Re: [R] How to remove rows based on frequency of factor and then difference date scores
Many thanks to you both. I have now filed away for future reference the 2 factor tapply as well as the extremely useful looking plyr library. And the code worked beautifully :-) On 24 Aug 2010, at 19:47, Abhijit Dasgupta, PhD aikidasgu...@gmail.com wrote: The paste-y argument is my usual trick in these situations. I forget that tapply can take multiple ordering arguments :) Abhijit On 08/24/2010 02:17 PM, David Winsemius wrote: On Aug 24, 2010, at 1:59 PM, Abhijit Dasgupta, PhD wrote: The only problem with this is that Chris's unique individuals are a combination of Type and ID, as I understand it. So Type=A, ID=1 is a different individual from Type=B,ID=1. So we need to create a unique identifier per person, simplistically by uniqueID=paste(Type, ID, sep=''). Then, using this new identifier, everything follows. I see your point. I agree that a tapply method should present both factors in the indices argument. new.df - txt.df[ -which( txt.df$nn =1), ] new.df - new.df[ with(new.df, order(Type, ID) ), ] # and possibly needs to be ordered? new.df$diffdays - unlist( tapply(new.df$dt2, list(new.df$ID, new.df$Type), function(x) x[1] -x) ) new.df Type ID Date Valuedt2 nn diffdays 1A 1 16/09/2020 8 2020-09-16 30 2A 1 23/09/2010 9 2010-09-23 3 3646 4B 1 13/5/2010 6 2010-05-13 30 But do not agree that you need, in this case at least, to create a paste()-y index. Agreed, however, such a construction can be useful in other situations. -- Abhijit Dasgupta, PhD Director and Principal Statistician ARAASTAT Ph: 301.385.3067 E: adasgu...@araastat.com W: http://www.araastat.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] how do R calculates the number of intervals between tick-marks
Hello, I want to know how do R calculates the number of intervals between tick-marks in the y axis in a plot. I'm making a three y-axes plot and this information would help me a lot. Thanks in advance. Antonio Olinto Webmail - iBCMG Internet http://www.ibcmg.com.br __ 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 all objects except a few specified objects?
Thanks Josh and Karl. function dnrm() works well for my purpose. -- View this message in context: http://r.789695.n4.nabble.com/How-to-remove-all-objects-except-a-few-specified-objects-tp2335651p2337398.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] Documenting S4 Methods
I'm in the process of converting some S3 methods to S4 methods.
I have this function :
setGeneric(enrichmentCalc, function(rs, organism, seqLen,
...){standardGeneric(enrichmentCalc)})
setMethod(enrichmentCalc, c(GenomeDataList, BSgenome), function(rs,
organism, seqLen, ...) {
... ... ...
})
setMethod(enrichmentCalc, c(GenomeData, BSgenome), function(rs, organism,
seqLen=NULL, do.warn=FALSE) {
...... ...
})
setMethod(enrichmentCalc, c(GRanges, BSgenome), function(rs, organism,
seqLen=NULL) {
...... ...
}
and a part of my Rd file is :
\name{enrichmentCalc}
\docType{methods}
\alias{enrichmentCalc,GenomeDataList,BSgenome-method}
\alias{enrichmentCalc,GenomeData,BSgenome-method}
\alias{enrichmentCalc,GRanges,BSgenome-method}
...... ...
\usage{
enrichmentCalc(rs, organism, seqLen, ...)
enrichmentCalc(rs, organism, seqLen=NULL, do.warn=FALSE)
enrichmentCalc(rs, organism, seqLen=NULL)
}
.........
Can anyone suggest why I'm seeing this error :
* checking for code/documentation mismatches ... WARNING
Codoc mismatches from documentation object 'enrichmentCalc':
enrichmentCalc
Code: function(rs, organism, seqLen, ...)
Docs: function(rs, organism, seqLen = NULL, do.warn = FALSE)
Argument names in code not in docs:
...
Argument names in docs not in code:
do.warn
Mismatches in argument names:
Position: 4 Code: ... Docs: do.warn
Mismatches in argument default values:
Name: 'seqLen' Code: Docs: NULL
enrichmentCalc
Code: function(rs, organism, seqLen, ...)
Docs: function(rs, organism, seqLen = NULL)
Argument names in code not in docs:
...
Mismatches in argument default values:
Name: 'seqLen' Code: Docs: NULL
* checking Rd \usage sections ... WARNING
Objects in \usage without \alias in documentation object 'enrichmentCalc':
enrichmentCalc
Also, what is the difference between
... ... ...
\docType{methods}
... ... ...
\alias{methodName,class-method}
... ... ...
\usage{methodName(arg1)}
... ... ...
and
... ... ...
\alias{methodName,class-method}
... ... ...
\usage
{
\S4method{methodName}{class}(arg1)
}
... ... ...
I've seen both ways used for S4 methods and don't know what is the underlying
difference.
I haven't been able to find any good tutorials for the new S4 architecture
(written post 2006), so I'm not sure where to start with S4.
Thanks,
Dario.
--
Dario Strbenac
Research Assistant
Cancer Epigenetics
Garvan Institute of Medical Research
Darlinghurst NSW 2010
Australia
__
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] Removing inter-bar spaces in barchart
Rhelpers: I'm trying to make a barchart of a 2-group dataset (barchart(x~y,data=data,groups=z,horizontal=FALSE)). My problem is that I can't, for the life of me, seem to get rid of the inter-bar space -- box.ratio set to 1 doesn't do much. Any ideas? I'd ideally want zero space between the bars. Thanks! --j -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of California, Davis One Shields Avenue Davis, CA 95616 Phone: 415-763-5476 AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 __ 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] modify a nomogram
Hi,
I would like to emphasize (zoom) the zone of a nomogram where the
probability are 0.01
(nomogram built with nomogram, Design).
