, 4, 5, NA, 2, 3, NA, 3, 4, NA, 2, 3, NA, 3, 4, NA), Freq = c(0, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 4, 5, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 5, 1, 3, 2, 3, 3, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 1, 1, 4, 1, 1, 4, 1, 4, 1, 1, 2, 1, 3, 2, 1, 2, 2, 2, 3, 3, 2, 1, 4, 1, 1, 4, 2, 3, 2, 1, 3, 2, 1, 3, 2, 3, 2, 3, 5, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 4, 1, 1, 3, 1, 1, 1, 2, 1, 4, 5, 3, 2, 3, 3, 3, 2, 2, 4, 1, 5, 1, 1, 3, 2, 3, 2, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 5, 5, 5, 5, 1, 4, 5, 5, 6, 2, 1, 2, 6, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 2, 1, 2, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 6, 2, 1, 2, 6, 1, 1, 4, 1, 1, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 2, 3, 3, 2, 2, 3, 1, 2, 2, 1, 1, 3, 1,
Re: [R] grep txt file names from html
they're all of the form http.*txt but the best way to grep them (by which I assume you mean extract the file names from the page source) depends on what you plan to do with them, and what sort of output you expect. It isn't even clear whether you plan to do this in R. Sarah On Wed, Oct 31, 2012 at 12:56 PM, chuck.01 charliethebrow...@gmail.comwrote: Sorry, I know I should read a little 1st about this, but I am actually just helping somebody really quick and need help too. I want to grep all of the names of the .txt files mentioned on this html web page: http://www.epa.gov/emap/remap/html/three/data/index.html Thanks ahead of time. -- Sarah Goslee http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pseudo R-squared for model generated with spgm (splm)
I am working with the splm package. I use the spgm function: general estimation of a panel data model. Based on this approach, I know it is possible to compute a R2, eg the ratio of variation explained by a given model. My model is : bivmod-spgm(logIKA~NBLITRE0+NBLITRE1,data=mydatap,listw=comsKnn.nbW,spatial.error=TRUE) I know that we can calculate the R^2 as the variance of the fitted values from the reduced form of the model (Yfitted) over the variance of y (here logIKA). Since I am using the option lag = FALSE; the fitted value for an error model are Yfitted = X\beta. If I well understood : *1) We can get X* (NT * k matrix of observations on the non-stochastic regressors. with N: spatial units , T: time unit et k : number of non-stochastic regressors) with ** bivmod$model[,c(2,3)] # in the matrix of the data used (bivmod$model), the first column corresponds to the response variable and the subsequent one to regressors). NBLITRE0 NBLITRE1 ADAM-LES-PASSAVANT-2004 0 0 ADAM-LES-PASSAVANT-2005 0 0 ADAM-LES-PASSAVANT-2006 0 0 ADAM-LES-PASSAVANT-2007 0 0 ADAM-LES-PASSAVANT-2008 0 0 ADAM-LES-VERCEL-2004 0 0 *2) Beta* corresponds to coefficients associated to each regressor : we can get them with ** bivmod$coefficients NBLITRE0 NBLITRE1 -0.001131662 -0.001083650 But I do not know what means the \ in the formula. I suppose it is different from / which means divide by . Any hint appreciated. Regards, Marion -- Marion Jacquot Laboratoire de Chrono-environnement UMR UFC/CNRS 6249 USC INRA Université de Franche-Comté Place Leclerc F-25030 Besançon cedex FRANCE Tel. : +33 (0)381 665 829 Fax : +33 (0)381 665 797 http://chrono-environnement.univ-fcomte.fr/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep txt file names from html
On Oct 31, 2012, at 9:56 AM, chuck.01 wrote:
Sorry, I know I should read a little 1st about this, but I am actually just
helping somebody really quick and need help too.
I want to grep all of the names of the .txt files mentioned on this html web
page:
http://www.epa.gov/emap/remap/html/three/data/index.html
This shows code that will identify lines in that source page containing URLs
that end in '.txt'
lines -
readLines(con=url(http://www.epa.gov/emap/remap/html/three/data/index.html;)
)
Warning message:
In readLines(con =
url(http://www.epa.gov/emap/remap/html/three/data/index.html;)) :
incomplete final line found on
'http://www.epa.gov/emap/remap/html/three/data/index.html'
# You can generally ignore that warning.
length(grep('\\http://([./A-Za-z]){1+}\\.txt', lines) )
[1] 11
Should be fairly straightforward to remove the preceding and trailing material.
sub('(^.*\\)(http://([./A-Za-z]){1+}\\.txt)(.*$)', \\2, lines[
grep('\\http://([./A-Za-z]){1+}\\.txt', lines) ] )
[1]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/benthic/benmet.txt;
[2]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/benthic/bencnt.txt;
[3]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/location/watchr.txt;
[4]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/location/habbest.txt;
[5]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/design/sdesign.txt;
[6]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;
[7]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/fish/fshmet.txt;
[8]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/fish/fshcnt.txt;
[9]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/fish/fshnam.txt;
[10]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/tissue/ftmet.txt;
[11]
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/tissue/ftorg.txt;
Thanks ahead of time.
--
View this message in context:
http://r.789695.n4.nabble.com/grep-txt-file-names-from-html-tp4648037.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Alameda, CA, USA
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Aggregate Table Data into Cell Frequencies
On Oct 31, 2012, at 11:38 AM, Edward Patzelt patze...@umn.edu wrote: R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the temp (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency (Freq | counts) of each time each delayValue occurs in the cells. I've tried this command without luck, it seems to be dropping a bunch of values. tmp - reshape(sampleData, direction = wide,timevar = delayValue, new.row.names = unique(sampleData$id) ) structure(list(delayValue = c(0, NA, 7, 8, 9, 10, NA, NA, 4, 5, 6, 7, 8, NA, NA, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, NA, NA, 5, 6, 7, 8, 9, NA, 4, 5, 6, 7, 8, 9, 10, NA, 9, 10, NA, NA, NA, 4, 5, 6, NA, 5, 6, 7, NA, 7, 8, 9, 10, 11, NA, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, NA, 6, NA, 4, 7, NA, 4, NA, 2, 3, 4, NA, 5, NA, 5, 6, NA, 6, NA, 6, NA, 6, NA, 2, 3, NA, 5, NA, 3, 4, NA, 4, 5, NA, 4, NA, 3, NA, NA, 2, 3, 4, 5, 6, 7, 8, 9, NA, 9, NA, NA, 2, NA, 3, 4, 5, 6, NA, 4, 5, 6, NA, 5, NA, 2, 3, NA, 4, 5, 6, NA, 3, NA, NA, 2, NA, 2, NA, 6, NA, 2, NA, 4, NA, 2, 3, NA, 2, NA, 2, 3, 4, NA, 4, 5, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, 22, 23, 24, 25, 26, NA, NA, NA, NA, NA, 4, NA, NA, NA, NA, 5, 6, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, 14, NA, 10, NA, 11, 12, 13, 14, 15, 16, NA, 10, 11, 12, NA, 13, 14, 15, 16, 17, NA, 2, NA, 2, 3, 4, NA, 2, NA, 3, NA, 2, 3, 4, 5, 6, 5, 6, 7, NA, 6, 8, NA, 7, 8, NA, 6, 7, 8, 9, NA, NA, 6, 7, NA, NA, 5, 6, NA, 4, 5, 6, 7, NA, 6, NA, 7, 8, NA, 6, 7, 8, NA, 5, NA, 2, 3, 4, 5, NA, 7, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, 8, NA, 7, 8, NA, 5, 6, NA, 4, NA, 7, NA, 4, NA, 7, NA, 7, 8, NA, 8, NA, 7, 8, NA, 7, NA, 7, NA, 7, NA, NA, 16, 17, 18, NA, 19, 20, 21, NA, 21, 22, 23, 24, 25, 25, 26, 27, 28, NA, NA, NA, NA, NA, NA, NA, 4, 6, NA, NA, 4, 5, 6, 7, NA, 7, 8, 9, NA, 10, 11, 12, NA, 12, 13, 14, NA, 13, 14, NA, 12, 13, 14, NA, 15, 16, NA, NA, 2, 3, NA, 2, 3, NA, 4, 5, 6, NA, 5, 6, 7, 8, 9, 10, 11, 12, NA, 9, 10, 11, NA, 12, 13, NA, 9, 10, 11, NA, 12, 13, NA, 13, 14, 15, 16, 17, NA, 6, 7, NA, NA, 6, 7, 8, NA, 9, 10, 11, NA, NA, 5, 7, NA, 5, NA, NA, 2, 3, 4, 5, NA, NA, 8, 9, NA, 8, 9, NA, NA, NA, NA, NA, NA, NA, NA, NA, 6, NA, NA, 6, 7, 8, NA, 8, 9, NA, 9, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 7, 2, 3, NA, 2, NA, 2, 3, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, 3, 4, NA, 3, 4, NA, 5, NA, 2, 3, 4, NA, 3, 5, NA, 3, 4, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, 4, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, NA, 21, 22, 23, 24, NA, 23, 25, NA, 23, 24, 25, 26, NA, NA, 22, 23, 24, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, NA, 14, 15, NA, 13, 14, 15, NA, 14, 15, 16, NA, 15, 16, 17, 18, NA, 16, 17, NA, 7, 20, 21, 22, 23, NA, 22, 23, 24, NA, 25, 26, 27, NA, 27, 28, 29, 30, NA, 31, NA, NA, NA, NA, NA, NA, 7, 8, NA, NA, 8, NA, 8, 9, NA, 10, 11, 12, 13, NA, 13, 14, 15, NA, 15, 16, 17, NA, 15, 16, NA, 17, 18, 19, NA, NA, 12, 13, 14, NA, 13, NA, 9, 10, 11, NA, 12, 13, 14, 15, NA, NA, 9, 10, NA, NA, NA, NA, NA, 6, NA, NA, 4, 5, 6, NA, 3, 4, 5, NA, 6, 7, 8, NA, 9, NA, 3, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, 13, NA, 7, 13, NA, 11, 12, NA, 13, 14, 15, 16, 17, 14, NA, 13, 15, NA, 14, 15, 16, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, NA, 7, NA, 8, NA, 8, 9, 10, NA, 10, 11, 12, 13, NA, 13, 14, 15, 16, NA, 14, NA, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 4, 5, NA, 5, 6, NA, 4, 5, NA, 4, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, NA, 4, 5, 6, NA, 6, 7, 8, NA, 8, NA, 8, NA, 3, NA, 2, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 6, NA, 2, NA, 3, 4, 5, 6, 7, NA, 4, 5, NA, NA, NA, 2, 3, NA, 4, NA, 3, 4, NA, 7, 2, 3, 4, 5, 6, 4, 5, NA, 2, 3, NA, 3, 4, NA, 5, 6, 7, NA, 8, 9, 10, NA, 11, NA, 10, 12, NA, 9, 10, NA, 10, 11, NA, 6, 7, NA, 8, 9, 10, NA, 4, 5, 6, NA, 6, NA, 3, 4, 5, NA, 2, 3, NA, 3, 4, NA, 2, 3, NA, 3, 4, NA), Freq = c(0, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 4, 5, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 5, 1, 3, 2, 3, 3, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 1, 1, 4, 1, 1, 4, 1, 4, 1, 1, 2, 1, 3, 2, 1, 2, 2, 2, 3, 3, 2, 1, 4, 1, 1, 4, 2, 3, 2, 1, 3, 2, 1, 3, 2, 3, 2, 3, 5, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 4, 1, 1, 3, 1, 1, 1, 2, 1, 4, 5, 3, 2, 3, 3, 3, 2, 2, 4, 1, 5, 1, 1, 3, 2, 3, 2, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 5, 5, 5, 5, 1, 4, 5, 5, 6, 2, 1, 2, 6, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 2, 1, 2, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1,
Re: [R] FW: replace repeated id in a pedigree list
I use kinship package of R and it doesn't create a kinship matrix with repeated id. The kinship package is out of date, use the kinship2 and coxme packages (it was split into two parts). Then the problem you describe no longer exists -- the kinship function no longer requires each subject id to be unique across all pedigrees. pedlist - with(mydata, pedgiree(id, fa_id, mo_id, sex, famid=famid)) kmat - kinship(pedlist) fit - coxme(Surv(time, status) ~ x1 + x2 +... + (1| famid/id), varlist=kmat) The pedlist object contains all the families, plot(pedlist[4]) would pull out and plot the 4th family. The column labels of kmat will be of the form famid/id To solve your original problem (though you don't need to) create a single unified id. uid - paste(famid, id, sep='/') idlist - unique(uid) newid - match(uid, idlist) newmom - match(paste(famid, mo_id, sep='/'), idlist) newdad - match(paste(famid, fa_id, sep='/'), idlist) Terry Therneau author of kinship and coxme, but not :-) the maintainer of kinship2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Aggregate Table Data into Cell Frequencies
Hello, I'm not sure if this is what you want. aggregate(delayValue ~ id + Freq, data = sampleData, length) Hope this helps, Rui Barradas Em 31-10-2012 16:38, Edward Patzelt escreveu: R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the temp (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency (Freq | counts) of each time each delayValue occurs in the cells. I've tried this command without luck, it seems to be dropping a bunch of values. tmp - reshape(sampleData, direction = wide,timevar = delayValue, new.row.names = unique(sampleData$id) ) structure(list(delayValue = c(0, NA, 7, 8, 9, 10, NA, NA, 4, 5, 6, 7, 8, NA, NA, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, NA, NA, 5, 6, 7, 8, 9, NA, 4, 5, 6, 7, 8, 9, 10, NA, 9, 10, NA, NA, NA, 4, 5, 6, NA, 5, 6, 7, NA, 7, 8, 9, 10, 11, NA, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, 6, NA, NA, 6, NA, 4, 7, NA, 4, NA, 2, 3, 4, NA, 5, NA, 5, 6, NA, 6, NA, 6, NA, 6, NA, 2, 3, NA, 5, NA, 3, 4, NA, 4, 5, NA, 4, NA, 3, NA, NA, 2, 3, 4, 5, 6, 7, 8, 9, NA, 9, NA, NA, 2, NA, 3, 4, 5, 6, NA, 4, 5, 6, NA, 5, NA, 2, 3, NA, 4, 5, 6, NA, 3, NA, NA, 2, NA, 2, NA, 6, NA, 2, NA, 4, NA, 2, 3, NA, 2, NA, 2, 3, 4, NA, 4, 5, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, 22, 23, 24, 25, 26, NA, NA, NA, NA, NA, 4, NA, NA, NA, NA, 5, 6, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, 14, NA, 10, NA, 11, 12, 13, 14, 15, 16, NA, 10, 11, 12, NA, 13, 14, 15, 16, 17, NA, 2, NA, 2, 3, 4, NA, 2, NA, 3, NA, 2, 3, 4, 5, 6, 5, 6, 7, NA, 6, 8, NA, 7, 8, NA, 6, 7, 8, 9, NA, NA, 6, 7, NA, NA, 5, 6, NA, 4, 5, 6, 7, NA, 6, NA, 7, 8, NA, 6, 7, 8, NA, 5, NA, 2, 3, 4, 5, NA, 7, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, NA, 7, 8, NA, 7, 8, NA, 5, 6, NA, 4, NA, 7, NA, 4, NA, 7, NA, 7, 8, NA, 8, NA, 7, 8, NA, 7, NA, 7, NA, 7, NA, NA, 16, 17, 18, NA, 19, 20, 21, NA, 21, 22, 23, 24, 25, 25, 26, 27, 28, NA, NA, NA, NA, NA, NA, NA, 4, 6, NA, NA, 4, 5, 6, 7, NA, 7, 8, 9, NA, 10, 11, 12, NA, 12, 13, 14, NA, 13, 14, NA, 12, 13, 14, NA, 15, 16, NA, NA, 2, 3, NA, 2, 3, NA, 4, 5, 6, NA, 5, 6, 7, 8, 9, 10, 11, 12, NA, 9, 10, 11, NA, 12, 13, NA, 9, 10, 11, NA, 12, 13, NA, 13, 14, 15, 16, 17, NA, 6, 7, NA, NA, 6, 7, 8, NA, 9, 10, 11, NA, NA, 5, 7, NA, 5, NA, NA, 2, 3, 4, 5, NA, NA, 8, 9, NA, 8, 9, NA, NA, NA, NA, NA, NA, NA, NA, NA, 6, NA, NA, 6, 7, 8, NA, 8, 9, NA, 9, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 9, 10, NA, 7, 2, 3, NA, 2, NA, 2, 3, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, 3, 4, NA, 3, 4, NA, 5, NA, 2, 3, 4, NA, 3, 5, NA, 3, 4, NA, 2, 3, 4, NA, 2, 3, NA, 2, 3, 4, NA, 2, NA, 3, 4, NA, NA, 18, 19, 20, 21, NA, 21, 22, 23, 24, NA, 23, 25, NA, 23, 24, 25, 26, NA, NA, 22, 23, 24, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, 8, 9, 10, NA, 11, 12, 13, NA, 14, 15, NA, 13, 14, 15, NA, 14, 15, 16, NA, 15, 16, 17, 18, NA, 16, 17, NA, 7, 20, 21, 22, 23, NA, 22, 23, 24, NA, 25, 26, 27, NA, 27, 28, 29, 30, NA, 31, NA, NA, NA, NA, NA, NA, 7, 8, NA, NA, 8, NA, 8, 9, NA, 10, 11, 12, 13, NA, 13, 14, 15, NA, 15, 16, 17, NA, 15, 16, NA, 17, 18, 19, NA, NA, 12, 13, 14, NA, 13, NA, 9, 10, 11, NA, 12, 13, 14, 15, NA, NA, 9, 10, NA, NA, NA, NA, NA, 6, NA, NA, 4, 5, 6, NA, 3, 4, 5, NA, 6, 7, 8, NA, 9, NA, 3, 4, 5, NA, 6, 7, 8, 9, 10, 11, 12, 13, NA, 7, 13, NA, 11, 12, NA, 13, 14, 15, 16, 17, 14, NA, 13, 15, NA, 14, 15, 16, NA, NA, NA, NA, NA, 6, 7, NA, NA, 7, NA, 7, NA, 8, NA, 8, 9, 10, NA, 10, 11, 12, 13, NA, 13, 14, 15, 16, NA, 14, NA, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 4, 5, NA, 5, 6, NA, 4, 5, NA, 4, NA, 2, NA, 2, NA, 2, 3, NA, 2, 3, NA, 3, NA, 2, NA, NA, 4, 5, 6, NA, 6, 7, 8, NA, 8, NA, 8, NA, 3, NA, 2, NA, 2, NA, 2, NA, 2, NA, 2, 3, NA, 6, NA, 2, NA, 3, 4, 5, 6, 7, NA, 4, 5, NA, NA, NA, 2, 3, NA, 4, NA, 3, 4, NA, 7, 2, 3, 4, 5, 6, 4, 5, NA, 2, 3, NA, 3, 4, NA, 5, 6, 7, NA, 8, 9, 10, NA, 11, NA, 10, 12, NA, 9, 10, NA, 10, 11, NA, 6, 7, NA, 8, 9, 10, NA, 4, 5, 6, NA, 6, NA, 3, 4, 5, NA, 2, 3, NA, 3, 4, NA, 2, 3, NA, 3, 4, NA), Freq = c(0, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 4, 5, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 5, 1, 3, 2, 3, 3, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 1, 1, 4, 1, 1, 4, 1, 4, 1, 1, 2, 1, 3, 2, 1, 2, 2, 2, 3, 3, 2, 1, 4, 1, 1, 4, 2, 3, 2, 1, 3, 2, 1, 3, 2, 3, 2, 3, 5, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 1, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 4, 1, 1, 3, 1, 1, 1, 2, 1, 4, 5, 3, 2, 3, 3, 3, 2, 2, 4, 1, 5, 1, 1, 3, 2, 3, 2, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 5, 5, 5, 5, 1, 4, 5, 5, 6, 2, 1, 2, 6, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1,
Re: [R] Mean Help
