[R] Odp: R not recognizing words
Hi Hello all I'm new to R and am experiencing a problem with a categorical variable. All the data of this variable are Low, High, or NA. When I put summary(x$y), it gives me the number of High, Low, and NA entries. However, when I try to subset by writing x$y==Low or x$y==High, R does not recognize the word and it writes FALSE for all the entries (but not the NA entries). Can anybody help me out? You are not telling the whole story x- sample(c(Low, High), 20, replace=T) x [1] Low Low Low High Low Low Low Low Low High [11] Low High High Low Low High Low Low High Low x[5:7] - NA x-data.frame(y=x) x$y==Low [1] TRUE TRUE TRUE FALSENANANA TRUE TRUE FALSE TRUE FALSE [13] FALSE TRUE TRUE FALSE TRUE TRUE FALSE TRUE x[x$y==Low,] [1] Low Low Low NA NA NA Low Low Low Low Low Low Low Low Levels: High Low Works as expected. So probably your expectation is wrong. Regards Petr Thanks -- View this message in context: http://r.789695.n4.nabble.com/R-not- recognizing-words-tp4268283p4268283.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] question about rev
Hi [R] question about rev Hi,all: I have a vector,and wanna get the opposite value via rev function. a [1] FALSE FALSE TRUE TRUE TRUE rev(a) [1] TRUE TRUE TRUE FALSE FALSE I don't know why the 3rd TRUE has not been reversed,while all other values are reversed? Misunderstanding what rev function does. see letters rev(letters) !a Regards Petr Thanks My best [[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.
[R] Reply on ggplot2 - tricky problem
That comes really close to what I had in mind.
Thanks a lot for helping out, Justin!
Good luck,
Mario
Re: [R] ggplot2 - tricky problem
how bout:
dat-data.frame(id=1:4,city=c('berlin','munich'),likeability=c(5,4,6,5),uniqueness=c(3,4,4,4))
ggplot(ddply(melt(dat,
id.vars=c('id','city')),
.(variable,city),
summarise,
value=mean(value)),
aes(x=factor(city),y=value)) +
geom_point() +
facet_wrap(~variable)
the line drawing is a bit more tricky... Since the x values are factors
rather than continuous, fitting a line to them is kind of nonsense. It
matters which order they are in for example. If instead you want to
plot something like:
ggplot(dat,aes(x=likeability,y=uniqueness,colour=city))+geom_point()+geom_smooth(aes(group=city),method='lm')
You could draw fit lines that make a bit more sense. Forgive me if I'm over
simplifying your problem!
Justin
On Thu, Jan 5, 2012 at 7:46 AM, Mario Giesel rr.gie...@yahoo.de wrote:
Hello, R friends,
I've been struggling quite a bit with ggplot2.
Having worked through Hadleys book twice I still wonder how to solve this task.
1. Short example Dataframe:
id city Likeability Uniqueness
1 Berlin 5 3
2 Munich 4 4
3 Berlin 6 4
4 Munich 5 4
2. Task:
a) Facetting plots for each attitude (1 plot for likeability and uniqueness
each, horizontally on one page)
b) Showing Berlin and Munich together on x axis
c) Showing the means of Berlin and Munich on y axis (means of cities in
likeability on first plot, means of cities in uniqueness on second plot)
d) Drawing a line through mean points on each plot
Hope I could explain it understandably. Any help is appreciated!
Thanks a lot,
Mario
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Antwort an: Antwort an Justin Haynes Senden
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[R] how to remove an element from list or character
HI I want to remove an element from a list or character. i can any of the class type like cat= [1] mac med if my input is 2 i need to remove med and if my input is one i need to remove mac from cat Please help - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/how-to-remove-an-element-from-list-or-character-tp4268556p4268556.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] Bayesian estimate of prevalence with an imperfect test
Thanks Bert I've been through my variables again and have managed to get the code working - shouldn't have tried to deal with it at the end of the day yesterday! all the best Lian -- View this message in context: http://r.789695.n4.nabble.com/Bayesian-estimate-of-prevalence-with-an-imperfect-test-tp4265595p4268526.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] question about rev
Yes£¬I find out later. rev is only reverse the order. What I use is ! Thanks At 2012-01-06 16:07:11,Petr PIKAL petr.pi...@precheza.cz wrote: Hi [R] question about rev Hi,all: I have a vector,and wanna get the opposite value via rev function. a [1] FALSE FALSE TRUE TRUE TRUE rev(a) [1] TRUE TRUE TRUE FALSE FALSE I don't know why the 3rd TRUE has not been reversed,while all other values are reversed? Misunderstanding what rev function does. see letters rev(letters) !a Regards Petr Thanks My best [[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. [[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] need help with connecting to SQL Server2008 Express with RJDBC
i have been trying RJDBC for many hours but still failed to connect to my sql server. platform: win7 professional; SQL server 2008 express; the code is : drv - JDBC(com.microsoft.sqlserver.jdbc.SQLServerDriver, C:/Users/myname/sqljdbc_3.0/enu/sqljdbc4.jar); dbConnect(drv,jdbc:sqlserver://mypc/MyDB:1433;databaseName=bkt1;) and I got this err message: jcall(drv@jdrv, Ljava/sql/Connection;, connect, as.character(url)[1], : com.microsoft.sqlserver.jdbc.SQLServerException: The TCP/IP connection to the host cvez-PC/MyDB, port 1433 has failed. Error: null. Verify the connection properties, check that an instance of SQL Server is running on the host and accepting TCP/IP connections at the port, and that no firewall is blocking TCP connections to the port.. I have turned down MS fire wall and I have opened port 1433, as you can see in this pic http://r.789695.n4.nabble.com/file/n4268704/sql_port_config.png Any help will be appreciated, thank you very much! -- View this message in context: http://r.789695.n4.nabble.com/need-help-with-connecting-to-SQL-Server2008-Express-with-RJDBC-tp4268704p4268704.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] question about rev
the function /rev / REVERSEs the order of the vector, as v= c(1,2,3) rev(v)= 3,2,1 if you want the opposite value , just add a exclamation in front of the vector: v=c(T,F,T,T); !v is F,T,F,F -- View this message in context: http://r.789695.n4.nabble.com/question-about-rev-tp4268518p4268716.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] how to remove an element from list or character
Le jeudi 05 janvier 2012 à 23:52 -0800, arunkumar a écrit : HI I want to remove an element from a list or character. i can any of the class type like cat= [1] mac med if my input is 2 i need to remove med and if my input is one i need to remove mac from cat This is really basic, please read the R intro or some tutorial. Basically, you can do: input - 1 (or input - 2) cat - cat[-input] Cheers __ 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] plot rq lm
I asked a while ago what diagnostic plots exists for quantile regression, here: http://stats.stackexchange.com/questions/19291/what-diagnostic-plots-exists-for-quantile-regression And had received no answer. I hope some more information will become available regarding this. Best, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Wed, Jan 4, 2012 at 7:52 PM, agent dunham crossp...@hotmail.com wrote: Dear Community, I'd like to plot an rq object the same way I do with a lm one, is it possible? Something like this plot(rqmodel , 1:4, id.labels=rownames(pga1)); where rqmodel - rq(log(vd) ~ v1 + log(v2) +log(v3) + v4 + v5 ,data =dat) Thanks in advance and apologies, I'm pretty newbie with this, u...@host.com -- View this message in context: http://r.789695.n4.nabble.com/plot-rq-lm-tp4262170p4262170.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 not recognizing words
On Jan 6, 2012, at 06:20 , arabarkev wrote: Hello all I'm new to R and am experiencing a problem with a categorical variable. All the data of this variable are Low, High, or NA. When I put summary(x$y), it gives me the number of High, Low, and NA entries. However, when I try to subset by writing x$y==Low or x$y==High, R does not recognize the word and it writes FALSE for all the entries (but not the NA entries). Can anybody help me out? Trailing whitespace? Check levels(x$y) Thanks -- View this message in context: http://r.789695.n4.nabble.com/R-not-recognizing-words-tp4268283p4268283.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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.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] add data to a file while doing a loop
Hi, I would like to know how can I keep adding data to a file while doing a loop and without deleting the data of the previous iteration. Thanks. __ 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] add data to a file while doing a loop
Without context, read reproducible code, it is hard to answer this question. What system are you on? Do you need to write one line of data or a data frame or a list out? Are trying to write a graphic out? It will be easier to answer your question with some context. Good luck! Stephen On Fri 06 Jan 2012 05:49:01 AM CST, Joao Fadista wrote: Hi, I would like to know how can I keep adding data to a file while doing a loop and without deleting the data of the previous iteration. Thanks. __ 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] Spatial data, rpoispp, using window with fixed radius?
Dear All, I was searching through the spatstat manual in order to find a function to simulate a Poisson pattern only within a fixed radius (circular moving window) around individual points. If points are distributed heterogeneously over a large area this may help to only assess deviation from CSR within the window and thus does not require additional information on a covariate. I could not find such a function in spatstat. Can please anyone help? Thanks, Herb -- __ 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] add data to a file while doing a loop
Thanks for the reply. Every iteration produces a text data.frame directly into a file without saving it in R. And every iteration overwrites the data produced from the previous iteration. So what I would like to do is being able not to overwrite the data of any iteration but adding it to the same file with maybe the functions paste or rbind. sessionInfo() R version 2.13.2 (2011-09-30) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=Swedish_Sweden.1252 LC_CTYPE=Swedish_Sweden.1252 LC_MONETARY=Swedish_Sweden.1252 LC_NUMERIC=C LC_TIME=Swedish_Sweden.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] MatrixEQTL_1.2.0 loaded via a namespace (and not attached): [1] tools_2.13.2 Best regards, João Fadista, Ph.D. Post Doc Lund University Diabetes Centre CRC, Malmö University Hospital Entrance 72, building 60, level 13 SE-205 02 Malmö, Sweden Tel: +46 (0)40 391237 e-mail: joao.fadi...@med.lu.se -Original Message- From: stephen sefick [mailto:ssef...@gmail.com] On Behalf Of Stephen Sefick Sent: den 6 januari 2012 14:14 To: Joao Fadista Cc: r-help@r-project.org Subject: Re: [R] add data to a file while doing a loop Without context, read reproducible code, it is hard to answer this question. What system are you on? Do you need to write one line of data or a data frame or a list out? Are trying to write a graphic out? It will be easier to answer your question with some context. Good luck! Stephen On Fri 06 Jan 2012 05:49:01 AM CST, Joao Fadista wrote: Hi, I would like to know how can I keep adding data to a file while doing a loop and without deleting the data of the previous iteration. Thanks. __ 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] [R-pkgs] The BRugs 0.7-4 interface to OpenBUGS released
After so many questions and a considerable delay (almost entirely my fault), the BRugs package version 0.7-4 has been released yesterday. The BRugs package is available from CRAN master already as source and in binary form for Windows and will be synced by the mirrors shortly. Note the following changes (more details in the NEWS file): BRugs works on Linux now BRugs now supports 64-bit R on Windows BRugs works with an existing installation of OpenBUGS (at least 3.2.1) instead of being distributed with the OpenBUGS library, hence the OpenBUGS installation is a requirement. Both the Linux version as well as the 64-bit version for 64-bit R on Windows are typically slower than the more native interface used for 32-bit R on Windows. Thanks to all who helped to get so far. Special thanks to Chris Jackson for many improvement and all his work to build the Linux part of the interface. Best, Uwe Ligges ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] Problem with axes in a plot of Kaplan-Meier
After changing involuntarily some of the graphics parameters with the command par() (I did not know that changes with this command are permanent), now when I made a plot of the survival Kaplan-Meier function, the Y axis does not start at 1, and the X axis does starts at 0. The commands that I use are: - Par command are cumulative PER graphics device. Your unfortunate par commands would not affect a new plot with pdf() for instance. I often resort to the simple solution of dev.off() #close the current window x11() # start a new one (If not on Unix replace x11 with the appropriate command for your system). David's solution is the more elegant one. Terry T. __ 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 not recognizing words
I've developed a preference for
x$y %in% Low
when subsetting.
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of arabarkev
Sent: Friday, January 06, 2012 12:20 AM
To: r-help@r-project.org
Subject: [R] R not recognizing words
Hello all
I'm new to R and am experiencing a problem with a categorical variable. All
the data of this variable are Low, High, or NA. When I put summary(x$y),
it gives me the number of High, Low, and NA entries. However,
when I try to subset by writing x$y==Low or x$y==High, R does not
recognize the word and it writes FALSE for all the entries (but not the NA
entries).
Can anybody help me out?
Thanks
--
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Re: [R] automatic SI prefixes as ticklabels on axis
Ben's solution is much better, I was using only Jonas' code.
There might be typo with large values of 'x',
getSIstring(1e309)
[1] Inf Y
I don't believe this will happen, the range upper limit is much smaller.
Anyway, it can be easily solved.