As a consequence, I don't need to draw the part of the Total points
axis with score 60 equivalent in my case to a linear predictor 4.5
- As far as I know, this is not possible with the arguments of the function.
- Changing the code of the function is beyond my abilities
-- can not even create a nomo.f function with the same body:
body(nomo) - expression({
conf.lp - match.arg(conf.lp) . rest of the function body
this does not even work
-- I am not sure to find the piece of code responsible for
defining the axis
nomogram(logistic.fit, fun = plogis, fun.at = c(seq(0.01,1,by=.2)),
lmgp=.2, lp = T,maxscale = 100, total.sep.page = T )
Thanks
David
__
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 bar lines like excel
Thanks a lot for the nice explanation. however, you mention that it will be nicer if there is data. yes you are right it be very kind of you if you include a simple example. where my data are: x1-c(11.5,9.38,9.3,9.8,9,9.06,9.42,8.8,9.05,8.14,8.2,7.59,6.92,6.47,7.12, 7.47,6.81,8.41,9.64,9.62,9.2,8.92,8,7.6,7.6,7.19,6.94,6.91,7,7.25,6.6,7.18,7.78, 7.37,7.29,6.71,7.05,6.69,6.05,6.38,6.18,7.67,7.34,7.13,6.95,6.8,6.45,6.81,6.27,6.35) x2-c(11.5,8.959535,9.728104,9.609262,8.755494,9.221545,8.244458,7.63944,7.92052,7.690449,7.853247, 7.239616,6.609007,5.92946,6.47822,5.906936,6.966004,8.630517,9.506582,9.57479,9.236638,9.581875,8.838992,8.368731,8.608884,7.998023,7.918123,7.832322,7.930177,7.479222,6.866978,7.454062,8.206185,7.344037,7.059774,6.547646,7.005803,6.623987,5.992691,6.313481,6.194613,6.224266,7.084932,6.568976,6.43866,5.70081,6.7593,6.6929,6.46806,5.964816) regards, -- View this message in context: http://r.789695.n4.nabble.com/Plot-bar-lines-like-excel-tp2337089p2337394.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] Odp: Finding pairs
Hi without other details it is probably impossible to give you any reasonable advice. Do you have your data already in R? What is their form? Are they in 2 columns in data frame? How did you get them paired? So without some more information probably nobody will invest his time as it seems no trivial to me. Regards Petr r-help-boun...@r-project.org napsal dne 24.08.2010 20:28:42: Dear R Helpers, I am a newbie and recently got introduced to R. I have a large database containing the names of bank branch offices along-with other details. I am into Operational Risk as envisaged by BASEL II Accord. I am trying to express my problem and I am using only an indicative data which comes in coded format. A (branch) B (controlled by) 144 145 146 147 144 148 145 149 147 151 146 .. ... .. ... where 144's etc are branch codes in a given city and B is subset of A. If a branch code appearing in A also appears in B (which is paired with some otehr element of A e.g. 144 appearing in A, also appears in B and is paired with 147 of A and likewise), then that means 144 is controlling operations of bank office 147. Again, 147 itself appears again in B and is paired with bank branch coded 149. Thus, 149 is controlled by 147 and 147 is controlled by 144. Likewise there are more than 700 hundred branch name codes available. My objective is to group them as follows - Bank Branch 144 147149 145 146 151 148 . or even the following output will do. 144 147 149 145 146 151 148 151 .. I understand I should be writing some R code to begin with which I had tried also but as of now I am helpless. Please guide me. Mike [[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] How to include trend (drift term) in arima.sim
Thanks to Mark Leeds and Dennis Murphy for their suggestions. The function arima.sim() only simulates stationary series without a trend, so the best approach appears to be to add the simulated stationary part to the trend as follows: Temp - arima.sim(n=N.Years.Forecast, list(ar=AR.Coef, ma=MA.Coef, intercept=Intercept), sd=SD) if (1 == D.) Simulated.Kappa - Trend + cumsum(Temp) if (2 == D.) Simulated.Kappa - Trend + cumsum(cumsum(Temp)) This appears to work well for d=1 in an ARIMA(p,d,q) model, but less well for d=2, where the results appear to be unstable. The problem is that if the forecast starts with two or three biggish values of the differences then these seem to take over the forecast of Kappa. -- View this message in context: http://r.789695.n4.nabble.com/How-to-include-trend-drift-term-in-arima-sim-tp2331581p2337729.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 plot y-axis on the right of x-axis
On Wed, 2010-08-25 at 07:12 +0800, elaine kuo wrote: Dear List, I have a richness data distributing across 20 N to 20 S latitude. (120 E-140 E longitude). I would like to draw the richness in the north hemisphere and a regression line in the plot (x-axis: latitude, y-axis: richness in the north hemisphere). The above demand is done using plot. Then, south hemisphere richness and regression are required to be generated using the same y-axis above but an x-axis on the left side of the y-axis. (The higher latitude in the south hemisphere, the left it would move) Elaine, It is not very clear what it is that you want here. Two plot panels on the same device like: layout(1:2) plot(1:100) plot(1:100) layout(1) The second panel for southern species richness, do you mean you want the plot to go like this: plot(1:100, xlim = c(100,1)) i.e. back to front to the normal axis limits? If so, you'll need to produce a regression model for southern richness just as you would for northern richness, but then predict from it at a sequence of latitude values that covers the range of your southern hemisphere observations, then plot these predictions versus the sequence of latitudes using lines(). For example, say you have: set.seed(123) rich - data.frame(richness = rpois(100, lambda = 5), latitude = sample(seq(0, 90, length = 1000), 100)) plot(richness ~ latitude, data = rich, xlim = c(90,0)) ## ignoring that this is a count and might be more meaningfully ## modelled using a Poisson GLM mod - lm(richness ~ latitude, data = rich) ## new prediction data, 100 equally spaced latitudes pdat - data.frame(latitude = seq(0, 90, length = 100)) pdat - within(pdat, richness - predict(mod, newdata = pdat)) lines(richness ~ latitude, data = pdat, col = red) If that doesn't help, we'll need more info. HTH G Please kindly share how to design the south plot and regression line for richness. Also, please advise if any more info is in need. Elaine [[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. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ 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] multi day R run crashed - is there a log
Dear all, I am using an R 2.10 installation on a Windows 203 server that I have no control over. After a multi-day run I found that it was terminated/crashed. Is there any log kept by R where I could see whether something/what happened? The same process has been run beofre on a smaller dataset (also at least a day of computing) without problems. Thanks Martin __ 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] Draw a perpendicular line?