Dear Michael ... You're a genius thak you very much really thank you !!! It worked!!! -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000p4648038.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mean Help
Sure. set.seed(1) dat1-data.frame(Height=sample(150:180,12,replace=TRUE),EyeColor=rep(c(Green,Blue,Brown),each=4)) dat1[,2]==Blue # [1] FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE # Returns a logical vector which checks whether each row of 2nd column of dat1 is Blue or not. dat1[,1][dat1[,2]==Blue] #gives me values of 1st column where the index (logical vector) is TRUE. #[1] 156 177 179 170 If I use, dat1[dat1[,2]==Blue,] #returns the subset of data with all the columns of dat1 # Height EyeColor #5 156 Blue #6 177 Blue #7 179 Blue #8 170 Blue #Finally, mean(dat1[,1][dat1[,2]==Blue]) gives the mean of those values I mentioned above. #[1] 170.5 - Original Message - From: Hard Core gi...@hotmail.it To: r-help@r-project.org Cc: Sent: Wednesday, October 31, 2012 11:19 AM Subject: Re: [R] Mean Help Thank you ... can you explain what you've done? -- View this message in context: http://r.789695.n4.nabble.com/Mean-Help-tp4648000p4648013.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] extracting information from txt file
Hello, Here is a link to some data: http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt I am trying to read this in, and want to use: chmval - read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, sep=,, skip= 84, header=T) the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt file # Header Records: 85 so, I need that # (-1) for input into the read.table statement above I've tried grep but that didn't work: (for this I downloaded the txt file and manually removed that hash mark!) grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements Any ideas? Can I just extract the 5th line? -- View this message in context: http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep txt file names from html
Sorry Sarah. I want to store them as a vector for use later. so, similar to this: links - c(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/benthic/benmet.txt;, http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/location/watchr.txt;, http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;) Sarah Goslee wrote they're all of the form http.*txt but the best way to grep them (by which I assume you mean extract the file names from the page source) depends on what you plan to do with them, and what sort of output you expect. It isn't even clear whether you plan to do this in R. Sarah On Wed, Oct 31, 2012 at 12:56 PM, chuck.01 lt; CharlieTheBrown77@ gt;wrote: Sorry, I know I should read a little 1st about this, but I am actually just helping somebody really quick and need help too. I want to grep all of the names of the .txt files mentioned on this html web page: http://www.epa.gov/emap/remap/html/three/data/index.html Thanks ahead of time. -- Sarah Goslee http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@ mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/grep-txt-file-names-from-html-tp4648037p4648043.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Snallball, rJava, and R 2.15.1
Found the problem on this. This is a new computer and does not have Java installed. Loaded JDK and everything is working. Thanks. -Original Message- From: Pascal Oettli [mailto:kri...@ymail.com] Sent: Wednesday, October 31, 2012 10:15 AM To: Ashton, Triss Cc: r-help@r-project.org Subject: Re: [R] Snallball, rJava, and R 2.15.1 Hello, It is said In addition: Warning messages: 1: package ‘lsa’ was built under R version 2.15.2 2: package ‘Snowball’ was built under R version 2.15.2 Install R 2.15.2 first. Regards, Pascal Le 31/10/12 23:29, Triss.Ashton a écrit : I just bought a new machine and installed the latest release of R 2.15.1 two days ago. Loaded libraries yesterday (all reported successful loads). While running scripts, rJava and Snowball would not load. Here is Snowball successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/Snowball_0.0-8.zip' Content type 'application/zip' length 386840 bytes (377 Kb) opened URL downloaded 377 Kb package ‘Snowball’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages Here is rJava' successful install: utils:::menuInstallPkgs() trying URL 'http://cran.cs.wwu.edu/bin/windows/contrib/2.15/rJava_0.9-3.zip' Content type 'application/zip' length 746108 bytes (728 Kb) opened URL downloaded 728 Kb package ‘rJava’ successfully unpacked and MD5 sums checked The downloaded binary packages are in C:\Users\Triss\AppData\Local\Temp\RtmpKsz7Qg\downloaded_packages The directories for rJava and Snowball are viewable in the Libraries directory. Tried to run my scripts but rJava and Snowball failed to load giving the following error when the libraries try to load: library(rJava) Error : .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry Error: package/namespace load failed for ‘rJava’ library(Snowball) Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: NULL error: .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: JAVA_HOME cannot be determined from the Registry In addition: Warning message: package ‘Snowball’ was built under R version 2.15.2 Error: package/namespace load failed for ‘Snowball’ I then found a different copy of rJava on R-Force last night. Installed it and it seemed to fix the rJava problem. However, Snowball continues to be a problem: Error : .onLoad failed in loadNamespace() for 'Snowball', details: call: get(Info[i, 1], envir = env) error: internal error -3 in R_decompress1 In addition: Warning messages: 1: package ‘lsa’ was built under R version 2.15.2 2: package ‘Snowball’ was built under R version 2.15.2 Error: package ‘Snowball’ could not be loaded To illustrate this problem, I reinstalled the rJava package from Cran this morning. After writing this up, went back to R-force to reinstall a working copy of rJava, but it seems they have now pulled the rJava library so I am now stuck with two broken libraries. Thought the problem might be with R 2.15.1. Installed 2.15.0 but same results. One other point: at one time last night there was a window popping up with a warning message - something to do with Rf_copyListMatrix. That window is not coming up this morning. Let me know if there is anything else you need to figure this out. Thanks, Triss -- View this message in context: http://r.789695.n4.nabble.com/Snallball-rJava-and-R-2-15-1-tp4648007.h tml Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting information from txt file
Hello, Use readLines instead. ?readLines # see argument 'n' readLines(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, n = 5)[5] Hope this helps, Rui Barradas Em 31-10-2012 16:46, chuck.01 escreveu: Hello, Here is a link to some data: http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt I am trying to read this in, and want to use: chmval - read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, sep=,, skip= 84, header=T) the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt file # Header Records: 85 so, I need that # (-1) for input into the read.table statement above I've tried grep but that didn't work: (for this I downloaded the txt file and manually removed that hash mark!) grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements Any ideas? Can I just extract the 5th line? -- View this message in context: http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] standard error for quantile
[see in-line below]
On 31-Oct-2012 10:26:14 PIKAL Petr wrote:
Hi Ted
-Original Message-
From: ted@deb [mailto:ted@deb] On Behalf Of Ted Harding
Sent: Tuesday, October 30, 2012 6:41 PM
To: r-help@r-project.org
snip
The general asymptotic result for the pth quantile (0p1) X.p of a
sample of size n is that it is asymptotically Normally distributed with
mean the pth quantile Q.p of the parent distribution and
var(X.p) = p*(1-p)/(n*f(Q.p)^2)
where f(x) is the probability density function of the parent
distribution.