(Ben's code, slightly changed.)
getSIstring - function(x){
lut - rev(c(1e24,1e21,1e18,1e15,1e12,1e9,1e6,1e3,1e0,
1e-3,1e-6,1e-9,1e-12,1e-15,1e-18,1e-21,1e-24))
pre - rev(c(Y, Z, E, P, T, G, M, k, ,
m, u, n, p, f, a, z, y))
ix - findInterval(x, lut)
ifelse(length(ix) 0, sistring - paste(x/lut[ix], pre[ix]),
sistring - as.character(x))
sistring[which(sistring == Inf Y)] - Inf # make it look better
return(sistring)
}
getSIstring(c(4.2e-3, 2e7))
x1 - .Machine$double.xmax
x2 - x1 + 10^(308 - 16)
getSIstring(c(x1, x2))
Rui Barradas
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[R] What is wrong with this plotting?
Hello I have a data frame, called input, like this: DateTime CO2_A1cont 1 2011-04-08 11:47:01 NA 2 2011-04-08 12:42:018.9 3 2011-04-08 13:07:01 NA 4 2011-04-08 13:32:01 NA 5 2011-04-08 13:57:01 7.556482 6 2011-04-08 14:22:01 NA 57 2011-04-09 16:52:01 4.961558 And like to plot this series with plot() and connected lines, no interruption by the NAs. I found that this code works: y-input[,2] times - DateTime plot(y~as.POSIXct(times, format=%d.%m. %H:%M), type=l, data=na.omit(data.frame(y,times))) whereas this plot command... plot(input[,2]~as.POSIXct(DateTime, format=%d.%m. %H:%M), type=l, data=na.omit(data.frame(input[,2],DateTime))) ... produces the error: Error in model.frame.default(formula = input[, 2] ~ as.POSIXct(DateTime, : variable lengths differ (found for 'as.POSIXct(DateTime, format = %d.%m. %H:%M)') I already checked the length of y, times, input[,2], DateTime, as.POSIXct(DateTime, format=%d.%m. %H:%M) which give all 57! nrow(data.frame(input[,2],DateTime)) and nrow(data.frame(y,times)) give both 57! Too. Thanks for any help. Best regards Raphael [[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] add data to a file while doing a loop
Hello, See ?open and ?capture.output or ?textConnection To open a connection (with 'open') then write to it is probably the solution. Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/add-data-to-a-file-while-doing-a-loop-tp4269086p4269396.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] lme model specification problem (Error in MEEM...)
Dear all, In lme, models in which a factor is fully contained in another lead to an error. This is not the case when using lm/aov. I understand that these factors are aliased, but believe that such models make sense when the factors are fitted sequentially. For example, I sometimes fit a factor first as linear term (continuous variable with discrete levels, e.g. 1,2,4,6), and then as factor, with the latter then accounting for the deviation from linearity. If x contains the values 1,2,4,6, the model formula then would be y ~ x + as.factor(x) A complete example is here: library(nlme) d - data.frame(x=rep(c(1,2,4,6),each=2),subj=factor(rep(1:16,each=2))) d$y - rnorm(32) + rnorm(length(levels(d$subj)))[d$subj] anova(lme(y~x,random=~1|subj,data=d)) ## works anova(lme(y~as.factor(x),random=~1|subj,data=d))## works anova(lme(y~x+as.factor(x),random=~1|subj,data=d)) ## fails: # Error in MEEM(object, conLin, control$niterEM) : # Singularity in backsolve at level 0, block 1 summary(aov(y~x+as.factor(x)+Error(subj),data=d)) ## works: # Error: subj # Df Sum Sq Mean Sq F value Pr(F) # x 1 8.434 8.434 4.780 0.0493 * # as.factor(x) 2 10.459 5.230 2.964 0.0900 . # Residuals12 21.176 1.765 # ... rest of output removed ... I understand I can to some extent work around this limitation by modifying the contrast encoding, but I then still don't get an overall test for as.factor(x) (in the example above, a test for deviation from linearity). d$xfac - factor(d$x) contrasts(d$xfac)-c(1,2,4,6) summary(lme(y~xfac,random=~1|subj,data=d)) Is there a way to work around this limitation of lme? Or do I mis-specify the model? Thanks for your advice. Pascal __ 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 not recognizing words
There was a trailing white space! Now I feel not so smart Thanks a lot for your help! -- View this message in context: http://r.789695.n4.nabble.com/R-not-recognizing-words-tp4268283p4269326.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] add data to a file while doing a loop
So you're looking for: write.table(dataset, file=x.txt, append=TRUE) Or do I understand your question wrongly? __ 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] add data to a file while doing a loop
On Jan 6, 2012, at 8:27 AM, Joao Fadista wrote: Thanks for the reply. Every iteration produces a text data.frame directly into a file without saving it in R. And every iteration overwrites the data produced from the previous iteration. So what I would like to do is being able not to overwrite the data of any iteration but adding it to the same file with maybe the functions paste or rbind. If you are using write.table, then pay more attention to the append parameter documented in the help file. -- David. sessionInfo() R version 2.13.2 (2011-09-30) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=Swedish_Sweden.1252 LC_CTYPE=Swedish_Sweden.1252 LC_MONETARY=Swedish_Sweden.1252 LC_NUMERIC=C LC_TIME=Swedish_Sweden.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] MatrixEQTL_1.2.0 loaded via a namespace (and not attached): [1] tools_2.13.2 Best regards, João Fadista, Ph.D. Post Doc Lund University Diabetes Centre CRC, Malmö University Hospital Entrance 72, building 60, level 13 SE-205 02 Malmö, Sweden Tel: +46 (0)40 391237 e-mail: joao.fadi...@med.lu.se -Original Message- From: stephen sefick [mailto:ssef...@gmail.com] On Behalf Of Stephen Sefick Sent: den 6 januari 2012 14:14 To: Joao Fadista Cc: r-help@r-project.org Subject: Re: [R] add data to a file while doing a loop Without context, read reproducible code, it is hard to answer this question. What system are you on? Do you need to write one line of data or a data frame or a list out? Are trying to write a graphic out? It will be easier to answer your question with some context. Good luck! Stephen On Fri 06 Jan 2012 05:49:01 AM CST, Joao Fadista wrote: Hi, I would like to know how can I keep adding data to a file while doing a loop and without deleting the data of the previous iteration. Thanks. __ 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. David Winsemius, MD West Hartford, CT __ 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] venneuler plots may be inaccurate
Hi, I just started playing w/ venneuler, and have a question about the last method listed for creating inputs. This is the method where the source is a matrix X, for which each column represents a set, and the co-occurrence is defined by the rows. So, for example: Rgames foo [,1] [,2] [,3] [,4] [,5] [1,]0110 NA [2,]0110 NA [3,]0110 NA [4,]0110 NA [5,]0110 NA [6,]0111 NA [7,]0111 NA [8,]0111 NA [9,]0111 NA [10,]0111 NA [11,]1011 NA [12,]1011 NA [13,]1011 NA [14,]1011 NA [15,]1011 NA [16,]1010 NA [17,]1010 NA [18,]1010 NA [19,]1010 NA [20,]1010 NA Rgames plot(venneuler(foo[,1:3])) The result I get is at http://home.comcast.net/~cgwcgw2/venndiagram.pdf I would not expect either A or B to extend outside C, and they shouldn't overlap either. Any ideas? So far as I can tell, the first method (define weights for each 'overlap' region, e.g. AB=.2, seems to work correctly). thanks Carl -- Sent from my Cray XK6 Pendeo-navem mei anguillae plena est. __ 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] ggplot using scale_x_date gives Error in seq.int(r1$year, to$year, by)
Dear all, ggplot gives me an error when trying to plot time series data using a date variable as the x axis. g-structure(list(Date = c(2011-12-23, 2011-12-30, 2012-01-06, 2011-12-23, 2011-12-30, 2012-01-06, 2011-12-23, 2011-12-30, 2012-01-06), variable = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c(Price, Yield, CDS Spread), class = factor), value = c(86.777, 86.037, 86.437, 9.737, 9.542, 9.683, 580.132, 576.866, 573.564)), .Names = c(Date, variable, value ), row.names = c(100L, 101L, 102L, 202L, 203L, 204L, 304L, 305L, 306L), class = data.frame) g Date variable value 100 2011-12-23 Price 86.8 101 2011-12-30 Price 86.0 102 2012-01-06 Price 86.4 202 2011-12-23 Yield 9.7 203 2011-12-30 Yield 9.5 204 2012-01-06 Yield 9.7 304 2011-12-23 CDS Spread 580.1 305 2011-12-30 CDS Spread 576.9 306 2012-01-06 CDS Spread 573.6 gp-ggplot(g,aes(x=Date,y=value,group=variable,lty=variable)) +geom_line() gp #THIS WORKS FINE (BUT AXIS LABELLING NOT VISIBLE WITH MORE DATA, HENCE I WOULD LIKE TO USE SCALE_X_DATE) gp-gp+ scale_x_date() gp Error in seq.int(r1$year, to$year, by) : 'from' must be finite In addition: Warning messages: 1: In get(x, envir = this, inherits = inh)(this, ...) : NAs introduced by coercion 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf 4: In min(x) : no non-missing arguments to min; returning Inf 5: In max(x) : no non-missing arguments to max; returning -Inf In previous help requests, the workaround of specifying the unit was suggested: gp-gp+ scale_x_date(major=years) but this doesn't work for me (same error). Any help is very much appreciated! Thanks in advance. Aidan R version 2.14.1 (2011-12-22) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=English_Ireland.1252 LC_CTYPE=English_Ireland.1252 LC_MONETARY=English_Ireland.1252 [4] LC_NUMERIC=C LC_TIME=English_Ireland.1252 attached base packages: [1] datasets tools grDevices grid splines graphics stats tcltk utils [10] methods base other attached packages: [1] micEcon_0.6-6 miscTools_0.6-12 np_0.40-11 cubature_1.0 boot_1.3-3 [6] RODBC_1.3-3sqldf_0.4-6.1 chron_2.3-41 gsubfn_0.5-7 DBI_0.2-5 [11] Haver_1.0 xtable_1.5-6 plm_1.2-7 sandwich_2.2-7 MASS_7.3-16 [16] Formula_1.0-1 nlme_3.1-102 bdsmatrix_1.0 RBloomberg_0.4-150 rJava_0.9-1 [21] gtools_2.6.2 gdata_2.8.2ggplot2_0.8.9 proto_0.3-9.2 zoo_1.7-4 [26] reshape_0.8.4 plyr_1.6 svSocket_0.9-52 TinnR_1.0.3R2HTML_2.2 [31] Hmisc_3.8-3survival_2.36-10 loaded via a namespace (and not attached): [1] cluster_1.14.1digest_0.5.0 lattice_0.20-0 RSQLite_0.10.0 [5] RSQLite.extfuns_0.0.1 svMisc_0.9-63 __ 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] What is wrong with this plotting?
On Fri, Jan 6, 2012 at 5:49 AM, raphael.fel...@art.admin.ch wrote: Hello I have a data frame, called input, like this: DateTime CO2_A1cont 1 2011-04-08 11:47:01 NA 2 2011-04-08 12:42:01 8.9 3 2011-04-08 13:07:01 NA 4 2011-04-08 13:32:01 NA 5 2011-04-08 13:57:01 7.556482 6 2011-04-08 14:22:01 NA 57 2011-04-09 16:52:01 4.961558 And like to plot this series with plot() and connected lines, no interruption by the NAs. I found that this code works: y-input[,2] times - DateTime plot(y~as.POSIXct(times, format=%d.%m. %H:%M), type=l, data=na.omit(data.frame(y,times))) whereas this plot command... plot(input[,2]~as.POSIXct(DateTime, format=%d.%m. %H:%M), type=l, data=na.omit(data.frame(input[,2],DateTime))) ... produces the error: Error in model.frame.default(formula = input[, 2] ~ as.POSIXct(DateTime, : variable lengths differ (found for 'as.POSIXct(DateTime, format = %d.%m. %H:%M)') I already checked the length of y, times, input[,2], DateTime, as.POSIXct(DateTime, format=%d.%m. %H:%M) which give all 57! nrow(data.frame(input[,2],DateTime)) and nrow(data.frame(y,times)) give both 57! Too. Try this reading DF into a zoo object. (tz= has the effect of setting its index to POSIXct using the current time zone -- you might need tz = GMT depending on what you want): library(zoo) z - read.zoo(DF, tz = ) plot(na.approx(z)) Assuming DF is: DF - structure(list(DateTime = structure(1:6, .Label = c(2011-04-08 11:47:01, 2011-04-08 12:42:01, 2011-04-08 13:07:01, 2011-04-08 13:32:01, 2011-04-08 13:57:01, 2011-04-08 14:22:01), class = factor), CO2_A1cont = c(NA, 8.9, NA, NA, 7.556482, NA)), .Names = c(DateTime, CO2_A1cont), class = data.frame, row.names = c(NA, -6L)) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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] VarCov matrix between coefficients across quantiles - summary.rq
Dear all, I am trying to design a Wald test to verify whether some coefficients are statistically equal. I would like to test not only whether they are jointly equal for a given quantile, but for a range of quantiles. Hence, I need to extract the variance-covariance matrix among the coefficients of a rq object: qf2 - summary(rq(wb2 ~ apv2 + vol2, tau=c(a1,a2)), cov=TRUE) I need not only the varcov matrix between the coefficients beta_0, beta_1, beta_2, but also between beta_0(a1) and beta_0(a2), and so on... (i.e., across quantiles) it is not a simple anova.rq test because I am comparing two different samples. The samples are independent, so, the varcov matrix between them is zero. However, I need to find the varcov between the coefficients across quantiles for each sample. Then, i will include the matrix in my Wald Test. I have tried the command qf2[[3]], but it does not work. Is there any other command that gives me the varcov between all those coefficients across quantiles? Thanks a lot! Julia [[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] Dropping columns from data frame
How does R do it, and should I ever be worried? I always remove
columns by index, and it works exactly as I would naively expect - but
HOW? The second illustration, which deletes non contiguous columns,
represents what I do all the time and have some trepidation about
because I don't know the mechanics (e.g. why doesn't the column
formerly-known-as-4 become 3 after column 1 is dropped: doesn't vector
removal from a df/list invoke a loop in C?). Can I delete a named
list of columns, which are examples 4 and 5 and which generate the
unary error' mesages, without resorting to orig.df$num1.10 - NULL?