Maybe perp.slope = -1/slope abline(cy - cx*perp.slope, perp.slope) where cx, cy are x- and y-coordinate of C, resp., and slope the slope you calculated for the line through A and B Am 24.Aug.2010 um 0:04 schrieb CZ: Hi, I am trying to draw a perpendicular line from a point to two points. Mathematically I know how to do it, but to program it, I encounter some problem and hope can get help. Thanks. I have points, A, B and C. I calculate the slope and intercept for line drawn between A and B. I am trying to check whether I can draw a perpendicular line from C to line AB and get the x,y value for the point D at the intersection. Assume I get the slope of the perpendicular line, I will have my point (D) using variable x and y which is potentially on line AB. My idea was using |AC|*|AC| = |AD|*|AD|+ |CD|*|CD|. I don't know what function I may need to call to calculate the values for point D (uniroot?). Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Draw-a-perpendicular-line-tp2335882p2335882.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] Odp: Finding pairs
Dear Mr Petr Pikal I am extremely sorry for the manner I have raised the query. Actually that was my first post to this R forum and in fact even I was also bit confused while drafting the query, for which I really owe sorry to all for consuming the precious time. Perhaps I will try to redraft my query in a better way as follows. I have two datasets A and B containing the names of branch offices of a particular bank say XYZ plc bank. The XYZ bank has number of main branch offices (say Parent) and some small branch offices falling under the purview of these main branch offices (say Child). The datalist A and B consists of these main branch office names as well as small branch office names. B is subset of A and these branch names are coded. Thus we have two datasets A and B as (again I am using only a portion of a large database just to have some idea) A B 144 145 146 147 144 148 145 149 147 151 148 Now the branch 144 appears in A as well as in B and in B it is mapped with 147. This means branch 147 comes under the purview of main branch 144. Again 147 is controlling the branch 149 (since 147 also has appeared in B and is mapped with 149 of A). Similarly, branch 145 is controlling branch 148 which further controls operations of bank branch 151 and like wise. So in the end I need an output something like - Main Branch Branch office1 Branch office2 144 147 149 145 148 151 146 NA NA ... .. I understand again I am not able to put forward my query properly. But I must thank all of you for giving a patient reading to my query and for reverting back earlier. Thanks once again. With warmest regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Odp: [R] Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 6:39 Hi without other details it is probably impossible to give you any reasonable advice. Do you have your data already in R? What is their form? Are they in 2 columns in data frame? How did you get them paired? So without some more information probably nobody will invest his time as it seems no trivial to me. Regards Petr r-help-boun...@r-project.org napsal dne 24.08.2010 20:28:42: Dear R Helpers, I am a newbie and recently got introduced to R. I have a large database containing the names of bank branch offices along-with other details. I am into Operational Risk as envisaged by BASEL II Accord. I am trying to express my problem and I am using only an indicative data which comes in coded format. A (branch) B (controlled by) 144 145 146 147 144 148 145 149 147 151 146 .. ... .. ... where 144's etc are branch codes in a given city and B is subset of A. If a branch code appearing in A also appears in B (which is paired with some otehr element of A e.g. 144 appearing in A, also appears in B and is paired with 147 of A and likewise), then that means 144 is controlling operations of bank office 147. Again, 147 itself appears again in B and is paired with bank branch coded 149. Thus, 149 is controlled by 147 and 147 is controlled by 144. Likewise there are more than 700 hundred branch name codes available. My objective is to group them as follows - Bank Branch 144 147 149 145 146 151 148 . or even the following output will do. 144 147 149 145 146 151 148 151 .. I understand I should be writing some R code to begin with which I had tried also but as of now I am helpless. Please guide me. Mike [[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