So if I understand correctly p*(1-p) is biggest when p=0.5 and decreases with
smaller or bigger p. The var(X.p) then depends on ratio to parent
distribution at this p probability. For lognorm distribution and 200 values
the resulting var is
(0.5*(1-.5))/(200*qlnorm(.5, log(200), log(2))^2)
[1] 3.125e-08
(0.1*(1-.1))/(200*qlnorm(.1, log(200), log(2))^2)
[1] 6.648497e-08
so 0.1 var is slightly bigger than 0.5 var. For different distributions this
can be reversed as Jim pointed out.
Did I manage to understand?
Thank you very much.
Regards
Petr
Yes, it looks as though you understand! As a further illustration,
here is some R code applied to examples where the parent distrbution
is uniform or Normal. For each case, the reraults are stated as
first: simulated; second: by the formula. It can be seen that for
n=200 the formula and the simulations are close.
Ted.
###
## Test of formula for var(quantile)
varQ - function(p,n,f.p) {
p*(1-p)/(n*(f.p^2))
}
## Test 1: Uniform (0,1), n = 200
n - 200
## Pick one of (a), (b), (c):
## (a)# p - 0.50 ; q - 100 ; f.p - 1
## (b)# p - 0.25 ; q - 50 ; f.p - 1
## (c)# p - 0.10 ; q - 25 ; f.p - 1
Nsim - 1000
Qs - numeric(Nsim)
for( i in (1:Nsim) ){
Qs[i] - sort(runif(n))[q]
}
var(Qs)
varQ(p,n,f.p)
## (a) 0.001239982
## 0.00125
## (b) 0.0008877879
## 0.0009375
## (c) 0.0005619348
## 0.00045
## Test 2: N(0,1), n = 200
n - 200
## Pick one of (a), (b), (c):
## (a)# p - 0.50 ; q - 100 ; f.p - dnorm(qnorm(0.50))
## (b)# p - 0.25 ; q - 50 ; f.p - dnorm(qnorm(0.25))
## (c)# p - 0.10 ; q - 20 ; f.p - dnorm(qnorm(0.10))
Nsim - 1000
Qs - numeric(Nsim)
for( i in (1:Nsim) ){
Qs[i] - sort(rnorm(n))[q]
}
var(Qs)
varQ(p,n,f.p)
## (a) 0.007633568
## 0.007853982
## (b) 0.009370099
## 0.009283837
## (c) 0.01420517
## 0.01461055
###
This is not necessarily very helpful for small sample sizes (depending
on the parent distribution).
However, it is possible to obtain a general result giving an exact
confidence interval for Q.p given the entire ordered sample, though
there is only a restricted set of confidence levels to which it
applies.
If you'd like more detail about the above, I could write up derivations
and make the write-up available.
Hoping this helps,
Ted.
-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 30-Oct-2012 Time: 17:40:55
This message was sent by XFMail
-
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and provide commented, minimal, self-contained, reproducible code.
-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 31-Oct-2012 Time: 18:10:16
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting information from txt file
This worked fine for me: x - read.csv(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, skip=84, as.is = TRUE) str(x) 'data.frame': 711 obs. of 75 variables: $ ALDI: chr . . . . ... $ ALDS: chr . S S S ... $ ALDSF : chr ... $ ALKCALC : chr 106.05 210.7 73.51 432.63 ... $ ALOR: chr . S S S ... $ ALORF : chr ... $ ALTD: chr 54 36 47 12 ... $ ALTDF : chr ... $ ANC : chr 115 207.2 82.2 435.2 ... On Wed, Oct 31, 2012 at 12:46 PM, chuck.01 charliethebrow...@gmail.com wrote: Hello, Here is a link to some data: http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt I am trying to read this in, and want to use: chmval - read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, sep=,, skip= 84, header=T) the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt file # Header Records: 85 so, I need that # (-1) for input into the read.table statement above I've tried grep but that didn't work: (for this I downloaded the txt file and manually removed that hash mark!) grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements Any ideas? Can I just extract the 5th line? -- View this message in context: http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting information from txt file
Using na.string works better: x - read.csv(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, skip=84, as.is = TRUE, na.string = '.') str(x) 'data.frame': 711 obs. of 75 variables: $ ALDI: int NA NA NA NA NA NA NA NA NA NA ... $ ALDS: chr NA S S S ... $ ALDSF : chr ... $ ALKCALC : num 106 210.7 73.5 432.6 38.7 ... $ ALOR: chr NA S S S ... $ ALORF : chr ... $ ALTD: int 54 36 47 12 19 10 12 5 8 6 ... $ ALTDF : chr ... $ ANC : num 115 207.2 82.2 435.2 37.4 ... $ ANCF: chr ... $ ANDEF : num 82.5 52.3 31.8 21.9 12.2 ... $ ANSUM : num 771 728 328 892 251 ... $ CA : num 303 529 182 392 124 ... On Wed, Oct 31, 2012 at 12:46 PM, chuck.01 charliethebrow...@gmail.com wrote: Hello, Here is a link to some data: http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt I am trying to read this in, and want to use: chmval - read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, sep=,, skip= 84, header=T) the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt file # Header Records: 85 so, I need that # (-1) for input into the read.table statement above I've tried grep but that didn't work: (for this I downloaded the txt file and manually removed that hash mark!) grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements Any ideas? Can I just extract the 5th line? -- View this message in context: http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting information from txt file
On Oct 31, 2012, at 9:46 AM, chuck.01 wrote: Hello, Here is a link to some data: http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt I am trying to read this in, and want to use: chmval - read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, sep=,, skip= 84, header=T) the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt file # Header Records: 85 so, I need that # (-1) for input into the read.table statement above That # (-1) is fairly cryptic to my reading, but it appears you are seeing the behavior of the 3 character in terminating input for comments. Changing the comment character in the call to read.table will allow input from that line. ?read.table You will need to read only the first 5 or 6 lines first, then execute a separate read.table while skipping input from those lines as well as the variable list that forms a secondary header. headfrm - read.table( file=url( http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;), nrows=6, sep=:, comment.char=) headfrm V1 V2 1 Dataset EMAP Stream Chemistry Data 2File Name chmval 3 Date Created 02/22/99 4 # Variables 75 5 # Header Records 85 6 # Data Records 711 I've tried grep but that didn't work: (for this I downloaded the txt file and manually removed that hash mark!) grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements Any ideas? Can I just extract the 5th line? -- View this message in context: http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting information from txt file
On Oct 31, 2012, at 11:11 AM, David Winsemius wrote: On Oct 31, 2012, at 9:46 AM, chuck.01 wrote: Hello, Here is a link to some data: http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt I am trying to read this in, and want to use: chmval - read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;, sep=,, skip= 84, header=T) the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt file # Header Records: 85 so, I need that # (-1) for input into the read.table statement above That # (-1) is fairly cryptic to my reading, but it appears you are seeing the behavior of the 3 character in terminating input for comments. That would be the shifted-3. Changing the comment character in the call to read.table will allow input from that line. ?read.table You will need to read only the first 5 or 6 lines first, then execute a separate read.table while skipping input from those lines as well as the variable list that forms a secondary header. headfrm - read.table( file=url( http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;), nrows=6, sep=:, comment.char=) headfrm V1 V2 1 Dataset EMAP Stream Chemistry Data 2File Name chmval 3 Date Created 02/22/99 4 # Variables 75 5 # Header Records 85 6 # Data Records 711 I've tried grep but that didn't work: (for this I downloaded the txt file and manually removed that hash mark!) grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements Any ideas? Can I just extract the 5th line? -- View this message in context: http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] boxplots of various levels