Thanks!
orig.df - data.frame(cbind(
1:10
,11:20
,letters[1:10]
,letters[11:20]
,LETTERS[1:10]
,LETTERS[11:20]
))
names(orig.df) - c(
'num1.10'
,'num11.20'
,'lc1.10'
,'lc11.20'
,'uc1.10'
,'uc11.20'
)
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)])
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)])
# Illustration 3: contiguous columns at end of data frame
head(orig.df[,-c(4:6)]) ## as expected
# Illustrations 4-5: unary errors
head(orig.df[,-c(as.list('num1.10', 'lc1.10', 'uc1.10'))])
head(orig.df[,-c('num1.10', 'lc1.10', 'uc1.10')])
Mike
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[R] how to use rgl to plot dynamic orbit
Dear all,
I have a question about the plotting of the dynamical orbit by using rgl.
My code is as follows:
open3d()
par3d(cex=2)
bg3d(white)
for(i in 1:out.length){
ind-which
plot3d(x[1:i],y[1:i],out[,z][1:i],xlim=c(-10,10),ylim=c(-10,10),zlim=c(-10,10),col=red,cex=2)
Sys.sleep(0.0001)
}
,where x,y,out[,z] are the position of the object in different time. I
would like to show the orbit evoluting with time. However, I can just get
animation with flash windows. I don't know how to use play3d to realise
this dynamical orbit, are there anyone could help me for this?
Thanks a lot!
Fabo
[[alternative HTML version deleted]]
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[R] data.frame: temporal complexity
Hello,
I created a data.frame which contains two columns: df$P (Power) et
df$DateTime (time). I'd like to add a new column df$diffP (difference of
Power between T and T-2).
I made a loop :
for (i in 3:length(df$DateTime)){
df$diffP[i] = df$P[i] - df$P[i-2]
}
execution time result is unaceptable: 24s !!
Is there any way to reduce complexity about O(n) ? for example 2 or 3s (10s
maxi)
Does anybody find better than ~24s ?
thanks for your help
--
View this message in context:
http://r.789695.n4.nabble.com/data-frame-temporal-complexity-tp4269585p4269585.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
[R] plots for residual analysis
Hello List: I'm writing in R some code to produce plots for residual analysis and diagnostics in linear regressions. An example of the plots produced is given for downloading at http://dl.dropbox.com/u/25445316/res_plots.png . Regarding the example plot, I'd like to point out that: 1) Tendency lines based on lowess estimations are drawn as green continuous curves, 2) Extra variables or components (variables not in the model formula, but in the original dataframe or otherwise specified by the user) are plotted and labeled with an extra: preffix. 3) Moderate and extreme atipical levels of residuals are drawn as dashed yellow or red horizontal lines. The dashed green lines represent the 0 mean residuals are supposed to have, according to model assumptions. This plotting function is part of a set of tools I'm developing for spanish-sepaking pre-grad students. For more information, please consult https://sites.google.com/site/unamatematicaseltigre/trabajos-de-estadistica . The questions I'd like to put forth to the list are: 1) This plot is based on one I saw with similar dashed lines for atipical value limits, but now I can't recall in which package I saw this. Does anyone know of any package that produces very similar residual plots? I need to know this in order to give due credit. 2) What do the experts think of the green trend lines? Are they effective for spotting residuals/predictor variables relationships or are they misleading? 3) In the future, I would like to plot a grey shaded region approximately corresponding to the dispersion cloud, in order to appreciate homocedasticity or the lack thereof. Perhaps this could be done by calculating upper and lower limit trend lines? Any ideas on how to do this? Thanks in advance, José Loreto Romero Palma [[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] data.frame: temporal complexity
1. You are doing exactly what one is recommended to avoid in R. Have
you read an Introduction to R? -- especially about vectorization?
2. To answer your question: ?diff
It will probably be an order or 2 of magnitude faster.
-- Bert
On Fri, Jan 6, 2012 at 6:39 AM, ikuzar raz...@hotmail.fr wrote:
Hello,
I created a data.frame which contains two columns: df$P (Power) et
df$DateTime (time). I'd like to add a new column df$diffP (difference of
Power between T and T-2).
I made a loop :
for (i in 3:length(df$DateTime)){
df$diffP[i] = df$P[i] - df$P[i-2]
}
execution time result is unaceptable: 24s !!
Is there any way to reduce complexity about O(n) ? for example 2 or 3s (10s
maxi)
Does anybody find better than ~24s ?
thanks for your help
--
View this message in context:
http://r.789695.n4.nabble.com/data-frame-temporal-complexity-tp4269585p4269585.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
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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] data.frame: temporal complexity
Try this:
df - data.frame(power = runif(1))
# add difference
diffGap - 2
df$diff - c(rep(NA, diffGap), head(df$power, -diffGap) - tail(df$power,
-diffGap))
head(df, 10)
powerdiff
1 0.86170585 NA
2 0.90672473 NA
3 0.96868367 -0.10697782
4 0.44199262 0.46473211
5 0.48593923 0.48274443
6 0.99409592 -0.55210330
7 0.72790728 -0.24196804
8 0.38070013 0.61339579
9 0.69913680 0.02877047
10 0.07902925 0.30167088
On Fri, Jan 6, 2012 at 9:39 AM, ikuzar raz...@hotmail.fr wrote:
Hello,
I created a data.frame which contains two columns: df$P (Power) et
df$DateTime (time). I'd like to add a new column df$diffP (difference of
Power between T and T-2).
I made a loop :
for (i in 3:length(df$DateTime)){
df$diffP[i] = df$P[i] - df$P[i-2]
}
execution time result is unaceptable: 24s !!
Is there any way to reduce complexity about O(n) ? for example 2 or 3s (10s
maxi)
Does anybody find better than ~24s ?
thanks for your help
--
View this message in context:
http://r.789695.n4.nabble.com/data-frame-temporal-complexity-tp4269585p4269585.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.
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Re: [R] add data to a file while doing a loop
Is this an example of what you want to do?
fileName - tempfile(fileext=.txt)
fileCon - file(fileName, wt) # a file connection, opened for writing text
for(i in 1:5) {
cat(file=fileCon, Line, i, \n) # write to connection, not file name
}
close(fileCon)
Then look to see what the file contains:
readLines(fileName)
[1] Line 1 Line 2 Line 3 Line 4 Line 5
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 Joao Fadista
Sent: Friday, January 06, 2012 5:27 AM
To: sas0...@auburn.edu
Cc: r-help@r-project.org
Subject: Re: [R] add data to a file while doing a loop
Thanks for the reply. Every iteration produces a text data.frame directly
into a file without saving
it in R. And every iteration overwrites the data produced from the previous
iteration. So what I would
like to do is being able not to overwrite the data of any iteration but
adding it to the same file
with maybe the functions paste or rbind.
sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=Swedish_Sweden.1252 LC_CTYPE=Swedish_Sweden.1252
LC_MONETARY=Swedish_Sweden.1252
LC_NUMERIC=CLC_TIME=Swedish_Sweden.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] MatrixEQTL_1.2.0
loaded via a namespace (and not attached):
[1] tools_2.13.2
Best regards,
João Fadista, Ph.D.
Post Doc
Lund University Diabetes Centre
CRC, Malmö University Hospital
Entrance 72, building 60, level 13
SE-205 02 Malmö, Sweden
Tel: +46 (0)40 391237
e-mail: joao.fadi...@med.lu.se
-Original Message-
From: stephen sefick [mailto:ssef...@gmail.com] On Behalf Of Stephen Sefick
Sent: den 6 januari 2012 14:14
To: Joao Fadista
Cc: r-help@r-project.org
Subject: Re: [R] add data to a file while doing a loop
Without context, read reproducible code, it is hard to answer this question.
What system are you on?
Do you need to write one line of data or a data frame or a list out? Are
trying to write a graphic
out?
It will be easier to answer your question with some context. Good luck!
Stephen
On Fri 06 Jan 2012 05:49:01 AM CST, Joao Fadista wrote:
Hi,
I would like to know how can I keep adding data to a file while doing a
loop and without deleting
the data of the previous iteration. Thanks.
__
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Re: [R] Dropping columns from data frame
On Jan 6, 2012, at 10:00 AM, Mike Harwood wrote:
How does R do it, and should I ever be worried? I always remove
columns by index, and it works exactly as I would naively expect - but
HOW? The second illustration, which deletes non contiguous columns,
represents what I do all the time and have some trepidation about
because I don't know the mechanics (e.g. why doesn't the column
formerly-known-as-4 become 3 after column 1 is dropped: doesn't vector
removal from a df/list invoke a loop in C?).
You are NOT removing columns. You are returning (to `head` and then
to `print`) an extract from the dataframe, but that does not change
the original dataframe. To effect a change you would need to assign
the value back to the same name as the original daatframe.
--
David
Can I delete a named
list of columns, which are examples 4 and 5 and which generate the
unary error' mesages, without resorting to orig.df$num1.10 - NULL?
Thanks!
orig.df - data.frame(cbind(
1:10
,11:20
,letters[1:10]
,letters[11:20]
,LETTERS[1:10]
,LETTERS[11:20]
))
names(orig.df) - c(
'num1.10'
,'num11.20'
,'lc1.10'
,'lc11.20'
,'uc1.10'
,'uc11.20'
)
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)])
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)])
# Illustration 3: contiguous columns at end of data frame
head(orig.df[,-c(4:6)]) ## as expected
# Illustrations 4-5: unary errors
head(orig.df[,-c(as.list('num1.10', 'lc1.10', 'uc1.10'))])
head(orig.df[,-c('num1.10', 'lc1.10', 'uc1.10')])
Mike
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Re: [R] lme model specification problem (Error in MEEM...)
Pascal A. Niklaus pascal.niklaus at ieu.uzh.ch writes: In lme, models in which a factor is fully contained in another lead to an error. This is not the case when using lm/aov. I understand that these factors are aliased, but believe that such models make sense when the factors are fitted sequentially. For example, I sometimes fit a factor first as linear term (continuous variable with discrete levels, e.g. 1,2,4,6), and then as factor, with the latter then accounting for the deviation from linearity. If x contains the values 1,2,4,6, the model formula then would be y ~ x + as.factor(x) A complete example is here: library(nlme) d - data.frame(x=rep(c(1,2,4,6),each=2),subj=factor(rep(1:16,each=2))) d$y - rnorm(32) + rnorm(length(levels(d$subj)))[d$subj] anova(lme(y~x,random=~1|subj,data=d)) ## works anova(lme(y~as.factor(x),random=~1|subj,data=d))## works anova(lme(y~x+as.factor(x),random=~1|subj,data=d)) ## fails: # Error in MEEM(object, conLin, control$niterEM) : # Singularity in backsolve at level 0, block 1 summary(aov(y~x+as.factor(x)+Error(subj),data=d)) ## works: # Error: subj # Df Sum Sq Mean Sq F value Pr(F) # x 1 8.434 8.434 4.780 0.0493 * # as.factor(x) 2 10.459 5.230 2.964 0.0900 . # Residuals12 21.176 1.765 # ... rest of output removed ... I understand I can to some extent work around this limitation by modifying the contrast encoding, but I then still don't get an overall test for as.factor(x) (in the example above, a test for deviation from linearity). d$xfac - factor(d$x) contrasts(d$xfac)-c(1,2,4,6) summary(lme(y~xfac,random=~1|subj,data=d)) Is there a way to work around this limitation of lme? Or do I mis-specify the model? Thanks for your advice. This question would probably be more appropriate for the r-sig-mixed-mod...@r-project.org mailing list. A short answer (if you re-post to r-sig-mixed-models I might answer at greater length) would be that you should be able to quantify the difference between y~x and y~factor(x) by an anova comparison of the two *models*, i.e. anova(m1,m2) Another thing to consider would be setting using orthogonal polynomial contrasts for factor(x), where summary() would give you a result for the constant, linear, quadratic ... terms Ben Bolker __ 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] how to use rgl to plot dynamic orbit
On 06/01/2012 9:13 AM, Phillip Von wrote:
Dear all,
I have a question about the plotting of the dynamical orbit by using rgl.