Good afternoon! It may seem trivial to some/most of You, but I found it difficult to properly include a C++-based .dll into a package that I want to build for usage in R. I read through the Writing R extensions... R administration ... instructions, but it seems I did not grasp the bigger picture. The way I figured out to use the .dll in my package finally worked for using that package from the R console, but it gives multiple error and warning messages when running the RCMD check pkg command in the shell. What I did: I created the package skeleton via package.skeleton() including a function that makes both (1) a library.dynam() call to a .dll and (2) a .C() call including the name of the function in that DLL. Next, I run RCMD INSTALL pkg to install the package and then created a /lib folder in the installed package containing the .dll file. I tried alternatives, e.g., (a) containing a /src folder in may package skeleton with the .dll in it and/or (b) including zzz.R with a .First.lib function including the library call but nothing worked out when running the package check. help! thank You very much, MArtin [[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] lmer() causes segfault
Ben Bolker bbolker at gmail.com writes: Bertolt Meyer bmeyer at sozpsy.uzh.ch writes: Hello lmer() - users, A call to the lmer() function causes my installation of R (2.11.1 on Mac OS X 10.5.8) to crash and I am trying to figure out the problem. [snip snip] detach(package:nlme) library(lme4) mod1 - lmer(performance ~ time + (time | GroupID/StudentNumber), data = dataset.long, na.action = na.omit) However, this call results in a segfault: *** caught segfault *** address 0x154c3000, cause 'memory not mapped' and a lengthy traceback. I can reproduce this error. It also occurs when I don't load nlme before lme4. Can someone tell me what I am doing wrong? Any help is greatly appreciated. This may well be a bug in lmer. There have been a number of fussy computational issues with the lme4 package on the Mac platform. Ben, thanks for your reply. I tried to replicate this issue with a small clean data set on a windows machine. You can find the code for the data frame (100 observations from my data) at the end of this mail. Very simple: four test scores per student over time, and students are nested in groups. On my Windows installation, lmer() throws an error that does not seem to get caught on the Mac, resulting in the segfault: library(lme4) mlmoded1.lmer - lmer(Score ~ Time + (Time | GroupID/StudentID), data = test.data) Error: length(f1) == length(f2) is not TRUE Addditional Warnings: 1: In StudentID:GroupID : numeric expression has 100 elements: only first one is used 2: In StudentID:GroupID : numeric expression has 100 elements: only first one is used It seems to me that I am committing a trivial error here and that I am too blind to see it. Any ideas? Regards, Bertolt If it is at all possible, please (1) post the results of sessionInfo() [which will in particular specify which version of lme4 you are using]; (2) possibly try this with the latest development version of lme4, from R-forge, if that's feasible (it might be necessary to build the package from source), and most importantly: (3) create a reproducible (for others) example -- most easily by posting your data on the web somewhere, but if that isn't possible by simulating data similar to yours (if it doesn't happen with another data set of similar structure, that's a clue -- it says it's some more particular characteristic of your data that triggers the problem) and (4) post to to *either* the R-sig-mac or the R-sig-mixed-models list, where the post is more likely to come to the attention of those who can help diagnose/fix ... good luck Ben Bolker test.data - data.frame(c(17370, 17370, 17370, 17370, 17379, 17379, 17379, 17379, 17387, 17387, 17387, 17387, 17391, 17391, 17391, 17391, 17392, 17392, 17392, 17392, 17394, 17394, 17394, 17394, 17408, 17408, 17408, 17408, 17419, 17419, 17419, 17419, 17429, 17429, 17429, 17429, 17432, 17432, 17432, 17432, 17436, 17436, 17436, 17436, 17439, 17439, 17439, 17439, 17470, 17470, 17470, 17470, 17220, 17220, 17220, 17220, 17348, 17348, 17348, 17348, 17349, 17349, 17349, 17349, 17380, 17380, 17380, 17380, 17398, 17398, 17398, 17398, 17400, 17400, 17400, 17400, 17402, 17402, 17402, 17402, 17403, 17403, 17403, 17403, 17413, 17413, 17413, 17413, 17416, 17416, 17416, 17416, 17420, 17420, 17420, 17420, 17421, 17421, 17421, 17421), c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), c(1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4), c(76.76, 81.83, 89.78, 92.82, 75.86, 81.84, 88.96, 92.28, 75.28, 80.68, 88.62, 92.29, 76.60, 84.59, 92.03, 94.05, 75.57, 79.94, 86.11, 90.25, 74.54, 81.42, 87.50, 90.71, 76.02, 83.68, 91.11, 94.14, 76.31, 83.76, 90.44, 94.58, 72.29, 80.51, 86.09, 90.41, 74.99, 82.28, 88.77, 92.26, 75.28, 81.92, 89.25, 92.64, 76.31, 83.93, 91.00, 94.60, 76.31, 82.44, 90.57, 95.17, 76.94, 82.21, 83.81, 85.00, 79.96, 81.92, 86.32, 90.05, 82.01, 84.81, 88.79, 93.10, 77.87, 82.94, 86.86, 90.31, 77.87, 79.64, 85.66, 86.97, 79.35, 80.44, 84.26, 83.62, 79.06, 81.56, 85.00, 87.43, 79.34, 81.47, 83.23, 86.86, 79.44, 80.37, 84.36, 89.11, 78.77, 81.02, 81.60, 87.21, 75.75, 79.35, 80.38, 86.87, 76.04, 80.57, 83.36, 86.31)) names(test.data) - c(StudentID, GroupID, Time, Score) __ 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,
Re: [R] Draw a perpendicular line?