ok this is close, thanks for your effort i will be more specific about my data set now, maybe that will help. there are 2 cats, labeled rs29 and rs30 there are a number of experimental results, i will make a graph set for each one. for example we can look at the results of BRCLNET. the tasks are devided into right middle and left the tasks are performed over time 1 to 10. i was a boxplot per cat per side (1 left 1 middle 1 right) over the time (1 to 10) at this point I have adjusted the code you sent me to this: mydata - data.frame(BRCLNET, cat, side, group) mydata then gives me a list of 4 columns: BRCLNET, cat, side, time that is perfect then: ggplot(mydata , aes( group, BRCLNET, fill = cat ))+ geom_boxplot() + + facet_wrap(~side) this gives me a figure divided into 3 columns: LEFT MIDDLE RIGHT with 2 boxes in each (RS29 and RS30). the yaxis is labeled BRCLNET and the xaxis is labeled time. but there is not a separate box for each time point. there are only 2 boxes per plot. i would like to have six separate plots: 1. LEFT RS29 BRCLNET over time (10 boxes) 2. MIDDLE RS29 BRCLNET overtime (10 boxes) . 6. RIGHT RS30 BRCLNET over time thanks for your help On Wed, Oct 31, 2012 at 8:16 AM, John Kane [via R] ml-node+s789695n4647989...@n4.nabble.com wrote: I am not sure I understand exactly what you want but does this do anything like what you want? library(ggplot2) mydata - data.frame(result = rnorm(100), group = rep(c(1, 2), each = 50), side = sample(c(L, R), 100, replace = TRUE) , dtime = rep(1:10, each=10)) p - ggplot(mydata , aes( group, result, fill = side ))+ geom_boxplot() + facet_wrap(~dtime) p John Kane Kingston ON Canada -Original Message- From: [hidden email]http://user/SendEmail.jtp?type=nodenode=4647989i=0 Sent: Tue, 30 Oct 2012 11:21:49 -0700 (PDT) To: [hidden email]http://user/SendEmail.jtp?type=nodenode=4647989i=1 Subject: [R] boxplots of various levels noob here trying to make boxplots of some data i would like to separate the boxplots according to conditons of various levels for example: i have group:1 and 2, each group performed tests consisting of condition A,B,C,D side: left and right time: 1 to 10 I would like separate boxplots of the results (x) of the tests (numeric) for each group under each condition on each side over time. so far i have set it up like this: boxplot(test$x~test$time) this gives me the plot for all vaues of x in each time bin. basicaly i would need a command that tells R to include only the data that agrees with the group, condition, and side I set. something like boxplot(test$x~test$time) where test$group=1,test$condition=A,test$side=left can this be done? -- View this message in context: http://r.789695.n4.nabble.com/boxplots-of-various-levels-tp4647917.html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] http://user/SendEmail.jtp?type=nodenode=4647989i=2mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Send any screenshot to your friends in seconds... Works in all emails, instant messengers, blogs, forums and social networks. TRY IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if2 for FREE __ [hidden email] http://user/SendEmail.jtp?type=nodenode=4647989i=3mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/boxplots-of-various-levels-tp4647917p4647989.html To unsubscribe from boxplots of various levels, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4647917code=ZHlzb25zcGhlcmUyM0BnbWFpbC5jb218NDY0NzkxN3wtMTkzNDk0ODEyOA== . NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml -- Kenneth Stephen Dyson Chercheur postdoctoral Groupe de recherche sur le système nerveux central (GRSNC) Université de Montréal Département de physiologie 2960, chemin de la tour, bureau 4130 Montréal (Québec) Canada H3T 1J4 Tel: (514) 343-6111 ext
[R] Problem Installing Packages
I run R as administrator but when i try to install packages i get this message: install.packages() Warning in install.packages : apertura non riuscita: stato HTTP '403 Forbidden' Warning in install.packages : apertura non riuscita: stato HTTP '403 Forbidden' Warning in install.packages : unable to access index for repository http://cran.mirror.garr.it/mirrors/CRAN/bin/windows/contrib/2.15 Error in install.packages : no packages were specified How can i solve this issue? -- View this message in context: http://r.789695.n4.nabble.com/Problem-Installing-Packages-tp4648051.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] party tree coordinates
I'm hoping that folks out there with expertise in working with the party package can help me out here. My team is trying to convert party tree output into a text file format that can be read by our image processing software. We are running into difficulties because the way the two different programs identify their nodes is different. R numbers it's nodes 1, 2, 3, 4, 5, 6 etc. down the yes answers in the tree until a terminal node is reached, at which point it backs up to the parent of that terminal node, gives its no child the next number available, follows any additional yes children down from there until it hits a terminal node, and repeats necessary. Our image processing system uses a coordinate system that labels each node according to its location in the tree (e.g. row 4, column 43). For instance, the first split in the tree is at location (1,1), the two daughter nodes are (2,1) and (2,2), and their daughters are (3,1), (3,2), (3,3), (3,4), etc. etc. on down the tree. It assigns these location coordinates referencing all possible node locations. In other words, if node (2,1) in the above scenario happened to be a terminal node, the daughters of node 2,2 would still be labeled (3,3) and (3,4), even though nodes (3,1) and (3,2) would not actually exist. They are still, from a labeling/ID perspective, retained, as it were, as phantom place holders. Given that the plot function in the party package is able to produce output that shows these column and row relationships in graphical format, it seems to me that there ought to be a way to extract this positional information from the package, but I haven't found a way to do it yet. The plot.tree info in the manual says it creates an (invisible) list with components x and y giving the coordinates of the tree nodes, and I'm wondering if it's talking about tree coordinates, as used by our image processor, or if it's refering to the x and y position on the printed plot. Either way, I don't know how to access those x/y numbers. At any rate, I'm pretty stumped. It seems to me that since the positional relationships are produced in the plots, there ought to be a way to get at them, programmatically. Any suggestions out there? -- View this message in context: http://r.789695.n4.nabble.com/party-tree-coordinates-tp4648047.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] predict glm() with offset
Dear R friends. I have a question about running a glm( family= 'binomial', *offset=T*), (I know offset is a vector of values) My doubt is about predicting the values on a new data. Does the predict() function considers the offset? o should I especified something? Here is the model I´m using: *model-stepAIC(glm(f_ocur~altitud+UTM_X+UTM_Y+j_sin+j_cos+temp_res+pp, offset=(log(1/offset)) data=data, family='binomial', ), direction='both')* Now, I want to get the estimated values, how should I do it considering that I´m stablishing an offset value in the model? If anyone can help me, i´d really appreciate it. Thank you. Lucas. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Stock Watson DOLS in R
Hi everyone, Longtime reader first time questioner. A colleague was showing me the EViews statistical package and demonstrating some of its regression capabilities--like DOLS--and commenting how easy it was to do certain kinds of regression. Being more of an R type, I went to try and replicate his regression results in R and to my surprise couldn't find a way--or anyone discussing how--to perform a dynamic ordinary least squares regression in R. Are there any packages or guidelines on how to do this? I've started messing with d() and L() in dynlm to see if I can recreate the leads/lags portions of the DOLS equation, but I'm having little success at the moment. I'm incredibly confused how to do this at the moment so any pointers would be appreciated! Sincerely, Stephen [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting information from txt file
This worked for the example you provided. Assumes the header count is the only
numeric value on the 5th line.
epa_extract - function(address){
doc - readLines(address, n = 5)[5]
head_count - as.numeric(gsub(\\D, , doc))
read.table(address, sep = ,, header = TRUE, skip = head_count)
}
foo -
epa_extract(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;)
Taimur Sajid
Research Development Analyst
Primatics Financial
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of chuck.01
Sent: Wednesday, October 31, 2012 12:47 PM
To: r-help@r-project.org
Subject: [R] extracting information from txt file
Hello,
Here is a link to some data:
http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt
I am trying to read this in, and want to use:
chmval -
read.table(http://www.epa.gov/emap/html/data/surfwatr/data/mastreams/9396/wchem/chmval.txt;,
sep=,, skip= 84, header=T)
the # 84, for 84 lines skipped needs to be derived from the 5th line of the txt
file # Header Records: 85
so, I need that # (-1) for input into the read.table statement above
I've tried grep but that didn't work:
(for this I downloaded the txt file and manually removed that hash mark!)
grep(Header Records:, read.table(chmval.txt, header=T)) Error in scan(file,
what, nmax, sep, dec, quote, skip, nlines, na.strings,
:
line 1 did not have 5 elements
Any ideas?