My code is as follows:
open3d()
par3d(cex=2)
bg3d(white)
for(i in 1:out.length){
ind-which
plot3d(x[1:i],y[1:i],out[,z][1:i],xlim=c(-10,10),ylim=c(-10,10),zlim=c(-10,10),col=red,cex=2)
Sys.sleep(0.0001)
}
,where x,y,out[,z] are the position of the object in different time. I
would like to show the orbit evoluting with time. However, I can just get
animation with flash windows. I don't know how to use play3d to realise
this dynamical orbit, are there anyone could help me for this?
Thanks a lot!
If you just want to add points to the plot, then I'd use points3d() to
add them one at a time. (Or if you want them joined,
use lines3d() or segments3d() to add a line from the previous point to
the next one.) For example,
plot3d(numeric(0), numeric(0), numeric(0),
xlim=c(-10,10),ylim=c(-10,10),zlim=c(-10,10))
for(i in 1:out.length){
ind-which
points3d(x[i],y[i],out[i,z],col=red,cex=2)
Sys.sleep(0.0001)
}
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[R] How to fit my data with a distribution?
Dear All, I have a bunch of data points as follows: x 100 y 200 z 300 ... where 100, 200, 300 are the values. I would like to know the distribution of my data? how can I fit my data into a distribution? Thanks a lot, Andra [[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] Dropping columns from data frame
Thank you, David. I was merely using head to limit the code/
output. My question remains, because a created data frame has the
same columns as was output from head:
head(orig.df,3)
num1.10 num11.20 lc1.10 lc11.20 uc1.10 uc11.20
1 1 11 a k A K
2 2 12 b l B L
3 3 13 c m C M
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)],2)
lc11.20 uc1.10 uc11.20
1 k A K
2 l B L
new.df - orig.df[,-c(1:3)]
head(new.df,2)
lc11.20 uc1.10 uc11.20
1 k A K
2 l B L
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)],2)
num11.20 lc11.20 uc11.20
1 11 k K
2 12 l L
new.df - orig.df[,-c(1,3,5)]
head(new.df,2)
num11.20 lc11.20 uc11.20
1 11 k K
2 12 l L
On Jan 6, 9:49 am, David Winsemius dwinsem...@comcast.net wrote:
On Jan 6, 2012, at 10:00 AM, Mike Harwood wrote:
How does R do it, and should I ever be worried? I always remove
columns by index, and it works exactly as I would naively expect - but
HOW? The second illustration, which deletes non contiguous columns,
represents what I do all the time and have some trepidation about
because I don't know the mechanics (e.g. why doesn't the column
formerly-known-as-4 become 3 after column 1 is dropped: doesn't vector
removal from a df/list invoke a loop in C?).
You are NOT removing columns. You are returning (to `head` and then
to `print`) an extract from the dataframe, but that does not change
the original dataframe. To effect a change you would need to assign
the value back to the same name as the original daatframe.
--
David
Can I delete a named
list of columns, which are examples 4 and 5 and which generate the
unary error' mesages, without resorting to orig.df$num1.10 - NULL?
Thanks!
orig.df - data.frame(cbind(
1:10
,11:20
,letters[1:10]
,letters[11:20]
,LETTERS[1:10]
,LETTERS[11:20]
))
names(orig.df) - c(
'num1.10'
,'num11.20'
,'lc1.10'
,'lc11.20'
,'uc1.10'
,'uc11.20'
)
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)])
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)])
# Illustration 3: contiguous columns at end of data frame
head(orig.df[,-c(4:6)]) ## as expected
# Illustrations 4-5: unary errors
head(orig.df[,-c(as.list('num1.10', 'lc1.10', 'uc1.10'))])
head(orig.df[,-c('num1.10', 'lc1.10', 'uc1.10')])
Mike
__
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PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
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PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
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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] Error using qda on the spambase training dataset
Hello, I am trying to run the code for quadratic discriminant analysis with the Spambase data set following the code example in the book R in a Nutshell by Joseph Adler (page 443). The line of code: spam.qda - qda(formula=is_spam~., data=spambase.training) gives me an error: Error in qda.default(x, grouping, ...) : rank deficiency in group 1 (There really is a tilde after is_spam in the code, only on this web-page it appears as a hyphen :) Can someone please tell me why it does not work with me but appears to do so in the book? I am using R version 2.13.0. Thanks, Ravi -- View this message in context: http://r.789695.n4.nabble.com/Error-using-qda-on-the-spambase-training-dataset-tp4269941p4269941.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] What is wrong with this plotting?
I think it's even simpler than that: plot(y~as.POSIXct(times, format=%d.%m. %H:%M), type=l, data=na.omit(data.frame(y,times))) This is using y from the data specification in the plot command (not the y in the base environment, or you'd be getting the same error), and data= is specifying the data with NAs omitted. whereas this plot command... plot(input[,2]~as.POSIXct(DateTime, format=%d.%m. %H:%M), type=l, data=na.omit(data.frame(input[,2],DateTime))) ... produces the error: Error in model.frame.default(formula = input[, 2] ~ as.POSIXct(DateTime, : variable lengths differ (found for 'as.POSIXct(DateTime, format = %d.%m. %H:%M)') This command is using input[,2] for its y varialble, which has *not* had the NA values removed. It's a matter of which data your plot() command is using, and it's not what you seem to think it is. Sarah -- Sarah Goslee http://www.functionaldiversity.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.
Re: [R] Dropping columns from data frame
On Fri, Jan 6, 2012 at 10:00 AM, Mike Harwood harwood...@gmail.com wrote:
How does R do it, and should I ever be worried?
You should be worried, but not about that.
I always remove
columns by index, and it works exactly as I would naively expect - but
HOW? The second illustration, which deletes non contiguous columns,
represents what I do all the time and have some trepidation about
because I don't know the mechanics (e.g. why doesn't the column
formerly-known-as-4 become 3 after column 1 is dropped: doesn't vector
removal from a df/list invoke a loop in C?).
The R programmers, and the S programmers before them, were smart
enough to make this a simultaneous rather than sequential operation.
If you really need the details, the source is out there.
Can I delete a named
list of columns, which are examples 4 and 5 and which generate the
unary error' mesages, without resorting to orig.df$num1.10 - NULL?
head(orig.df[, !colnames(orig.df) %in% c('num1.10', 'lc1.10', 'uc1.10')])
Sarah
Thanks!
orig.df - data.frame(cbind(
1:10
,11:20
,letters[1:10]
,letters[11:20]
,LETTERS[1:10]
,LETTERS[11:20]
))
names(orig.df) - c(
'num1.10'
,'num11.20'
,'lc1.10'
,'lc11.20'
,'uc1.10'
,'uc11.20'
)
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)])
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)])
# Illustration 3: contiguous columns at end of data frame
head(orig.df[,-c(4:6)]) ## as expected
# Illustrations 4-5: unary errors
head(orig.df[,-c(as.list('num1.10', 'lc1.10', 'uc1.10'))])
head(orig.df[,-c('num1.10', 'lc1.10', 'uc1.10')])
Mike
--
Sarah Goslee
http://www.functionaldiversity.org
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fit my data with a distribution?
MASS::fitdistr Michael On Jan 6, 2012, at 10:47 AM, Andra Isan andra_i...@yahoo.com wrote: Dear All, I have a bunch of data points as follows: x� 100 y� 200 z� 300 ... where 100, 200, 300 are the values. I would like to know the distribution of my data? how can I fit my data into a distribution? Thanks a lot, Andra [[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] Dropping columns from data frame
On Jan 6, 2012, at 11:43 AM, Mike Harwood wrote:
Thank you, David. I was merely using head to limit the code/
output. My question remains, because a created data frame has the
same columns as was output from head:
head(orig.df,3)
num1.10 num11.20 lc1.10 lc11.20 uc1.10 uc11.20
1 1 11 a k A K
2 2 12 b l B L
3 3 13 c m C M
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)],2)
lc11.20 uc1.10 uc11.20
1 k A K
2 l B L
new.df - orig.df[,-c(1:3)]
head(new.df,2)
lc11.20 uc1.10 uc11.20
1 k A K
2 l B L
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)],2)
num11.20 lc11.20 uc11.20
1 11 k K
2 12 l L
new.df - orig.df[,-c(1,3,5)]
head(new.df,2)
num11.20 lc11.20 uc11.20
1 11 k K
2 12 l L
I guess my short attention span got the better of me. (But calling
them unary errors was somewhat cryptic and not a particularly
helpful description of what you were actually seeing.) Here are more
constructive responses:
Negative indexing is not accepted for character vectors, so you need
to convert to either numeric or logical and then negativize:
orig.df[ !names(orig.df) %in% c('num1.10', 'lc1.10', 'uc1.10')]
These are equivalent:
orig.df[ , !names(orig.df) %in% c('num1.10', 'lc1.10', 'uc1.10')]
orig.df[,-match( c(num1.10, lc1.10, uc1.10), names(orig.df))]
orig.df[ , -sapply(c('num1.10', 'lc1.10', 'uc1.10'), grep,
x=names(orig.df)) ]
And when there is a pattern, such as with your not wanting any of the .
10 names, then grep can be quite efficient:
orig.df[ , -grep(.10, names(orig.df), fixed=TRUE)]
--
David
On Jan 6, 9:49 am, David Winsemius dwinsem...@comcast.net wrote:
On Jan 6, 2012, at 10:00 AM, Mike Harwood wrote:
How does R do it, and should I ever be worried? I always remove
columns by index, and it works exactly as I would naively expect -
but
HOW? The second illustration, which deletes non contiguous columns,
represents what I do all the time and have some trepidation about
because I don't know the mechanics (e.g. why doesn't the column
formerly-known-as-4 become 3 after column 1 is dropped: doesn't
vector
removal from a df/list invoke a loop in C?).
You are NOT removing columns. You are returning (to `head` and then
to `print`) an extract from the dataframe, but that does not change
the original dataframe. To effect a change you would need to assign
the value back to the same name as the original daatframe.
--
David
Can I delete a named
list of columns, which are examples 4 and 5 and which generate the
unary error' mesages, without resorting to orig.df$num1.10 -
NULL?
Thanks!
orig.df - data.frame(cbind(
1:10
,11:20
,letters[1:10]
,letters[11:20]
,LETTERS[1:10]
,LETTERS[11:20]
))
names(orig.df) - c(
'num1.10'
,'num11.20'
,'lc1.10'
,'lc11.20'
,'uc1.10'
,'uc11.20'
)
# Illustration 1: contiguous columns at beginning of data frame
head(orig.df[,-c(1:3)])
# Illustration 2: non-contiguous columns
head(orig.df[,-c(1,3,5)])
# Illustration 3: contiguous columns at end of data frame
head(orig.df[,-c(4:6)])## as expected
# Illustrations 4-5: unary errors
head(orig.df[,-c(as.list('num1.10', 'lc1.10', 'uc1.10'))])
head(orig.df[,-c('num1.10', 'lc1.10', 'uc1.10')])
Mike
__
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PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/
listinfo/r-help
PLEASE do read the posting guidehttp://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
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David Winsemius, MD
West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] data.frame: temporal complexity
Ok, thanks, it works ! -- View this message in context: http://r.789695.n4.nabble.com/data-frame-temporal-complexity-tp4269585p4270073.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] add data to a file while doing a loop
Look at the documentation for whatever function you are using to write data to the file. It should be pretty obvious (look for an append argument). Otherwise you'll have to provide more information, such as a short simple example of what you have tried. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/6/12 3:49 AM, Joao Fadista joao.fadi...@med.lu.se wrote: Hi, I would like to know how can I keep adding data to a file while doing a loop and without deleting the data of the previous iteration. Thanks. __ 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] cbind alternate
I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- - Mary Kindall Yorktown Heights, NY 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] RODBC installation: error message
Hi Gregory, See below. On Jan 5, 2012, at 4:20 PM, gregory benison wrote: As Duncan noted, the message is pretty clear in that the ODBC header files are missing, which are required to compile RODBC from source. On RH based Linuxen, this requires the installation of the unixODBC-devel RPM, much as one would need to have other *-devel RPMs (eg. readline-devel) installed for compiling many applications from source. Since a lot of R users may not be familiar with these header files, I think the configure script could go a step further in explaining why it was looking for these headers (to find an installed ODBC driver manager), and that it didn't find one. To illustrate what I mean, here is the result of running R CMD INSTALL on a system lacking an installed dsm where I swapped the order of the tests in configure.ac (see patch below) such that the library tests come before the header tests: * installing *source* package 'RODBC' ... checking for library containing SQLTables... no configure: error: no ODBC driver manager found ERROR: configuration failed for package 'RODBC' To me, that is a more helpful error message, because it makes it clear that you need to install an ODBC driver manager. snipped other content To split hairs, there are two issues at play here, irrespective of the ordering of the tests: 1. If you do not have unixODBC installed, you will get an error message regarding a missing ODBC driver manager, if as you have done here, that test is first rather than second. 2. If you have unixODBC installed (so you pass the first test in your scenario), but do not have the 'devel' part, which is a separate install on most Linuxen with package managers (RHEL/CentOS, Fedora, Debian and derivatives like Ubuntu and Mint, etc.), you will still get the error message regarding the missing header files. That is an entirely different error than missing the actual ODBC driver manager. So you can argue about the ordering of the tests, but the end result is the same. Only if you have some knowledge about compiling source code (on ANY operating system), will you understand the meaning of missing header files AND only if you have some lower level knowledge of development on Linuxen specifically, will you know that the header files are contained in a separate binary package that needs to be installed as well, the details of which will be specific to the Linux distribution that you have chosen to install. On top of that, if as with most Linuxen today, you don't have the software development packages installed (you are a naive user of Linux and only did a default install of the distribution), you won't have compilers and other associated tools (eg. make, etc.) installed either. At some point, the user has to be responsible for obtaining the requisite knowledge of their operating system, independent of the use of R. Despite the protestations by some on the web that their grandmothers are running Linux, it is still largely a geek's platform (with a 1% to 1.5% marketshare on the desktop) and due to the marketplace fragmentation of the plethora of distributions and desktop environments, further requires distribution and even version specific knowledge as well. As I noted earlier, if you are going to use Linux, you are generally going to be expected to have a more technical level of knowledge than somebody who is running Windows or OSX, where for the most part, one can install pre-compiled binary packages from CRAN, thanks in large measure to folks like Kurt, Uwe, Duncan, Simon and Prof. Ripley (I may be missing others) who have taken the time over the years to put the testing and build infrastructure in place to automate much of that process. Since there are over 3,500 packages on CRAN and some proportion (I don't know the number) have C/C++ and/or FORTRAN code in them, which will require compilation when installing source packages using either install.packages() or R CMD INSTALL on Linuxen, addressing RODBC in isolation is ultimately not helpful, since like most CRAN packages, it has a finite audience of users (me being very thankful to Prof. Ripley to be one of them). I cannot speak for Prof. Ripley here, but since you seem to have the requisite skills, he may be open to accepting a patch against his package that would narrowly address the issue that you raise. However, that does not change the defacto behavior for all of the other CRAN packages that will have similar requirements when being installed from source and will, as is the case with Linux generally, output relatively terse error messages when faced with these kinds of issues. Also as Duncan noted, perhaps there may be a better mechanism in gaining access to install time requirements for CRAN packages. In the case of RODBC, Prof. Ripley has taken the time to create a vignette which covers much of this material. That is directly available from the CRAN page for the package,
Re: [R] Spatial data, rpoispp, using window with fixed radius?