hi, also, make sure you have set the aspect ratio to 1:1 when plotting (asp=1). HTH, baptiste On 25 August 2010 10:20, Benno Pütz pu...@mpipsykl.mpg.de wrote: Maybe perp.slope = -1/slope abline(cy - cx*perp.slope, perp.slope) where cx, cy are x- and y-coordinate of C, resp., and slope the slope you calculated for the line through A and B Am 24.Aug.2010 um 0:04 schrieb CZ: Hi, I am trying to draw a perpendicular line from a point to two points. Mathematically I know how to do it, but to program it, I encounter some problem and hope can get help. Thanks. I have points, A, B and C. I calculate the slope and intercept for line drawn between A and B. I am trying to check whether I can draw a perpendicular line from C to line AB and get the x,y value for the point D at the intersection. Assume I get the slope of the perpendicular line, I will have my point (D) using variable x and y which is potentially on line AB. My idea was using |AC|*|AC| = |AD|*|AD|+ |CD|*|CD|. I don't know what function I may need to call to calculate the values for point D (uniroot?). Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Draw-a-perpendicular-line-tp2335882p2335882.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. -- Dr. Baptiste Auguié Departamento de Química Física, Universidade de Vigo, Campus Universitario, 36310, Vigo, Spain tel: +34 9868 18617 http://webs.uvigo.es/coloides __ 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] percentage sign in expression
On 24/08/2010, David Winsemius dwinsem...@comcast.net wrote:
On Aug 24, 2010, at 9:37 AM, e-letter wrote:
Readers,
According to the documentation for the function 'plotmath' there is no
apparent possibility to add the percent sign (%) to a plot function,
Where did you see an assertion made???
Within R I entered the command:
?plotmath
Also accessed using:
help.start(browser=opera)
Navigating the web browser page:
packages
packages in /usr/lib/R/library
grdevices
plotmath
In the list headed 'syntax' and 'meaning' within the section 'details'.
e.g.
plot(a[,1]~b[,2],ylab=expression(x~%),xlab=expression(z))
How to achieve this please?
Read the plotmath helo page more carefully. The section immediatedly
below the plotmath expressions points you to the use of the symbol()
expression-function and to the points help page where generation of
the available glyphs proceeds according to the advice on help(plotmath):
In my system the paragraph immediately after the list of features
(i.e. 'syntax','meaning') describes a note to TeX users. I cannot see
reference to 'symbol()'.
TestChars - function(sign=1, font=1, ...)
+ {
+if(font == 5) { sign - 1; r - c(32:126, 160:254)
+} else if (l10n_info()$MBCS) r - 32:126 else r - 32:255
+if (sign == -1) r - c(32:126, 160:255)
+par(pty=s)
+plot(c(-1,16), c(-1,16), type=n, xlab=, ylab=,
+ xaxs=i, yaxs=i)
+grid(17, 17, lty=1)
+for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
+ }
TestChars(font=5)
Notice that the % sign is three characters to the right (i.e.
higher) of the forall symbol that is produced by the example code
I can't see 'forall' in the code above.
they offer. (The numbering proceeds from bottom up which confused me
at first.)
What numbering?
The documentation makes reference to the command:
demo(plotmath)
I applied this command and could not see an instruction to produce the
percent (%) symbol.
__
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] Rotate x-axis label on log scale
Yes, I do want the labels on the x-axis but the problem arises when the
y-axis is logarithmic presumably because the position labels on the x-axis
still needs to be defined in terms of x and y coordinates.
The simplified examples below should make the problem clearer.
I can't see a way of altering the staxlab code to fix this but would be very
grateful for any suggestions.
#Example 1
#plotting on a linear scale the labels are rotated as desired
library(plotrix)
x - y - 1:4
labels= c('label1', 'label2', 'label3', 'label4')
plot(x,y, xaxt='n')
staxlab(side=1,at=1:4,labels=labels, srt=45)
#Example 2
#plotting on a y-axis log scale the labels don't appear
plot(x,y, xaxt='n', log='y')
staxlab(side=1,at=1:4,labels=labels, srt=45)
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, August 25, 2010 11:46 AM
To: tesutton
Cc: 'Jim Lemon'; r-help@r-project.org
Subject: Re: [R] Rotate x-axis label on log scale
On Aug 24, 2010, at 11:37 PM, Tim Elwell-Sutton wrote:
Hi Jim
Thanks for this. The staxlab function seems very useful.
Unfortunately, the
rotation option doesn't seem to work for me when the y-axis is on a
log
scale.
What part of:
If srt is not NA, the labels will be rotated srt degrees and placed
below the plot. This method will only place labels at the bottom.
... is unclear? You did say you wanted the rotation to be on the x-
axis, did you not?
You could, of course, look at Lemon's code and hack it to do the y-axis:
else {
xylim - par(usr)
ypos - xylim[3] - ticklen * (xylim[4] - xylim[3])
par(xpd = TRUE)
text(at, ypos, labels, srt = srt, adj = 1, ...)
par(xpd = FALSE)
}
--
david.
It will stagger the labels but not rotate them. There's no error
message. On a linear axis the rotation works nicely. Any ideas?