Can I just extract the 5th line?
--
View this message in context:
http://r.789695.n4.nabble.com/extracting-information-from-txt-file-tp4648033.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] HELP!! how to remove 10% of data randomly in R
HI, May be this helps. dat1-read.table(text= TDate TTime O3 No2 Temp Sun Wspeed Wdirect Hum Indicator 1 19980101 2400 0.065 0.036 31.4 765 9.9 351 NA 1 2 19980102 2400 0.053 0.025 31.8 624 7.7 351 NA 1 3 19980103 2400 0.027 0.033 31.5 852 8.8 331 NA 2 4 19980104 2400 0.034 0.023 30.7 679 7.0 338 NA 2 5 19980105 2400 0.019 0.016 28.1 376 9.6 354 NA 1 6 19980106 2400 0.021 0.018 29.9 603 9.3 356 NA 1 7 19980107 2400 0.026 0.047 31.2 857 10.7 336 NA 1 8 19980108 2400 0.024 0.014 31.1 635 7.8 330 NA 1 9 19980109 2400 0.058 0.033 32.5 742 10.7 334 NA 1 10 19980110 2400 0.026 0.032 33.9 923 10.6 347 NA 2 11 19980111 2400 0.064 0.034 32.5 751 6.3 355 NA 2 12 19980112 2400 0.066 0.034 33.3 697 8.5 319 NA 1 13 19980113 2400 0.026 0.030 33.4 992 12.5 341 NA 1 14 19980114 2400 0.101 0.028 33.8 705 8.7 349 NA 1 15 19980115 2400 0.069 0.030 33.3 718 11.4 348 NA 1 16 19980116 2400 0.054 0.026 33.4 639 10.9 354 NA 1 17 19980117 2400 0.090 0.039 33.1 653 13.2 342 NA 2 18 19980118 2400 0.048 0.017 33.2 825 10.8 323 NA 2 19 19980119 2400 0.038 0.027 33.7 984 10.3 353 NA 1 20 19980120 2400 0.026 0.032 34.2 994 15.0 357 NA 1 21 19980121 2400 0.065 0.044 33.8 999 17.5 343 NA 1 22 19980122 2400 0.046 0.024 33.5 931 10.1 332 NA 1 23 19980123 2400 0.050 0.041 33.9 881 11.3 353 NA 1 24 19980124 2400 0.036 0.027 33.8 877 9.1 328 NA 2 25 19980125 2400 0.043 0.021 33.2 777 10.5 340 NA 2 26 19980126 2400 0.029 0.016 33.1 999 14.1 341 NA 1 27 19980127 2400 0.033 0.030 33.9 943 12.9 344 NA 1 28 19980128 2400 0.040 0.022 33.7 805 12.6 354 NA 1 29 19980129 2400 0.029 0.015 30.2 512 7.4 356 NA 1 30 19980130 2400 0.027 0.013 31.7 656 13.9 349 NA 1 ,sep=,header=TRUE,stringsAsFactors=FALSE) #creating NA for 10% of data in the specified columns (deviant of David's method). is.na(dat1[sample(1:nrow(dat1),0.1*nrow(dat1)),3:7])-TRUE tail(dat1) # TDate TTime O3 No2 Temp Sun Wspeed Wdirect Hum Indicator #25 19980125 2400 NA NA NA NA NA 340 NA 2 #26 19980126 2400 0.029 0.016 33.1 999 14.1 341 NA 1 #27 19980127 2400 0.033 0.030 33.9 943 12.9 344 NA 1 #28 19980128 2400 0.040 0.022 33.7 805 12.6 354 NA 1 #29 19980129 2400 0.029 0.015 30.2 512 7.4 356 NA 1 #30 19980130 2400 0.027 0.013 31.7 656 13.9 349 NA 1 #If you need to create NA for individual columns randomly res-do.call(cbind,lapply(lapply(dat1[,3:7],function(x) data.frame(x)),function(x) x[sample(1:nrow(x),0.1*nrow(x)),])) dat1[,3][dat1[,3]%in%res[,1]]-NA dat1[,4][dat1[,4]%in%res[,2]]-NA dat1[,5][dat1[,5]%in%res[,3]]-NA dat1[,6][dat1[,6]%in%res[,4]]-NA dat1[,7][dat1[,7]%in%res[,5]]-NA head(dat1) # TDate TTime O3 No2 Temp Sun Wspeed Wdirect Hum Indicator #1 19980101 2400 0.065 0.036 31.4 765 9.9 351 NA 1 #2 19980102 2400 0.053 0.025 31.8 624 7.7 351 NA 1 #3 19980103 2400 0.027 0.033 31.5 852 8.8 331 NA 2 #4 19980104 2400 NA NA 30.7 679 7.0 338 NA 2 #5 19980105 2400 0.019 0.016 28.1 376 9.6 354 NA 1 #6 19980106 2400 0.021 0.018 29.9 603 NA 356 NA 1 A.K. - Original Message - From: Eugenie leemean...@hotmail.com To: r-help@r-project.org Cc: Sent: Wednesday, October 31, 2012 8:42 AM Subject: Re: [R] HELP!! how to remove 10% of data randomly in R tDate tTime O3 No2 Temp Sun Wspeed Wdirect Hum Indicator 1 19980101 2400 0.065 0.036 31.4 765 9.9 351 NA 1 2 19980102 2400 0.053 0.025 31.8 624 7.7 351 NA 1 3 19980103 2400 0.027 0.033 31.5 852 8.8 331 NA 2 4 19980104 2400 0.034 0.023 30.7 679 7.0 338 NA 2 5 19980105 2400 0.019 0.016 28.1 376 9.6 354 NA 1 6 19980106 2400 0.021 0.018 29.9 603 9.3 356 NA 1 7 19980107 2400 0.026 0.047 31.2 857 10.7 336 NA 1 8 19980108 2400 0.024 0.014 31.1 635 7.8 330 NA 1 9 19980109 2400 0.058 0.033 32.5 742 10.7 334 NA 1 10 19980110 2400 0.026 0.032 33.9 923 10.6 347 NA 2 11 19980111 2400 0.064 0.034 32.5 751 6.3 355 NA 2 12 19980112 2400 0.066 0.034 33.3 697 8.5 319 NA 1 13 19980113 2400 0.026 0.030 33.4 992 12.5 341 NA 1 14 19980114 2400 0.101 0.028 33.8 705 8.7 349 NA 1 15 19980115 2400 0.069 0.030 33.3 718 11.4 348 NA 1 16 19980116 2400 0.054 0.026 33.4 639 10.9
Re: [R] boxplots of various levels
Hello, Like this? library(ggplot2) set.seed(3101) mydata - data.frame( BRCLNET = rnorm(1000), group = rep(c(1, 2), each = 500), side = sample(c(Left, Middle, Right), 1000, replace = TRUE) , dtime = rep(1:10, 100)) p - ggplot(mydata , aes(factor(dtime), BRCLNET, fill = side)) + geom_boxplot() + facet_grid(group~side) p Note that it's factor(dtime) and facet_grid(), not facet_wrap() Hope this helps, Rui Barradas Em 31-10-2012 17:33, dysonsphere escreveu: ok this is close, thanks for your effort i will be more specific about my data set now, maybe that will help. there are 2 cats, labeled rs29 and rs30 there are a number of experimental results, i will make a graph set for each one. for example we can look at the results of BRCLNET. the tasks are devided into right middle and left the tasks are performed over time 1 to 10. i was a boxplot per cat per side (1 left 1 middle 1 right) over the time (1 to 10) at this point I have adjusted the code you sent me to this: mydata - data.frame(BRCLNET, cat, side, group) mydata then gives me a list of 4 columns: BRCLNET, cat, side, time that is perfect then: ggplot(mydata , aes( group, BRCLNET, fill = cat ))+ geom_boxplot() + + facet_wrap(~side) this gives me a figure divided into 3 columns: LEFT MIDDLE RIGHT with 2 boxes in each (RS29 and RS30). the yaxis is labeled BRCLNET and the xaxis is labeled time. but there is not a separate box for each time point. there are only 2 boxes per plot. i would like to have six separate plots: 1. LEFT RS29 BRCLNET over time (10 boxes) 2. MIDDLE RS29 BRCLNET overtime (10 boxes) . 6. RIGHT RS30 BRCLNET over time thanks for your help On Wed, Oct 31, 2012 at 8:16 AM, John Kane [via R] ml-node+s789695n4647989...@n4.nabble.com wrote: I am not sure I understand exactly what you want but does this do anything like what you want? library(ggplot2) mydata - data.frame(result = rnorm(100), group = rep(c(1, 2), each = 50), side = sample(c(L, R), 100, replace = TRUE) , dtime = rep(1:10, each=10)) p - ggplot(mydata , aes( group, result, fill = side ))+ geom_boxplot() + facet_wrap(~dtime) p John Kane Kingston ON Canada -Original Message- From: [hidden email]http://user/SendEmail.jtp?type=nodenode=4647989i=0 Sent: Tue, 30 Oct 2012 11:21:49 -0700 (PDT) To: [hidden email]http://user/SendEmail.jtp?type=nodenode=4647989i=1 Subject: [R] boxplots of various levels noob here trying to make boxplots of some data i would like to separate the boxplots according to conditons of various levels for example: i have group:1 and 2, each group performed tests consisting of condition A,B,C,D side: left and right time: 1 to 10 I would like separate boxplots of the results (x) of the tests (numeric) for each group under each condition on each side over time. so far i have set it up like this: boxplot(test$x~test$time) this gives me the plot for all vaues of x in each time bin. basicaly i would need a command that tells R to include only the data that agrees with the group, condition, and side I set. something like boxplot(test$x~test$time) where test$group=1,test$condition=A,test$side=left can this be done? -- View this message in context: http://r.789695.n4.nabble.com/boxplots-of-various-levels-tp4647917.html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] http://user/SendEmail.jtp?type=nodenode=4647989i=2mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Send any screenshot to your friends in seconds... Works in all emails, instant messengers, blogs, forums and social networks. TRY IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if2 for FREE __ [hidden email] http://user/SendEmail.jtp?type=nodenode=4647989i=3mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/boxplots-of-various-levels-tp4647917p4647989.html To unsubscribe from boxplots of various levels, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4647917code=ZHlzb25zcGhlcmUyM0BnbWFpbC5jb218NDY0NzkxN3wtMTkzNDk0ODEyOA== .