What's wrong with rpoispp in spatstat? It can simulate over a polygon, which can of course be used to closely approximate a circle. There is also spsample in the sp package. I'd also suggest asking this question on r-sig-geo. -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/6/12 5:17 AM, herbert8...@gmx.de herbert8...@gmx.de wrote: Dear All, I was searching through the spatstat manual in order to find a function to simulate a Poisson pattern only within a fixed radius (circular moving window) around individual points. If points are distributed heterogeneously over a large area this may help to only assess deviation from CSR within the window and thus does not require additional information on a covariate. I could not find such a function in spatstat. Can please anyone help? Thanks, Herb -- __ 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] (Edited) cbind alternate for data frames
I have two dataframes and want to perform cbind and then write into a file. The number of entries are more than a million in both frames. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1:4] y = table2[1:100,3:6] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- - Mary Kindall Yorktown Heights, NY 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.
[R] Assign and cmpfun
Hi All,
I've just recently discovered the cmpfun function, and was wanting to to
create a function to assign this to all the functions i have created, but
without explicitly naming them.
I've achieved this with:
foo - function(x) { print(x)}
bar - function(x) { print(x + 1)}
foo - function(x) { print(x)}
foo
function(x) { print(x)}
cmpfun(foo)
function(x) { print(x)}
bytecode: 0x26e3d40
find.all.functions - ls.str(mode = 'function')
for(i in seq_along(find.all.functions)) {
assign(find.all.functions[i], cmpfun(get(find.all.functions[i])))
}
But remember told that using assign is generally a bad idea, and ideally i
want to functionalize this to say something like:
CreateCompiledFunctions - function() {
find.all.functions - ls.str(mode = 'function')
for(i in seq_along(find.all.functions)) {
assign(find.all.functions[i], cmpfun(get(find.all.functions[i])))
}
}
Does anyone have a better solution?
Thanks in advance
Mike
[[alternative HTML version deleted]]
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Re: [R] data.frame: temporal complexity
Something like this (not tested)?
df$diffP - c(NA, NA, diff(df$P,2))
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/6/12 6:39 AM, ikuzar raz...@hotmail.fr wrote:
Hello,
I created a data.frame which contains two columns: df$P (Power) et
df$DateTime (time). I'd like to add a new column df$diffP (difference of
Power between T and T-2).
I made a loop :
for (i in 3:length(df$DateTime)){
df$diffP[i] = df$P[i] - df$P[i-2]
}
execution time result is unaceptable: 24s !!
Is there any way to reduce complexity about O(n) ? for example 2 or 3s
(10s
maxi)
Does anybody find better than ~24s ?
thanks for your help
--
View this message in context:
http://r.789695.n4.nabble.com/data-frame-temporal-complexity-tp4269585p426
9585.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] How to fit my data with a distribution?
That MASS::fitdistr is used for the case when I have some sense about the distribution of my data. When I do not know anything about my data, is there any function that can I use to tell what distribution of my data is? Thanks a lot, Andra From: R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com Cc: r-help@r-project.org r-help@r-project.org Sent: Friday, January 6, 2012 11:18 AM Subject: Re: [R] How to fit my data with a distribution? MASS::fitdistr Michael Dear All, I have a bunch of data points as follows: x� 100 y� 200 z� 300 ... where 100, 200, 300 are the values. I would like to know the distribution of my data? how can I fit my data into a distribution? Thanks a lot, Andra   [[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. [[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] (Edited) cbind alternate for data frames
For me, this example runs in a fraction of a second: t1 - data.frame(matrix(rnorm(3e6),ncol=3)) t2 - data.frame(matrix(rnorm(3e6),ncol=3)) t3 - cbind(t1,t2) dim(t3) [1] 100 6 Maybe it takes longer if your data frames have other classes of objects in them. If some of your data frame columns are factors, try converting to character first. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/6/12 9:56 AM, Mary Kindall mary.kind...@gmail.com wrote: I have two dataframes and want to perform cbind and then write into a file. The number of entries are more than a million in both frames. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1:4] y = table2[1:100,3:6] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- - Mary Kindall Yorktown Heights, NY 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. __ 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] cbind alternate
You could break the data into chunks, so you cbind and save 50,000
observations at a time. That should be less taxing on your machine and
memory.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mary Kindall
Sent: Friday, January 06, 2012 12:43 PM
To: r-help@r-project.org
Subject: [R] cbind alternate
I have two one dimensional list of elements and want to perform cbind
and then write into a file. The number of entries are more than a
million in both lists. R is taking a lot of time performing this
operation.
Is there any alternate way to perform cbind?
x = table1[1:100,1]
y = table2[1:100,5]
z = cbind(x,y) //hanging the machine
write.table(z,'out.txt)
--
-
Mary Kindall
Yorktown Heights, NY
USA
[[alternative HTML version deleted]]
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***
This message is for the named person's use only. It may\...{{dropped:11}}
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Re: [R] How to fit my data with a distribution?
On Jan 6, 2012, at 1:02 PM, Andra Isan wrote: That MASS::fitdistr is used for the case when I have some sense about the distribution of my data. When I do not know anything about my data, is there any function that can I use to tell what distribution of my data is? ?density From: R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com Cc: r-help@r-project.org r-help@r-project.org Sent: Friday, January 6, 2012 11:18 AM Subject: Re: [R] How to fit my data with a distribution? MASS::fitdistr Michael I have a bunch of data points as follows: x� 100 y� 200 z� 300 ... where 100, 200, 300 are the values. I would like to know the distribution of my data? how can I fit my data into a distribution? Thanks a lot, Andra David Winsemius, MD West Hartford, CT __ 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] add data to a file while doing a loop
Using append=TRUE in many functions will work, but
more slowly than opening a connection once, writing
to the connection many times, then closing it.
(Opening a file is a pretty expensive operation,
while writing to it is much cheaper.) Some more
recently written functions do not have an append=
argument because you can achieve the same with connections.
E.g., I wrote functions that used
cat(file=fileName, append=TRUE) repeatedly
and the open/cat-repeatedly/close method:
f0 - function (n, fileName)
{
unlink(fileName)
system.time(for (i in seq_len(n)) cat(Line, i, \n, file = fileName,
append = TRUE))
}
f1 - function (n, fileName) {
unlink(fileName)
system.time({
fileConn - file(fileName, wt)
on.exit(close(fileConn))
for (i in seq_len(n)) cat(Line, i, \n, file = fileConn)
})
}
and recorded the time they took to write
1000, 2000, and 2 lines on my Window XP
laptop:
tf0 - tempfile()
f0(1*10^3, tf0)
user system elapsed
0.160.458.25
f0(2*10^3, tf0)
user system elapsed
0.360.98 17.86
f0(20*10^3, tf0)
user system elapsed
5.03 10.64 393.95
tf1 - tempfile()
f1(1*10^3, tf1)
user system elapsed
0.050.090.15
f1(2*10^3, tf1)
user system elapsed
0.020.080.11
f1(20*10^3, tf1)
user system elapsed
0.300.700.98
Note that they produced identical output files:
identical(readLines(tf0), readLines(tf1))
[1] TRUE
and the connection-oriented version is still usable
for a million or two iterations:
f1(1e6, tf1)
user system elapsed
15.40 30.05 45.19
f1(2e6, tf1)
user system elapsed
31.95 60.29 91.42
Any of the standard functions with a file= (or con=)
argument will accept a connction object instead of
a file name. If you use the connection object you
don't need to restrict yourself to functions with
an append= argument to append to a file.
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 MacQueen, Don
Sent: Friday, January 06, 2012 9:32 AM
To: Joao Fadista; r-help@r-project.org
Subject: Re: [R] add data to a file while doing a loop
Look at the documentation for whatever function you are using to write
data to the file.
It should be pretty obvious (look for an append argument).
Otherwise you'll have to provide more information, such as a short simple
example of what you have tried.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/6/12 3:49 AM, Joao Fadista joao.fadi...@med.lu.se wrote:
Hi,
I would like to know how can I keep adding data to a file while doing a
loop and without deleting the data of the previous iteration. Thanks.
__
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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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] add data to a file while doing a loop
Thank you all for your useful replies. I got it to work!
Best regards,
João Fadista, Ph.D.
Post Doc
Lund University Diabetes Centre
CRC, Malmö University Hospital
Entrance 72, building 60, level 13
SE-205 02 Malmö, Sweden
Tel: +46 (0)40 391237
e-mail: joao.fadi...@med.lu.se
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: den 6 januari 2012 19:27
To: MacQueen, Don; Joao Fadista; r-help@r-project.org
Subject: RE: [R] add data to a file while doing a loop
Using append=TRUE in many functions will work, but more slowly than opening a
connection once, writing to the connection many times, then closing it.
(Opening a file is a pretty expensive operation, while writing to it is much
cheaper.) Some more recently written functions do not have an append= argument
because you can achieve the same with connections.
E.g., I wrote functions that used
cat(file=fileName, append=TRUE) repeatedly and the open/cat-repeatedly/close
method:
f0 - function (n, fileName)
{
unlink(fileName)
system.time(for (i in seq_len(n)) cat(Line, i, \n, file = fileName,
append = TRUE))
}
f1 - function (n, fileName) {
unlink(fileName)
system.time({
fileConn - file(fileName, wt)
on.exit(close(fileConn))
for (i in seq_len(n)) cat(Line, i, \n, file = fileConn)
})
}
and recorded the time they took to write 1000, 2000, and 2 lines on my
Window XP
laptop:
tf0 - tempfile()
f0(1*10^3, tf0)
user system elapsed
0.160.458.25
f0(2*10^3, tf0)
user system elapsed
0.360.98 17.86
f0(20*10^3, tf0)
user system elapsed
5.03 10.64 393.95
tf1 - tempfile()
f1(1*10^3, tf1)
user system elapsed
0.050.090.15
f1(2*10^3, tf1)
user system elapsed
0.020.080.11
f1(20*10^3, tf1)
user system elapsed
0.300.700.98
Note that they produced identical output files:
identical(readLines(tf0), readLines(tf1))
[1] TRUE
and the connection-oriented version is still usable for a million or two
iterations:
f1(1e6, tf1)
user system elapsed
15.40 30.05 45.19
f1(2e6, tf1)
user system elapsed
31.95 60.29 91.42
Any of the standard functions with a file= (or con=) argument will accept a
connction object instead of a file name. If you use the connection object you
don't need to restrict yourself to functions with an append= argument to append
to a file.
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 MacQueen, Don
Sent: Friday, January 06, 2012 9:32 AM
To: Joao Fadista; r-help@r-project.org
Subject: Re: [R] add data to a file while doing a loop
Look at the documentation for whatever function you are using to write
data to the file.
It should be pretty obvious (look for an append argument).