The example below works if you omit log='y' or srt=45
Thanks very much
Tim
#Create plot with log-scale on the y-axis
par(mar = c(7, 4, 4, 2) + 0.1)
plot(1, type='n', bty='n',
xlab=,
ylab='Odds Ratio',
xlim= c(0.5,4.5),
ylim= c(0.75, 2),
cex=2, xaxt='n', yaxt='n', cex.lab=1.3,
log='y')
#Estimates and confidence intervals
points(c(1:4),c(1.1,1.32,1.14,1.36), pch=17, cex=1.5, col='blue')
segments (c(1:4),c(0.93,1.11,0.94,1.15),c(1:4),c(1.3,1.58,1.37,1.61),
col='blue', lwd=2)
#Add x- and y-axes
axis(1,c(1:4), labels= F)
axis(2, at=seq(0.75,2, by=0.25), labels=seq(0.75,2, by=0.25), las=1)
# Add x-axis labels
labels - paste(Label, 1:4, sep = )
staxlab(side=1, at=1:4, labels, srt=45)
-Original Message-
From: Jim Lemon [mailto:j...@bitwrit.com.au]
Sent: Tuesday, August 24, 2010 7:48 PM
To: tesutton
Cc: r-help@r-project.org
Subject: Re: [R] Rotate x-axis label on log scale
On 08/24/2010 11:44 AM, Tim Elwell-Sutton wrote:
Hi
I'd appreciate some help with plotting odds ratios. I want to
rotate the
labels on the x-axis by 45 degrees.
The usual way of doing this, using text - e.g. text(1,
par('usr')[3]-2.25..)
- gives no result when the y-axis is a log scale.
I guess this is because, as the par help says, for a logarithmic y-
axis:
y-limits will be 10 ^ par(usr)[3:4]
Hi Tim,
If you know where to put the labels, try:
library(plotrix)
staxlab(1,at=...,labels=...,srt=45)
Jim
--
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] Odp: Finding pairs
Hm r-help-boun...@r-project.org napsal dne 25.08.2010 09:43:26: Dear Mr Petr Pikal I am extremely sorry for the manner I have raised the query. Actually that was my first post to this R forum and in fact even I was also bit confused while drafting the query, for which I really owe sorry to all for consuming the precious time. Perhaps I will try to redraft my query in a better way as follows. I have two datasets A and B containing the names of branch offices of a particular bank say XYZ plc bank. The XYZ bank has number of main branch offices (say Parent) and some small branch offices falling under the purview of these main branch offices (say Child). The datalist A and B consists of these main branch office names as well as small branch office names. B is subset of A and these branch names are coded. Thus we have two datasets A and B as (again I am using only a portion of a large database just to have some idea) A B 144 what is here in B? Empty space?, 145 146 147 144 How do you know that 144 from B relates to 147 in A? Is it according to its positions? I.e. 4th item in B belongs to 4.th item in A? 148 145 149 147 151 148 Now the branch 144 appears in A as well as in B and in B it is mapped with 147. This means branch 147 comes under the purview of main branch 144. Again 147 is controlling the branch 149 (since 147 also has appeared in B and is mapped with 149 of A). Similarly, branch 145 is controlling branch 148 which further controls operations of bank branch 151 and like wise. Well as you did not say anything about structure of your data A-144:151 B-144:148 data.frame(A,B) A B 1 144 NA 2 145 NA 3 146 NA 4 147 144 5 148 145 6 149 146 7 150 147 8 151 148 DF-data.frame(A,B) main-DF$A[is.na(DF$B)] branch1-DF[!is.na(DF$B),] selected.branch1-branch1$A[branch1$B%in%main] branch2-branch1[!branch1$B%in%main,] selected.branch2-branch2$A[branch2$B%in%selected.branch1] and for cbinding your data which has uneven number of values see Jim Holtman's answer to this How to cbind DF:s with differing number of rows? Regards Petr So in the end I need an output something like - Main Branch Branch office1 Branch office2 144 147 149 145 148 151 146 NA NA ... .. I understand again I am not able to put forward my query properly. But I must thank all of you for giving a patient reading to my query and for reverting back earlier. Thanks once again. With warmest regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Odp: [R] Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 6:39 Hi without other details it is probably impossible to give you any reasonable advice. Do you have your data already in R? What is their form? Are they in 2 columns in data frame? How did you get them paired? So without some more information probably nobody will invest his time as it seems no trivial to me. Regards Petr r-help-boun...@r-project.org napsal dne 24.08.2010 20:28:42: Dear R Helpers, I am a newbie and recently got introduced to R. I have a large database containing the names of bank branch offices along-with other details. I am into Operational Risk as envisaged by BASEL II Accord. I am trying to express my problem and I am using only an indicative data which comes in coded format. A (branch) B (controlled by) 144 145 146 147 144 148 145 149 147 151 146 .. ... .. ... where 144's etc are branch codes in a given city and B is subset of A. If a branch code appearing in A also appears in B (which is paired with some otehr element of A e.g. 144 appearing in A, also appears in B and is paired with 147 of A and likewise), then that means 144 is controlling operations of bank office 147. Again, 147 itself appears again
[R] (no subject)
Hi I am using repeated meaturement data for my project and I want to use quantile regression with multilevel or panel data in R. I dont find the basic version of software in R, so I have difficulty in using it. I would also appreciate if anyone more proficient in R could help me how to run this. best wishes M.A-Vakili Department of Biostatistics Faculty of Medicine Shiraz/Iran P.O.Box: 71348-45794 E-Mail: vakil...@sums.ac.ir or mavak...@yahoo.com [[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] Documenting S4 Methods
Dario Strbenac d.strbe...@garvan.org.au
on Wed, 25 Aug 2010 13:00:03 +1000 (EST) writes:
I'm in the process of converting some S3 methods to S4 methods.
I have this function :
setGeneric(enrichmentCalc, function(rs, organism, seqLen,
...){standardGeneric(enrichmentCalc)})
setMethod(enrichmentCalc, c(GenomeDataList, BSgenome), function(rs,
organism, seqLen, ...) {
... ... ...