Re: [R] Problem Installing Packages
On Oct 31, 2012, at 10:46 AM, Hard Core wrote: I run R as administrator but when i try to install packages i get this message: install.packages() Warning in install.packages : apertura non riuscita: stato HTTP '403 Forbidden' Warning in install.packages : apertura non riuscita: stato HTTP '403 Forbidden' Warning in install.packages : unable to access index for repository http://cran.mirror.garr.it/mirrors/CRAN/bin/windows/contrib/2.15 Error in install.packages : no packages were specified How can i solve this issue? What were you expecting to happen when you ran install.packages() with no arguments? -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lavaan model
On 10/31/2012 02:47 PM, sylvain.giroud wrote: Dear R-users, Does somebody know what does the Estimate reported by the Lavaan model tell us? I assume this tells the relative strength of the dyadic relations. The 'Estimate' column contains the estimated model parameters. There are many different kinds of model parameters. Under the heading 'Latent variables', you may have factor loadings. Under the heading 'Regressions' you may have regression coefficients. Other parameters are (residual) variances and (residual) covariances. Yves Rosseel. http://lavaan.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] expand.grip for permutations
if i were to have a block size of 4 people and i want to assign a treatment
combination to the entire block, there would be 16 different treatment
combinations (, TTTP, TTPP, PTTP, etc.)
i am trying to get all 16 permutations and i am able to use this code below.
drugs=c('P','T');
comb=expand.grid(drugs,drugs,drugs,drugs)
for a block size of 3 the code would be comb=expand.grid(drugs,drugs,drugs)
and for a block size of 2 it would be comb=expand.grid(drugs,drugs).
my question is whether there is a way to automatically create the comb
variable. i tried using expand.grid(rep(drugs, block.size) but that didn't
work.
any help on how i can proceed? thanks
--
View this message in context:
http://r.789695.n4.nabble.com/expand-grip-for-permutations-tp4648067.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Webinar signup: Gradient Boosting and Classification Trees: A Winning Combination. November 9, 10-11 a.m., PST
Webinar signup: Tree Ensembles and Classification Trees: A Winning Combination November 9, 10-11 a.m., PST. Webinar Registration: http://2.salford-systems.com/gradientboosting/ Understand major shortcomings of using only decision trees and how tree ensembles can help overcome these challenges and improve your model building. Combine all of the advantages of using classification and regression trees with the power-house interpretability of stochastic gradient boosting. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Strange compiling behaviour
Compiling this little function gets me some strange behaviour
.initDataDir - function(){
if(file.exists(LOCATION)) {
if(as.logical(file.info(LOCATION)[isdir]))return
stop(LOCATION, exists but is not a directory)
}
Z - dir.create(LOCATION)
if(!Z){
stop(geterrmessage())
##stop(Juggel)
}
return(Z)
}
.initDataDir - function(){
+ if(file.exists(LOCATION)) {
+ if(as.logical(file.info(LOCATION)[isdir]))return
+ stop(LOCATION, exists but is not a directory)
+ }
+ Z - dir.create(LOCATION)
+ if(!Z){
+ stop(geterrmessage())
+ ##stop(Juggel)
+ }
+ return(Z)
+ }
LOCATION
[1] /xxx
Q - .initDataDir()
Error in .initDataDir() :
Error in .initDataDir() : Error in .initDataDir() :
Error in .initDataDir() : Error in .initDataDir() : Juggel
In addition: Warning message:
In dir.create(LOCATION) :
cannot create dir '/xxx', reason 'Permission denied'
I commented out the stop(Juggel) line
I made a mistake when I first created the function
stop(geterrormessage())
Which failed of course.
Here are some more attempts
.initDataDir - function(){
+ if(file.exists(LOCATION)) {
+ if(as.logical(file.info(LOCATION)[isdir]))return
+ stop(LOCATION, exists but is not a directory)
+ }
+ Z - dir.create(LOCATION)
+ if(!Z){
+ stop(geterrmessage())
+ stop(Juggel)
+ }
+ return(Z)
+ }
LOCATION
[1] /xxx
Q - .initDataDir()
Error in .initDataDir() : Error in .initDataDir() : Juggel
In addition: Warning message:
In dir.create(LOCATION) :
cannot create dir '/xxx', reason 'Permission denied'
.initDataDir - function(){
+ if(file.exists(LOCATION)) {
+ if(as.logical(file.info(LOCATION)[isdir]))return
+ stop(LOCATION, exists but is not a directory)
+ }
+ Z - dir.create(LOCATION)
+ if(!Z){
+ stop(geterrormessage())
+ stop(Juggel)
+ }
+ return(Z)
+ }
LOCATION
[1] /xxx
Q - .initDataDir()
Error in stop(geterrormessage()) :
could not find function geterrormessage
In addition: Warning message:
In dir.create(LOCATION) :
cannot create dir '/xxx', reason 'Permission denied'
I tried killing the R session and restarting and then got this
.initDataDir - function(){
+ if(file.exists(LOCATION)) {
+ if(as.logical(file.info(LOCATION)[isdir]))return
+ stop(LOCATION, exists but is not a directory)
+ }
+ Z - dir.create(LOCATION)
+ if(!Z){
+ stop(geterrmessage())
+ ##stop(Juggel)
+ }
+ return(Z)
+ }
LOCATION - /
Q - .initDataDir()
Error in .initDataDir() : Error in stop(geterrormessage()) :
could not find function geterrormessage
In addition: Warning message:
In dir.create(LOCATION) :
cannot create dir '/', reason 'Permission denied'
This is very strange, but not a show stopper.