Otherwise you'll have to provide more information, such as a short
simple example of what you have tried.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/6/12 3:49 AM, Joao Fadista joao.fadi...@med.lu.se wrote:
Hi,
I would like to know how can I keep adding data to a file while doing
a loop and without deleting the data of the previous iteration. Thanks.
__
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
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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] data.frame: temporal complexity
I did not test this way, I 'll do it later, I think that df$diffP - c(NA, NA, diff(df$P,2)) is the best way to compute the difference !! I 'll post the result here -- View this message in context: http://r.789695.n4.nabble.com/data-frame-temporal-complexity-tp4269585p4270256.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] cbind alternate
On Jan 6, 2012, at 11:43 AM, Mary Kindall wrote: I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) The issue is not the use of cbind(), but that write.table() can be slow with data frames, where each column may be a different class (data type) and requires separate formatting for output. This is referenced in the Note section of ?write.table: write.table can be slow for data frames with large numbers (hundreds or more) of columns: this is inevitable as each column could be of a different class and so must be handled separately. If they are all of the same class, consider using a matrix instead. I suspect in this case, while you don't have a large number of columns, you do have a large number of rows, so that there is a tradeoff. If all of the columns in your source tables are of the same type (eg. all numeric), coerce 'z' to a matrix and then try using write.table(). z - matrix(rnorm(100 * 6), ncol = 6) str(z) num [1:100, 1:6] -0.713 0.79 -0.538 0.945 1.621 ... system.time(write.table(z, file = test.txt)) user system elapsed 12.664 0.292 13.029 The resultant file is about 118 Mb on my system. HTH, Marc Schwartz __ 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] How to fit my data with a distribution?
except, of course, that the distributions of all finite sets of data are the same: discrete. I think Andra would do well to seek help from his/her local statistician, as his/her query seems to indicate considerable confusion about basic concepts. Please excuse me if I have misjudged. -- Bert On Fri, Jan 6, 2012 at 10:13 AM, David Winsemius dwinsem...@comcast.net wrote: On Jan 6, 2012, at 1:02 PM, Andra Isan wrote: That MASS::fitdistr is used for the case when I have some sense about the distribution of my data. When I do not know anything about my data, is there any function that can I use to tell what distribution of my data is? ?density From: R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com Cc: r-help@r-project.org r-help@r-project.org Sent: Friday, January 6, 2012 11:18 AM Subject: Re: [R] How to fit my data with a distribution? MASS::fitdistr Michael I have a bunch of data points as follows: x� 100 y� 200 z� 300 ... where 100, 200, 300 are the values. I would like to know the distribution of my data? how can I fit my data into a distribution? Thanks a lot, Andra David Winsemius, MD West Hartford, CT __ 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] Correlated count data technique advice
Please excuse me for having posted a similar question on ecolog, but thus far I have received few useful answers there. I am looking for some advice concerning techniques in R that are appropriate for correlated count data. Specifically, I have some freezing days data, which is a count of the number of days each spring that were below freezing. The counts were taken at the same location over a period of years. The data set is highly zero inflated and over-dispersed; glm with a quasipoisson error structure would seem to be appropriate, except that there is a high degree of correlation at lags of 1 making something like a corAR1 structure appropriate. My difficulty is that glm() does not take an argument for correlation. I could use lmer() to fit a model like: freezing days~years+(1|years), family=quasipoisson, correlation=corAR1 but lmer (and glmer) don't seem to be operating on quasi families anymore; I've found plenty of old posts here where lmer seems to have accepted quasi families in the past, but I get an error message that indicates lmer does not in fact accept quasi families. I should note that I have run the following model: freeze.glmmPQL3-glmmPQL(num.freeze.days~years, random= ~1|years, family=quasipoisson,correlation=corAR1()) My gut says this is not the correct approach and I am unconvinced by the tiny p values that have been returned, especially as specification of poisson vs quasipoisson and the specification of corAR1() seem to make no difference to parameter estimation or p vals for said pars--it would seem that the random term for varying intercept by year is dominant. Maybe this is OK, but my above glm models return non-significant results and I expected handling the correlation to increase my p vals rather than decrease them. Perhaps an incorrect assumption. Therefore I need some alternative to look at trends in this data over time that allows for quasipoisson error and something along the lines of a corAR1() structure (or a mixed model that handles temporal pseudo-replication, but I am hesitant here). Thank you in advance, Lee [[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] cbind alternate
On Jan 6, 2012, at 12:43 PM, Mary Kindall wrote:
I have two one dimensional list of elements and want to perform
cbind and
then write into a file. The number of entries are more than a
million in
both lists. R is taking a lot of time performing this operation.
Is there any alternate way to perform cbind?
x =
y = table2[1:100,5]
z = cbind(x,y) //hanging the machine
You should have been able to bypass the intermediate steps with just:
z = cbind( table1[1:100,1] table2[1:100,5])
Whether you will have sufficient contiguous memory for that object at
the moment or even after rm(x), rm(y) is in doubt, but had you not
created the unneeded x and y, you _might_ have succeeded in your
limited environment. (Real answer: Buy more RAM.)
I speculate that you are on Windows and so refer your to the R-Win FAQ
for further reading about memory limits.
write.table(z,'out.txt)
I do not know of a way to bypass the requirement of a named object to
pass to write.table, but testing suggests that you could try:
write( t(cbind( table1[1:100,1] table2[1:100,5])).
test.txt, 2)
write() does not require a named object but is less inquisitive than
write table and will give you a transposed matrix with 5 columns by
default which will really mess up things, so you need to transpose and
specify the number of columns. (And that may not save any space over
creating a z object.)
So there is another thread today to which master R programmer Bill
Dunlap has offered this strategy (with minor modifications to your
situation by me):
###
f1 - function (n, fileName) {
unlink(fileName)
system.time({
fileConn - file(fileName, wt)
on.exit(close(fileConn))
for (i in seq_len(n)) cat( table1[i, 1], ,
table2[i, 5],
\n, file = fileConn)
})
}
f1(100, 'out.txt')
#
--
David Winsemius, MD
West Hartford, CT
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Re: [R] cbind alternate
On Jan 6, 2012, at 12:39 PM, Marc Schwartz wrote: On Jan 6, 2012, at 11:43 AM, Mary Kindall wrote: I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) Apologies, I mis-read where the hang up was. It is in the use of cbind() prior to calling write.table(), not in write.table() itself. Not sure why that part is taking a long time, unless as already mentioned, you are short on memory available. This runs quickly for me: x - matrix(rnorm(100 * 3), ncol = 3) y - matrix(rnorm(100 * 3), ncol = 3) system.time(z - cbind(x, y)) user system elapsed 0.039 0.025 0.065 str(z) num [1:100, 1:6] -0.5102 1.8776 2.4635 0.2982 0.0901 ... To give an example with two data frames containing differing data types, let's use the built-in 'iris' data set, which has 5 columns and 150 rows by default. Let's create a new version with over a million rows: iris.new - iris[rep(seq(nrow(iris)), 7000), ] str(iris.new) 'data.frame': 105 obs. of 5 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1 1 1 1 ... system.time(iris.new2 - cbind(iris.new, iris.new)) user system elapsed 5.289 0.282 5.658 str(iris.new2) 'data.frame': 105 obs. of 10 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1 1 1 1 ... $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1 1 1 1 ... You might verify the structures of your 'x' and 'y' to be sure that there is not something amiss with either one. HTH, Marc Schwartz __ 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] How to properly re-set a saved seed? I've got the answer, but no explanation
Hello, happy new year. I've come back to a problem from last spring. I need to understand what what is the technically correct method to save a seed and then re-set it. Although this example arises in the context of MT19937, I'm needing to do this with other generators as well, including L'Ecuyer streams. The puzzle is this comment in ?Random: ‘set.seed’ is the recommended way to specify seeds. What I did not understand before, and can't make sense of now, is that set.seed does not re-set a saved seed. Here's my working example: ## Paul Johnson ## April 20, 2011 ## If you've never dug into the R random number generator and its use of the ## default MT19937, this might be informative. It will also help you ## avoid a time-consuming mistake that I've made recently. ## I've been developing a simulation and I want to save and restore random ## streams exactly. I got that idea from John Chambers Software for ## Data Analysis and I studied his functions to see how he did it. ## The problem, as we see below, is that set.seed() doesn't do what I expected, ## and I feel lucky to have noticed it now, rather than later. ## I wish set.seed did work the way I want, though, and that's why I'm ## writing here, to see if anybody else wishes the same. ## Here's a puzzle. Try this: set.seed(444) s1 - .Random.seed runif(1) rnorm(1) set.seed(444) runif(1) rnorm(1) ## Those matched. Good ## Re-insert the saved seed s1 set.seed(s1) runif(1) rnorm(1) ## Why don't the draws match? I put back the seed, didn't I? ## Hm. Try again set.seed(s1) runif(1) rnorm(1) ## Why did those match? But neither matches cases 1 and 2. ## I was baffled discouraged. ## The help page for random numbers says: ## ‘set.seed’ is the recommended way to specify seeds. ## But, it doesn't say something like # set.seed(s1) ## is supposed to work. Ah. My mistake. Here's the fix and the puzzle ### ## To re-set the generator to its position at s1, it is required ## to do what the help page seems to say we should not. .Random.seed - s1 runif(1) # -- Paul E. Johnson Professor, Political Science 1541 Lilac Lane, Room 504 University of Kansas __ 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] cbind alternate
Hello,
I believe this function can handle a problem of that size, or bigger.
It does NOT create the full matrix, just writes it to a file, a certain
number of lines at a time.
write.big.matrix - function(x, y, outfile, nmax=1000){
if(file.exists(outfile)) unlink(outfile)
testf - file(outfile, at) # or wt - write text
on.exit(close(testf))
step - nmax # how many at a time
inx - seq(1, length(x), by=step) # index into 'x' and 'y'
mat - matrix(0, nrow=step, ncol=2) # create a work matrix
# do it 'nmax' rows per iteration
for(i in inx){
mat - cbind(x[i:(i+step-1)], y[i:(i+step-1)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
}
# and now the remainder
mat - NULL
mat - cbind(x[(i+1):length(x)], y[(i+1):length(y)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
# return the output filename
outfile
}
x - 1:1e6 # a numeric vector
y - sample(letters, 1e6, replace=TRUE) # and a character vector
length(x);length(y) # of the same length
fl - test.txt# output file
system.time(write.big.matrix(x, y, outfile=fl))
On my system it takes (sample output)
user system elapsed
1.590.041.65
and can handle different types of data. In the example, numeric and
character.
If you also need the matrix, try to use 'cbind' first, without writing to a
file.
If it's still slow, adapt the code above to keep inserting chunks in an
output matrix.
Rui Barradas
--
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http://r.789695.n4.nabble.com/cbind-alternate-tp4270188p4270444.html
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__
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Re: [R] How to properly re-set a saved seed? I've got the answer, but no explanation
Hi Paul, Did you try looking at s1, or ?.Random.seed s1 - 444 set.seed(s1) will give you the same result as set.seed(444) but .Random.seed does not contain the seed (444), but the state of the RNG, and is an integer vector. You can save that state using s1 - .Random.seed and restore it using .Random.seed - s1 but that's not the same thing that set.seed() does, or the information the latter requires. So you have two options: 1. save the integer seed and use set.seed() to assign it. 2. save the RNG state from .Random.seed and restore it later. The latter, by the way, is entirely reasonable to do. What the help warns you against is altering part of .Random.seed Unless I've completely misinterpreted your question, always possible, you're mixing two distinct but related types of information. Sarah On Fri, Jan 6, 2012 at 2:05 PM, Paul Johnson pauljoh...@gmail.com wrote: Hello, happy new year. I've come back to a problem from last spring. I need to understand what what is the technically correct method to save a seed and then re-set it. Although this example arises in the context of MT19937, I'm needing to do this with other generators as well, including L'Ecuyer streams. The puzzle is this comment in ?Random: ‘set.seed’ is the recommended way to specify seeds. What I did not understand before, and can't make sense of now, is that set.seed does not re-set a saved seed. Here's my working example: ## Paul Johnson ## April 20, 2011 ## If you've never dug into the R random number generator and its use of the ## default MT19937, this might be informative. It will also help you ## avoid a time-consuming mistake that I've made recently. ## I've been developing a simulation and I want to save and restore random ## streams exactly. I got that idea from John Chambers Software for ## Data Analysis and I studied his functions to see how he did it. ## The problem, as we see below, is that set.seed() doesn't do what I expected, ## and I feel lucky to have noticed it now, rather than later. ## I wish set.seed did work the way I want, though, and that's why I'm ## writing here, to see if anybody else wishes the same. ## Here's a puzzle. Try this: set.seed(444) s1 - .Random.seed runif(1) rnorm(1) set.seed(444) runif(1) rnorm(1) ## Those matched. Good ## Re-insert the saved seed s1 set.seed(s1) runif(1) rnorm(1) ## Why don't the draws match? I put back the seed, didn't I? ## Hm. Try again set.seed(s1) runif(1) rnorm(1) ## Why did those match? But neither matches cases 1 and 2. ## I was baffled discouraged. ## The help page for random numbers says: ## ‘set.seed’ is the recommended way to specify seeds. ## But, it doesn't say something like # set.seed(s1) ## is supposed to work. Ah. My mistake. Here's the fix and the puzzle ### ## To re-set the generator to its position at s1, it is required ## to do what the help page seems to say we should not. .Random.seed - s1 runif(1) # -- Sarah Goslee http://www.functionaldiversity.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.