})
setMethod(enrichmentCalc, c(GenomeData, BSgenome), function(rs,
organism, seqLen=NULL, do.warn=FALSE) {
...... ...
})
setMethod(enrichmentCalc, c(GRanges, BSgenome), function(rs,
organism, seqLen=NULL) {
...... ...
}
and a part of my Rd file is :
\name{enrichmentCalc}
\docType{methods}
\alias{enrichmentCalc,GenomeDataList,BSgenome-method}
\alias{enrichmentCalc,GenomeData,BSgenome-method}
\alias{enrichmentCalc,GRanges,BSgenome-method}
...... ...
\usage{
enrichmentCalc(rs, organism, seqLen, ...)
enrichmentCalc(rs, organism, seqLen=NULL, do.warn=FALSE)
enrichmentCalc(rs, organism, seqLen=NULL)
}
.........
Can anyone suggest why I'm seeing this error :
* checking for code/documentation mismatches ... WARNING
Codoc mismatches from documentation object 'enrichmentCalc':
enrichmentCalc
Code: function(rs, organism, seqLen, ...)
Docs: function(rs, organism, seqLen = NULL, do.warn = FALSE)
Argument names in code not in docs:
...
Argument names in docs not in code:
do.warn
Mismatches in argument names:
Position: 4 Code: ... Docs: do.warn
Mismatches in argument default values:
Name: 'seqLen' Code: Docs: NULL
enrichmentCalc
Code: function(rs, organism, seqLen, ...)
Docs: function(rs, organism, seqLen = NULL)
Argument names in code not in docs:
...
Mismatches in argument default values:
Name: 'seqLen' Code: Docs: NULL
* checking Rd \usage sections ... WARNING
Objects in \usage without \alias in documentation object 'enrichmentCalc':
enrichmentCalc
Also, what is the difference between
... ... ...
\docType{methods}
... ... ...
\alias{methodName,class-method}
... ... ...
\usage{methodName(arg1)}
... ... ...
and
... ... ...
\alias{methodName,class-method}
... ... ...
\usage
{
\S4method{methodName}{class}(arg1)
}
... ... ...
the last one is the one you should use.
I don't think that you can easily use \usage{..} without
\S4method{.}
and not get a warning (/error) from R CMD check.
I've seen both ways used for S4 methods and don't know what is the
underlying difference.
BTW: You should probably reread (parts of) the Writing R Extensions manual;
It's a good exercise ... even for me as an R core team member.
I haven't been able to find any good tutorials for the new S4
architecture (written post 2006), so I'm not sure where to start with S4.
In R, ?Methods - References (s r in ESS)
lists John Chambers book (with extra hints).
Chambers, John M. (2008) _Software for Data Analysis: Programming
with R_ Springer. (For the R version: see section 10.6 for method
selection and section 10.5 for generic functions).
Many people recommend learning from good examples.
As the Bioconductor project has made heavy use of S4 classes and
methods, many of their packages are good examples to follow.
To find all - locally installed - packages which depend directly
on methods and hence are probably using S4 methods / classes,
I do
library(tools)
str(meth.pkgs.direct - dependsOnPkgs(methods, recursive=FALSE))
chr [1:313] BufferedMatrix externalVector gdistancetest Matrix ...
sort(meth.pkgs.direct)
[1] AcceptanceSamplingaccuracy adegenet
[4] adephylo amer aod
[7] apcluster apsrtable archetypes
[10] ArDec arulesarulesNBMiner
[13] arulesSequences asuR automap
[16] aws aylmerbackfitRichards
[19] backtest bbmle bcp
[22] bdsmatrix biclust bifactorial
[25] biganalytics biglm bigmemory
[28] bigtabulate bild BLCOP
[31] Brobdingnag bsBufferedMatrix
[34] calib catnetcelsius
[37] ChainLadder classGraphclue
[40]
Re: [R] how to plot y-axis on the right of x-axis
On 08/25/2010 09:12 AM, elaine kuo wrote: Dear List, I have a richness data distributing across 20 N to 20 S latitude. (120 E-140 E longitude). I would like to draw the richness in the north hemisphere and a regression line in the plot (x-axis: latitude, y-axis: richness in the north hemisphere). The above demand is done using plot. Then, south hemisphere richness and regression are required to be generated using the same y-axis above but an x-axis on the left side of the y-axis. (The higher latitude in the south hemisphere, the left it would move) Please kindly share how to design the south plot and regression line for richness. Also, please advise if any more info is in need. Hi Elaine, Changing the position of the axes is easily done by changing the side and pos arguments of the axis function. If you want to move the y-axis to the right (or as you state it, the x axis to the left): # y axis on the left plot(1:5,axes=FALSE) axis(1) axis(2) # add a y axis one third of the way to the right xylim-par(usr) axis(2,pos=xylim[1]+diff(xylim[1:2])/3) # add another y axis two thirds of the way axis(4,pos=xylim[2]-diff(xylim[1:2])/3) # add one more on the right axis(4) You can move the x axis up and down using the same tricks. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] percentage sign in expression
On Wed, 2010-08-25 at 09:32 +0100, e-letter wrote:
On 24/08/2010, David Winsemius dwinsem...@comcast.net wrote:
On Aug 24, 2010, at 9:37 AM, e-letter wrote:
Readers,
According to the documentation for the function 'plotmath' there is no
apparent possibility to add the percent sign (%) to a plot function,
Where did you see an assertion made???