.version
_
platform x86_64-pc-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
major 2
minor 14.1
year 2011
month 12
day22
svn rev57956
language R
version.string R version 2.14.1 (2011-12-22)
cheers
Worik
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem Installing Packages
On Wed, Oct 31, 2012 at 8:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Oct 31, 2012, at 10:46 AM, Hard Core wrote: I run R as administrator but when i try to install packages i get this message: install.packages() Warning in install.packages : apertura non riuscita: stato HTTP '403 Forbidden' Warning in install.packages : apertura non riuscita: stato HTTP '403 Forbidden' Warning in install.packages : unable to access index for repository http://cran.mirror.garr.it/mirrors/CRAN/bin/windows/contrib/2.15 Error in install.packages : no packages were specified How can i solve this issue? What were you expecting to happen when you ran install.packages() with no arguments? Fo me it brings up an interactive menu, so I don't think that's the problem. Have you tried a different mirror? Italy/Milano seems down but Italy/Padua works for me. Use the chooseCRANmirror() function. Cheers, Michael __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] expand.grip for permutations
Instead of
rep(drugs, block.size)
use
rep(list(drugs), block.size)
as the argument to expand.grid.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of trekvana
Sent: Wednesday, October 31, 2012 1:18 PM
To: r-help@r-project.org
Subject: [R] expand.grip for permutations
if i were to have a block size of 4 people and i want to assign a treatment
combination to the entire block, there would be 16 different treatment
combinations (, TTTP, TTPP, PTTP, etc.)
i am trying to get all 16 permutations and i am able to use this code below.
drugs=c('P','T');
comb=expand.grid(drugs,drugs,drugs,drugs)
for a block size of 3 the code would be comb=expand.grid(drugs,drugs,drugs)
and for a block size of 2 it would be comb=expand.grid(drugs,drugs).
my question is whether there is a way to automatically create the comb
variable. i tried using expand.grid(rep(drugs, block.size) but that didn't
work.
any help on how i can proceed? thanks
--
View this message in context: http://r.789695.n4.nabble.com/expand-grip-for-
permutations-tp4648067.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cannot rescale a constant/zero column error.
Sounds like one of your data columns is constant. The variance of a constant is 0, and scaling would then divide by 0, which is impossible. Kevin Wright On Wed, Oct 31, 2012 at 7:47 AM, fillay89 jasonf...@gmail.com wrote: I am trying to run the R Script below, I have actually simplified it to just this part that is causing issues. When I run this script I continue to get an error that says cannot rescale a constant/zero column to a unit variance. I cannot figure out what is going on here. I have stripped down my data file so it is more manageable so I can try to figure this out. The data.txt file that is being read looks like this: I have made this file very basic on purpose to see if I could get this to work, but it is not working. Of course once I get this to actually work I will expand the data file to match the data I am actually using. If I change the attribute in the prcomp function to scale=FALSE of course I can run my script. But if it is scaling...which is causing the issues, it errors. Any help would be GREATLY appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Cannot-rescale-a-constant-zero-column-error-tp4647996.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kevin Wright [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] party tree coordinates
On Wed, 31 Oct 2012, aelmore wrote: I'm hoping that folks out there with expertise in working with the party package can help me out here. My team is trying to convert party tree output into a text file format that can be read by our image processing software. We are running into difficulties because the way the two different programs identify their nodes is different. R numbers it's nodes 1, 2, 3, 4, 5, 6 etc. down the yes answers in the tree until a terminal node is reached, at which point it backs up to the parent of that terminal node, gives its no child the next number available, follows any additional yes children down from there until it hits a terminal node, and repeats necessary. And that's essentially also how the visualization works. A suitably sized grid is set up first and then grid viewports are recursively created and traversed. Our image processing system uses a coordinate system that labels each node according to its location in the tree (e.g. row 4, column 43). For instance, the first split in the tree is at location (1,1), the two daughter nodes are (2,1) and (2,2), and their daughters are (3,1), (3,2), (3,3), (3,4), etc. etc. on down the tree. It assigns these location coordinates referencing all possible node locations. In other words, if node (2,1) in the above scenario happened to be a terminal node, the daughters of node 2,2 would still be labeled (3,3) and (3,4), even though nodes (3,1) and (3,2) would not actually exist. They are still, from a labeling/ID perspective, retained, as it were, as phantom place holders. Given that the plot function in the party package is able to produce output that shows these column and row relationships in graphical format, it seems to me that there ought to be a way to extract this positional information from the package, but I haven't found a way to do it yet. I think we haven't got anything in the package that delivers this out of the box. (Also note that the tree is often sparser than all binary splits that could occur.) The plot.tree info in the manual says it creates an (invisible) list with components x and y giving the coordinates of the tree nodes, and I'm wondering if it's talking about tree coordinates, as used by our image processor, or if it's refering to the x and y position on the printed plot. Either way, I don't know how to access those x/y numbers. That's not from the party package anyways. plot.tree is from the tree package whose code bases are not related. At any rate, I'm pretty stumped. It seems to me that since the positional relationships are produced in the plots, there ought to be a way to get at them, programmatically. Any suggestions out there? I would suggest that you take a loot at the partykit package. It contains a reimplementation of ctree() based on acleaner design of the recursive tree structure. It also offers more functions for user interaction. It comes with a vignette that is still somewhat rough but should give you a good idea how things work. hth, Z __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Σχετ: Σχετ: Swap rows and columns in a matrix
Okay; I think I get it. One (rather kludgy) way of accomplishing your goal is as follows: Let the two matrices which you want to check for being isomorphic be A and B. Do: rA - apply(A,1,paste,collapse= ) rB - apply(B,1,paste,collapse= ) cA - apply(A,2,paste,collapse= ) cB - apply(B,2,paste,collapse= ) iso - all(sort(rA) == sort(rB)) | all(sort(cA) == sort(cB)) Then iso will be TRUE if A and B are isomorphic and FALSE otherwise. As you can see, this works by converting the rows and columns of your matrices to character strings. This should be OK --- I think that the base method duplicated.data.frame() works along similar lines --- but it does seem kludgy to me. Perhaps someone else will come up with a sexier idea. cheers, Rolf On 31/10/12 19:57, Haris Rhrlp wrote: i want to check for isomorphish between matrices i will give you an example to see what i want. 1 -1 -1 1 1 1 1 1 11 -1 -1 -1 -1 1 -1 -1 1 -1 1 -1 -1 1 -1 this 2 matrices are isomorphic beacause if i swap the first 2 rows the matrices will be intetical next example 1 -1 -1 -1 1 -1 1 1 1 1 1 1 -1 -1 1 -1 -1 1 -1 1 -1 1 -1 -1 this 2 matrices are isomorphic beacuse if i swap the first 2 columns the matrices will be intetical i want to write an algorithm to check all the swaps of the rows and all the swaps of the columns and then if the matrices are indetical then will be isomorphic *Áðï:* Rolf Turner rolf.tur...@xtra.co.nz *Ðñïò:* Haris Rhrlp haris_r_h...@yahoo.com *Êïéí.:* R-help@r-project.org R-help@r-project.org *ÓôÜëèçêå:* 10:53 ì.ì. Ôñßôç, 30 Ïêôùâñßïõ 2012 *Èåìá:* Re: Ó÷åô: [R] Swap rows and columns in a matrix Unless there is a good reason not to, you should keep discussions on-list. On 31/10/12 08:29, Haris Rhrlp wrote: thank you for your answer but i dont want the transpose of matrix. I want to swap rows seperatly and columns the same Then I am afraid that your question is all Greek to me. :-) You will have to explain --- with a simple example or two --- exactly what it is that you want to accomplish. cheers, Rolf Turner *Áðï:* Rolf Turner rolf.tur...@xtra.co.nz mailto:rolf.tur...@xtra.co.nz *Ðñïò:* Haris Rhrlp haris_r_h...@yahoo.com mailto:haris_r_h...@yahoo.com *Êïéí.:* R-help@r-project.org mailto:R-help@r-project.org R-help@r-project.org mailto:R-help@r-project.org *ÓôÜëèçêå:* 9:17 ì.ì. Ôñßôç, 30 Ïêôùâñßïõ 2012 *Èåìá:* Re: [R] Swap rows and columns in a matrix On 31/10/12 07:59, Haris Rhrlp wrote: Dear R users, I want a help to write an algorithm for swapping rows and columns in a matrix thanks in advance ?t (???) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] expand.grip for permutations
thank you both! that worked! -- View this message in context: http://r.789695.n4.nabble.com/expand-grip-for-permutations-tp4648067p4648076.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] gauss fit with outlier removal
I have distribution that are gaussian to a good approximation. I fit a gaussian to these distributons. Once in a while there is an outlier. Could someone suggest a robust method (R package already?) that removes those outliers and redoes the gaussian fit to get a better fit? Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Clustering groups according to multiple variables
Dear R help, I am trying to cluster my data according to group in a data frame such as the following: df=data.frame(group=rep(c(a,b,c,d),10),(replicate(100,rnorm(40 I'm not sure how to tell hclust() that I want to cluster according to the group variable. For example: dfclust=hclust(dist(df),ave) plot(dfclust) Clusters according to each individual row. What I'm looking for is an unrooted tree that will show similarity/dissimilarity among groups according to the data set as a whole. I appreciate the help, MO [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