Re: [R] Assign and cmpfun
On 12-01-06 12:18 PM, Michael Pearmain wrote:
Hi All,
I've just recently discovered the cmpfun function, and was wanting to to
create a function to assign this to all the functions i have created, but
without explicitly naming them.
I've achieved this with:
foo- function(x) { print(x)}
bar- function(x) { print(x + 1)}
foo- function(x) { print(x)}
foo
function(x) { print(x)}
cmpfun(foo)
function(x) { print(x)}
bytecode: 0x26e3d40
find.all.functions- ls.str(mode = 'function')
for(i in seq_along(find.all.functions)) {
assign(find.all.functions[i], cmpfun(get(find.all.functions[i])))
}
But remember told that using assign is generally a bad idea, and ideally i
want to functionalize this to say something like:
CreateCompiledFunctions- function() {
find.all.functions- ls.str(mode = 'function')
for(i in seq_along(find.all.functions)) {
assign(find.all.functions[i], cmpfun(get(find.all.functions[i])))
}
}
Does anyone have a better solution?
Put your functions in a package, and install it with the --byte-compile
option. This is better in two ways:
1. You have a package, which is *much* better than having a bunch of
functions sitting around in an .Rdata file.
2. You don't recompile functions that are already compiled.
Duncan Murdoch
Thanks in advance
Mike
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Re: [R] How to properly re-set a saved seed? I've got the answer, but no explanation
In S+ you can do orig - .Random.seed x - runif(10) set.seed(orig) identical(x, runif(10)) # return TRUE because S+'s set.seed interprets a non-scalar input as a copy of an old .Random.seed and assigns it in the proper location. (It throws an error if it is not a valid value for .Random.seed.) In R you use set.seed to set a proxy seed (an integer that maps into the real seed) and, I think, .Random.seed - orig to set the real seed. 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 Sarah Goslee Sent: Friday, January 06, 2012 11:21 AM To: Paul Johnson Cc: R-help Subject: Re: [R] How to properly re-set a saved seed? I've got the answer, but no explanation Hi Paul, Did you try looking at s1, or ?.Random.seed s1 - 444 set.seed(s1) will give you the same result as set.seed(444) but .Random.seed does not contain the seed (444), but the state of the RNG, and is an integer vector. You can save that state using s1 - .Random.seed and restore it using .Random.seed - s1 but that's not the same thing that set.seed() does, or the information the latter requires. So you have two options: 1. save the integer seed and use set.seed() to assign it. 2. save the RNG state from .Random.seed and restore it later. The latter, by the way, is entirely reasonable to do. What the help warns you against is altering part of .Random.seed Unless I've completely misinterpreted your question, always possible, you're mixing two distinct but related types of information. Sarah On Fri, Jan 6, 2012 at 2:05 PM, Paul Johnson pauljoh...@gmail.com wrote: Hello, happy new year. I've come back to a problem from last spring. I need to understand what what is the technically correct method to save a seed and then re-set it. Although this example arises in the context of MT19937, I'm needing to do this with other generators as well, including L'Ecuyer streams. The puzzle is this comment in ?Random: 'set.seed' is the recommended way to specify seeds. What I did not understand before, and can't make sense of now, is that set.seed does not re-set a saved seed. Here's my working example: ## Paul Johnson ## April 20, 2011 ## If you've never dug into the R random number generator and its use of the ## default MT19937, this might be informative. It will also help you ## avoid a time-consuming mistake that I've made recently. ## I've been developing a simulation and I want to save and restore random ## streams exactly. I got that idea from John Chambers Software for ## Data Analysis and I studied his functions to see how he did it. ## The problem, as we see below, is that set.seed() doesn't do what I expected, ## and I feel lucky to have noticed it now, rather than later. ## I wish set.seed did work the way I want, though, and that's why I'm ## writing here, to see if anybody else wishes the same. ## Here's a puzzle. Try this: set.seed(444) s1 - .Random.seed runif(1) rnorm(1) set.seed(444) runif(1) rnorm(1) ## Those matched. Good ## Re-insert the saved seed s1 set.seed(s1) runif(1) rnorm(1) ## Why don't the draws match? I put back the seed, didn't I? ## Hm. Try again set.seed(s1) runif(1) rnorm(1) ## Why did those match? But neither matches cases 1 and 2. ## I was baffled discouraged. ## The help page for random numbers says: ## 'set.seed' is the recommended way to specify seeds. ## But, it doesn't say something like # set.seed(s1) ## is supposed to work. Ah. My mistake. Here's the fix and the puzzle ### ## To re-set the generator to its position at s1, it is required ## to do what the help page seems to say we should not. .Random.seed - s1 runif(1) # -- Sarah Goslee http://www.functionaldiversity.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-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] Bootstrapping nlme models
Hi, Let me start my saying that I am new to R hence my grasp of the appropriate used of R coding is undoubtedly way behind many on this forum. I am trying to use boostrapping to derive errors around my parameter estimate for the fixed effects in the following model. It is simply estimating the number of times an animal might cross a road based on the road's distance from a stream. I have tried several different methods but have been unsuccessful. I have tried both using the boot package and simply developing a bit of code that resamples my data and drops the parameter estimates into a new vector. I would be interested in running both a parametric and non-parametric version. model2-nlme(nwfcross~a*exp(-b*nwfdist), fixed=a+b~1, random=a+b~1|nwfid, start=c(a=300,b=0.016)) Any help would be greatly appreciated. If it is helpful I can add in some of the code I have tried and the resulting error messages. Lynn -- View this message in context: http://r.789695.n4.nabble.com/Bootstrapping-nlme-models-tp4270557p4270557.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] cbind alternate
Sorry Mary,
My function would write the remainder twice, I had only tested it
with multiples of the chunk size.
(And without looking at the lenghty output correctly.)
Now checked:
write.big.matrix - function(x, y, outfile, nmax=1000){
if(file.exists(outfile)) unlink(outfile)
testf - file(outfile, at) # or wt - write text
on.exit(close(testf))
step - nmax # how many at a time
inx - seq(1, length(x)-step, by=step) # index into 'x' and 'y'
mat - matrix(0, nrow=step, ncol=2) # create a work matrix
# do it 'nmax' rows per iteration
for(i in inx){
mat - cbind(x[i:(i+step-1)], y[i:(i+step-1)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
}
# and now the remainder
if(i+step length(x)){
mat - NULL
mat - cbind(x[(i+step):length(x)], y[(i+step):length(y)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
}
# return the output filename
outfile
}
x - 1:(1e6 + 1234) # a numeric vector
y - sample(letters, 1e6 + 1234, replace=TRUE) # and a character vector
length(x);length(y) # of the same length
fl - test.txt# output file
system.time(write.big.matrix(x, y, outfile=fl, nmax=100))
user system elapsed
3.040.063.09
system.time(write.big.matrix(x, y, outfile=fl))
user system elapsed
1.640.121.76
Rui Barradas
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[R] Please help!! How do I set graphical parameters for ploting ctree()
I'm trying to understand how to set graphical parameters for trees created with the party package. For example take the following code: library(party) data(airquality) airq - subset(airquality, !is.na(Ozone)) airct - ctree(Ozone ~ ., data = airq, controls = ctree_control(maxsurrogate = 3)) plot(airct) My problem is, I've got a ctree that has 14 terminal nodes, and as a result of the default graphical paramters of plot(ctree), all of the text is completely undreadable (decison node text is too large and as a result, these nodes overlap one another, terminal node barplot text overlaps hideously). What I would like to do is learn how to customize the elements of these plots. For the text, I have tried par(cex=.4, cex.main=.4, cex.axis=.2, cex.lab=.2)... to no avail. As far as reducing the size of the node ovals, I'm completely lost there... Anyone have any suggestions? Thanks in advance! Chris [[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] cbind alternate
What is it you want to do with the data after you save it? Are you just going to read it back into R? If so, consider using save/load. On Fri, Jan 6, 2012 at 12:43 PM, Mary Kindall mary.kind...@gmail.com wrote: I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- - Mary Kindall Yorktown Heights, NY 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. -- 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.
[R] Sweave Chunks
Hi, is there a way to collapse/expand the Sweave chunks? Best Riccardo __ 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] Spatial data, rpoispp, using window with fixed radius?
On 07/01/12 02:17, herbert8...@gmx.de wrote: Dear All, I was searching through the spatstat manual in order to find a function to simulate a Poisson pattern only within a fixed radius (circular moving window) around individual points. If points are distributed heterogeneously over a large area this may help to only assess deviation from CSR within the window and thus does not require additional information on a covariate. I could not find such a function in spatstat. Can please anyone help? It's not clear to me just what you want to do, but it *sounds* like you want to simulate a cluster process with each cluster being a Poisson pattern in a disk of fixed radius. If so, the function rMatClust() does just what you want. It seems to me also possible that you want to pre-specify the cluster centres, or parent points. Such a capability is *not* currently built into spatstat but would not be hard to code up. Let me know if you indeed want to pre-specify the cluster centres. cheers, Rolf Turner __ 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] RODBC installation: error message
1. If you do not have unixODBC installed, you will get an error message
regarding a missing ODBC driver manager, if as you have done here, that test
is first rather than second.
If the test for the headers comes first, and no driver manager is
installed, configure exits with an error before the library test is
run (hence the error message regarding a missing driver manager is not
displayed).
I cannot speak for Prof. Ripley here, but since you seem to have the
requisite skills, he may be open to accepting a patch against his package
that would narrowly address the issue that you raise.
OK, here's my shot at such a patch (below), together with the
resulting configure output in cases where the library or the headers
are missing. I realize that there are matters of style preference
here, that this approach errs on the side of verbosity rather than
terseness, and that it's entirely up to the R package maintainers
which style to adopt.
Thanks for the replies!
Greg
Case 1, no ODBC driver manager installed =
checking for stdint.h... yes
checking for unistd.h... yes
checking sql.h usability... no
checking sql.h presence... no
checking for sql.h... no
checking sqlext.h usability... no
checking sqlext.h presence... no
checking for sqlext.h... no
checking for library containing SQLTables... no
configure: error:
No ODBC driver manager found;
see section Installation in inst/doc/RODBC.pdf
(also at http://cran.r-project.org/web/packages/RODBC/vignettes/RODBC.pdf)
ERROR: configuration failed for package 'RODBC'
== Case 2, driver manager installed but no headers
checking for unistd.h... yes
checking sql.h usability... no
checking sql.h presence... no
checking for sql.h... no
checking sqlext.h usability... no
checking sqlext.h presence... no
checking for sqlext.h... no
checking for library containing SQLTables... -lodbc
configure: error:
Found ODBC driver manager library (-lodbc),
but not required ODBC headers sql.h and sqlext.h.
(Need -devel version of ODBC driver manager?)
ERROR: configuration failed for package 'RODBC'
= patch follows
diff --git a/configure.ac b/configure.ac
index 8f9fc6a..8661e8f 100644
--- a/configure.ac
+++ b/configure.ac
@@ -64,10 +64,6 @@ CPPFLAGS=${CPPFLAGS} ${RODBC_CPPFLAGS}
dnl Check the headers can be found
AC_CHECK_HEADERS(sql.h sqlext.h)
-if test ${ac_cv_header_sql_h} = no ||
- test ${ac_cv_header_sqlext_h} = no; then
- AC_MSG_ERROR(ODBC headers sql.h and sqlext.h not found)
-fi
dnl search for a library containing an ODBC function
if test [ -n ${odbc_mgr} ] ; then
@@ -75,9 +71,23 @@ if test [ -n ${odbc_mgr} ] ; then
AC_MSG_ERROR(ODBC driver manager '${odbc_mgr}' not found))
else
AC_SEARCH_LIBS(SQLTables, odbc odbc32 iodbc, ,
-AC_MSG_ERROR(no ODBC driver manager found))
+AC_MSG_ERROR([
+No ODBC driver manager found;
+see section Installation in inst/doc/RODBC.pdf
+(also at http://cran.r-project.org/web/packages/RODBC/vignettes/RODBC.pdf)]))
+fi
+
+dnl Exit with error if ODBC headers were not found
+if test ${ac_cv_header_sql_h} = no ||
+ test ${ac_cv_header_sqlext_h} = no; then
+ AC_MSG_ERROR([
+Found ODBC driver manager library ($ac_cv_search_SQLTables),
+but not required ODBC headers sql.h and sqlext.h.
+(Need -devel version of ODBC driver manager?)
+])
fi
+
dnl for 64-bit ODBC need SQL[U]LEN, and it is unclear where they are defined.
AC_CHECK_TYPES([SQLLEN, SQLULEN], , , [# include sql.h])
dnl for unixODBC header
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[R] Getting a list of unique gene names from a list with semi-colons
Hello, I have one column in my dataframe that has gene names of interest. Unfortunately, due to the fact that some probes lie between two genes or two transcripts of a gene, it looks something like this - FAM81A LOC283050;LOC283050;LOC283050;ZMIZ1 PINK1;PINK1 MRPL12;MRPL12 C1orf114 MMS19;UBTD1 I would like to know how to get a list with all the names with no semi-colons and removing the replicates. I would like the end result to look like - FAM81A LOC283050 ZMIZI PINK1 MRPL12 C1orf114 MMS19 UBTD1 Thanks a lot for your help! Kurinji [[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] Getting a list of unique gene names from a list with semi-colons
I think you can do this with something like this (untested): unique(unlist(strsplit(XXX, ,))) Michael On Jan 6, 2012, at 8:05 PM, Kurinji Pandiyan kurinji.pandi...@gmail.com wrote: Hello, I have one column in my dataframe that has gene names of interest. Unfortunately, due to the fact that some probes lie between two genes or two transcripts of a gene, it looks something like this - FAM81A LOC283050;LOC283050;LOC283050;ZMIZ1 PINK1;PINK1 MRPL12;MRPL12 C1orf114 MMS19;UBTD1 I would like to know how to get a list with all the names with no semi-colons and removing the replicates. I would like the end result to look like - FAM81A LOC283050 ZMIZI PINK1 MRPL12 C1orf114 MMS19 UBTD1 Thanks a lot for your help! Kurinji [[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] Getting a list of unique gene names from a list with semi-colons
On Fri, Jan 6, 2012 at 9:05 PM, Kurinji Pandiyan kurinji.pandi...@gmail.com wrote: Hello, I have one column in my dataframe that has gene names of interest. Unfortunately, due to the fact that some probes lie between two genes or two transcripts of a gene, it looks something like this - FAM81A LOC283050;LOC283050;LOC283050;ZMIZ1 PINK1;PINK1 MRPL12;MRPL12 C1orf114 MMS19;UBTD1 I would like to know how to get a list with all the names with no semi-colons and removing the replicates. I would like the end result to look like - FAM81A LOC283050 ZMIZI PINK1 MRPL12 C1orf114 MMS19 UBTD1 Thanks a lot for your help! Kurinji This uses strapply in gsubfn: x - FAM81A LOC283050;LOC283050;LOC283050;ZMIZ1 PINK1;PINK1 library(gsubfn) unique(strapply(x, \\w+, c)[[1]]) If x is very long then there is a high speed version of strapply specialized to using c called strapplyc in the development version of gsubfn. For example, see this example of extracting 275,000 words from a novel: https://groups.google.com/group/corpling-with-r/msg/b85f7ff917cccb5d?dmode=sourceoutput=gplainnoredirectpli=1 -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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] Getting a list of unique gene names from a list with semi-colons
Sorry. - that should be a semi-colon below. Michael Weylandt On Jan 6, 2012, at 8:17 PM, R. Michael Weylandt michael.weyla...@gmail.com michael.weyla...@gmail.com wrote: I think you can do this with something like this (untested): unique(unlist(strsplit(XXX, ,))) Michael On Jan 6, 2012, at 8:05 PM, Kurinji Pandiyan kurinji.pandi...@gmail.com wrote: Hello, I have one column in my dataframe that has gene names of interest. Unfortunately, due to the fact that some probes lie between two genes or two transcripts of a gene, it looks something like this - FAM81A LOC283050;LOC283050;LOC283050;ZMIZ1 PINK1;PINK1 MRPL12;MRPL12 C1orf114 MMS19;UBTD1 I would like to know how to get a list with all the names with no semi-colons and removing the replicates. I would like the end result to look like - FAM81A LOC283050 ZMIZI PINK1 MRPL12 C1orf114 MMS19 UBTD1 Thanks a lot for your help! Kurinji [[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] Please help!! How do I set graphical parameters for ploting ctree()
On Jan 6, 2012, at 6:09 PM, Chris Conner wrote: I'm trying to understand how to set graphical parameters for trees created with the party package. For example take the following code: library(party) data(airquality) airq - subset(airquality, !is.na(Ozone)) airct - ctree(Ozone ~ ., data = airq, controls = ctree_control(maxsurrogate = 3)) plot(airct) My problem is, I've got a ctree that has 14 terminal nodes, and as a result of the default graphical paramters of plot(ctree), all of the text is completely undreadable (decison node text is too large and as a result, these nodes overlap one another, terminal node barplot text overlaps hideously). What I would like to do is learn how to customize the elements of these plots. For the text, I have tried par(cex=.4, cex.main=.4, cex.axis=.2, cex.lab=.2)... to no avail. As far as reducing the size of the node ovals, I'm completely lost there... After looking at: party:::plot.BinaryTree and party:::node_inner ... I've decided that you cannot. Neither of those function appear to accept parameters for shrinking the text. You could of course modify the code but if you couldn't find this answer on your own, then I have doubts that you possess the skills. So what are you to do? My suggestion would be to make a bigger plot on a vector graphics device (ps, pdf) and then shrink it. -- David Winsemius, MD West Hartford, CT __ 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] Putting an index explicitly into function code --- a curiosity.
I want to create a list of functions in a for loop, with the index
of the loop appearing explicitly in the function code.
After quite a bit of thrashing around I figured out how to do it.
Here is a toy example:
junk - vector(list,4)
for(i in 1:4) {
itmp - i
junk[[i]] - eval(bquote(function(x){42 + .(itmp)*x}))
}
So I'm *basically* happy, but there's something I don't understand:
Why do I need itmp?
That is, if I do
junk - vector(list,4)
for(i in 1:4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
}
then every entry of junk is equal to
function (x)
{
42 + 4L * x
}
i.e. I seem to get the *last* value of the index always substituted,
rather than
the current value. Something (subtle?) is going on that I don't
understand.
And than makes me feel not quite comfy. Can anyone enlighten me?
Ta.
cheers,
Rolf Turner
P. S. Also: Is there a *better* way of accomplishing my objective than
what I came up with?
R. T.
__
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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] Putting an index explicitly into function code --- a curiosity.
I imagine the answer will involve lazy evaluation and require you use force()
but I'm not quite qualified to pronounce and not at a computer to test.
Michael
On Jan 6, 2012, at 8:43 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
I want to create a list of functions in a for loop, with the index
of the loop appearing explicitly in the function code.
After quite a bit of thrashing around I figured out how to do it.
Here is a toy example:
junk - vector(list,4)
for(i in 1:4) {
itmp - i
junk[[i]] - eval(bquote(function(x){42 + .(itmp)*x}))
}
So I'm *basically* happy, but there's something I don't understand:
Why do I need itmp?
That is, if I do
junk - vector(list,4)
for(i in 1:4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
}
then every entry of junk is equal to
function (x)
{
42 + 4L * x
}
i.e. I seem to get the *last* value of the index always substituted, rather
than
the current value. Something (subtle?) is going on that I don't understand.
And than makes me feel not quite comfy. Can anyone enlighten me?
Ta.
cheers,
Rolf Turner
P. S. Also: Is there a *better* way of accomplishing my objective than what
I came up with?
R. T.
__
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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] Putting an index explicitly into function code --- a curiosity.
On Jan 6, 2012, at 9:51 PM, R. Michael Weylandt michael.weyla...@gmail.com
wrote:
I imagine the answer will involve lazy evaluation and require you
use force() but I'm not quite qualified to pronounce and not at a
computer to test.
Your theory passes the experimental test:
for(i in 1:4) {force(i)
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))}
junk
--
David.
Michael
On Jan 6, 2012, at 8:43 PM, Rolf Turner rolf.tur...@xtra.co.nz
wrote:
I want to create a list of functions in a for loop, with the index
of the loop appearing explicitly in the function code.
After quite a bit of thrashing around I figured out how to do it.
Here is a toy example:
junk - vector(list,4)
for(i in 1:4) {
itmp - i
junk[[i]] - eval(bquote(function(x){42 + .(itmp)*x}))
}
So I'm *basically* happy, but there's something I don't understand:
Why do I need itmp?
That is, if I do
junk - vector(list,4)
for(i in 1:4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
}
then every entry of junk is equal to
function (x)
{
42 + 4L * x
}
i.e. I seem to get the *last* value of the index always
substituted, rather than
the current value. Something (subtle?) is going on that I don't
understand.
And than makes me feel not quite comfy. Can anyone enlighten me?
Ta.
cheers,
Rolf Turner
P. S. Also: Is there a *better* way of accomplishing my objective
than what I came up with?
R. T.
__
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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-help@r-project.org mailing list
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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.
David Winsemius, MD
West Hartford, CT
__
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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] Putting an index explicitly into function code --- a curiosity.
On 07/01/12 15:51, R. Michael Weylandt michael.weyla...@gmail.com wrote:
I imagine the answer will involve lazy evaluation and require you use force()
but I'm not quite qualified to pronounce and not at a computer to test.
I think you've got it; I tried
junk - vector(list,4)
for(i in 1:4) {
junk[[i]] - eval(bquote(function(x){42 + .(force(i))*x}))
}
and got the result that I wanted. Still don't completely understand, but
it at least makes vague sense and makes me a bit more comfy.
Thanks.
cheers,
Rolf
__
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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] Putting an index explicitly into function code --- a curiosity.
On Fri, Jan 6, 2012 at 9:43 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
I want to create a list of functions in a for loop, with the index
of the loop appearing explicitly in the function code.
After quite a bit of thrashing around I figured out how to do it.
Here is a toy example:
junk - vector(list,4)
for(i in 1:4) {
itmp - i
junk[[i]] - eval(bquote(function(x){42 + .(itmp)*x}))
}
So I'm *basically* happy, but there's something I don't understand:
Why do I need itmp?
That is, if I do
junk - vector(list,4)
for(i in 1:4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
}
then every entry of junk is equal to
function (x)
{
42 + 4L * x
}
i.e. I seem to get the *last* value of the index always substituted, rather
than
the current value. Something (subtle?) is going on that I don't
understand.
And than makes me feel not quite comfy. Can anyone enlighten me?
This seems quite strange. For example, if we replace the for with a
while then we don't need to force i:
junk - vector(list,4)
i - 1
while(i = 4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
i - i + 1
}
junk
I am using:
R.version.string
[1] R version 2.14.1 Patched (2011-12-26 r58001)
win.version()
[1] Windows Vista (build 6002) Service Pack 2
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at 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.
Re: [R] Putting an index explicitly into function code --- a curiosity.
Presumably because the i = 4 has to be re-evaluated at the start of
each iteration of the while-loop which implicitly force()s it?
Though, I don't know if it might not be a bad idea to put an implicit
force() in the internal code for `for` to prevent these sorts of
things. I can't immediately think of a scenario where this sort of
behavior would be useful. It might also simplify the internals to pass
a value rather than what I presume is actually a promise (but I
haven't looked yet and I'm still not in a position to verify)
Are there scenarios where this sort of behavior is preferable for a
loop (I understand the pros of lazy evaluation in general)
Michael
On Fri, Jan 6, 2012 at 9:46 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Fri, Jan 6, 2012 at 9:43 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
I want to create a list of functions in a for loop, with the index
of the loop appearing explicitly in the function code.
After quite a bit of thrashing around I figured out how to do it.
Here is a toy example:
junk - vector(list,4)
for(i in 1:4) {
itmp - i
junk[[i]] - eval(bquote(function(x){42 + .(itmp)*x}))
}
So I'm *basically* happy, but there's something I don't understand:
Why do I need itmp?
That is, if I do
junk - vector(list,4)
for(i in 1:4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
}
then every entry of junk is equal to
function (x)
{
42 + 4L * x
}
i.e. I seem to get the *last* value of the index always substituted, rather
than
the current value. Something (subtle?) is going on that I don't
understand.
And than makes me feel not quite comfy. Can anyone enlighten me?
This seems quite strange. For example, if we replace the for with a
while then we don't need to force i:
junk - vector(list,4)
i - 1
while(i = 4) {
junk[[i]] - eval(bquote(function(x){42 + .(i)*x}))
i - i + 1
}
junk
I am using:
R.version.string
[1] R version 2.14.1 Patched (2011-12-26 r58001)
win.version()
[1] Windows Vista (build 6002) Service Pack 2
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.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
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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] Putting an index explicitly into function code --- a curiosity.
On Fri, Jan 6, 2012 at 11:13 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: Presumably because the i = 4 has to be re-evaluated at the start of each iteration of the while-loop which implicitly force()s it? Though, I don't know if it might not be a bad idea to put an implicit force() in the internal code for `for` to prevent these sorts of things. I can't immediately think of a scenario where this sort of behavior would be useful. It might also simplify the internals to pass a value rather than what I presume is actually a promise (but I haven't looked yet and I'm still not in a position to verify) It does not appear to be a promise. Using a C routine that returns TRUE for a promise and FALSE otherwise: # test ispromise # is a promise delayedAssign(a, x+1) .Call(ispromise, a, .GlobalEnv) [1] TRUE # not a promise xx - 3 .Call(ispromise, xx, .GlobalEnv) [1] FALSE # seems index of for is not a promise # for(i in 1:3) print(.Call(ispromise, i, .GlobalEnv)) [1] FALSE [1] FALSE [1] FALSE -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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.