Within R I entered the command:
?plotmath
Surely you don't expect the R help pages to document everything that
*isn't* possible with a given function? They'd be infinitely long! ;-)
What you meant to say was, ?plotmath doesn't show me how to add a %
symbol to a plot. How do I do it?
Also accessed using:
help.start(browser=opera)
Navigating the web browser page:
packages
packages in /usr/lib/R/library
grdevices
plotmath
In the list headed 'syntax' and 'meaning' within the section 'details'.
snip /
TestChars - function(sign=1, font=1, ...)
+ {
+if(font == 5) { sign - 1; r - c(32:126, 160:254)
+} else if (l10n_info()$MBCS) r - 32:126 else r - 32:255
+if (sign == -1) r - c(32:126, 160:255)
+par(pty=s)
+plot(c(-1,16), c(-1,16), type=n, xlab=, ylab=,
+ xaxs=i, yaxs=i)
+grid(17, 17, lty=1)
+for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
+ }
TestChars(font=5)
Notice that the % sign is three characters to the right (i.e.
higher) of the forall symbol that is produced by the example code
I can't see 'forall' in the code above.
You need to run it in R. Then you will see a plot of the glyphs
available in the symbol font. 'forall' is a mathematical symbol:
http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
they offer. (The numbering proceeds from bottom up which confused me
at first.)
What numbering?
The numbering of the glyphs so you can use the number to draw the symbol
you want. They are on the plot. You did run the code provided by David?
Anyway, is there a problem with this:
plot(1:10, xlab = expression(foo == 15*%))
That way you don't need to worry about finding the right glyph.
HTH
G
The documentation makes reference to the command:
demo(plotmath)
I applied this command and could not see an instruction to produce the
percent (%) symbol.
__
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.
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
__
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] Cran-packages-ProfessoR- how to Automatically email a file to an address using the perl program.
Hi Jonathan lees How to use this code and after installing ProfessR package autoemail(eadd, sfile, hnote = Exam Results) thanks veepsirtt -- View this message in context: http://r.789695.n4.nabble.com/Cran-packages-ProfessoR-how-to-Automatically-email-a-file-to-an-address-using-the-perl-program-tp2335184p2337924.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] help
Martin, On Wed, Aug 25, 2010 at 10:17 AM, Loos, Martin martin.l...@eawag.ch wrote: Good afternoon! It may seem trivial to some/most of You, but I found it difficult to properly include a C++-based .dll into a package that I want to build for usage in R. I read through the Writing R extensions... R administration ... instructions, but it seems I did not grasp the bigger picture. The way I figured out to use the .dll in my package finally worked for using that package from the R console, but it gives multiple error and warning messages when running the RCMD check pkg command in the shell. What I did: I created the package skeleton via package.skeleton() including a function that makes both (1) a library.dynam() call to a .dll and (2) a .C() call including the name of the function in that DLL. Next, I run RCMD INSTALL pkg to install the package and then created a /lib folder in the installed package containing the .dll file. As you say below, the way to go here is to add a /src folder in the package skeleton which contains your C/C++ code and header files and then run R CMD INSTALL, which will compile your code into a .dll file and place it in the appropriate directory, ie the /lib directory. It should not be necessary to do this manually. hth, Ingmar PS: It's hard to tell what is going wrong without the actual error messages. I tried alternatives, e.g., (a) containing a /src folder in may package skeleton with the .dll in it and/or (b) including zzz.R with a .First.lib function including the library call but nothing worked out when running the package check. help! thank You very much, MArtin [[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] Rotate x-axis label on log scale
On 08/25/2010 01:37 PM, Tim Elwell-Sutton wrote:
Hi Jim
Thanks for this. The staxlab function seems very useful. Unfortunately, the
rotation option doesn't seem to work for me when the y-axis is on a log
scale. It will stagger the labels but not rotate them. There's no error
message. On a linear axis the rotation works nicely. Any ideas?
The example below works if you omit log='y' or srt=45
Hi Tim,
Thanks for letting me know about this problem. I never did get around to
making staxlab work with log axes, but I think this new version:
staxlab-function(side=1,at,labels,nlines=2,top.line=0.5,
line.spacing=0.8,srt=NA,ticklen=0.03,...) {
if(missing(labels)) labels-at
nlabels-length(labels)
if(missing(at)) at-1:nlabels
if(is.na(srt)) {
linepos-rep(top.line,nlines)
for(i in 2:nlines) linepos[i]-linepos[i-1]+line.spacing
linepos-rep(linepos,ceiling(nlabels/nlines))[1:nlabels]
axis(side=side,at=at,labels=rep(,nlabels))
mtext(text=labels,side=side,line=linepos,at=at,...)
}
else {
xylim-par(usr)
if(side == 1) {
xpos-at
if(par(ylog)) ypos-10^(xylim[3]-ticklen*(xylim[4]-xylim[3]))
else ypos-xylim[3]-ticklen*(xylim[4]-xylim[3])
}
else {
ypos-at
if(par(xlog)) xpos-10^(xylim[1]-ticklen*(xylim[2]-xylim[1]))
else xpos-xylim[1]-ticklen*(xylim[2]-xylim[1])
}
par(xpd=TRUE)
text(xpos,ypos,labels,srt=srt,adj=1,...)
par(xpd=FALSE)
}
}
will do what you want.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot bar lines like excel
Hi:
Here's a ggplot2 version of your graph. Since you didn't include the dates,
I had to make them up, and it may have some consequence with respect to your
times because yours don't appear to be exactly equally spaced.
I created two data frames - an 'original' one that contains both series in
separate columns, and another that 'melts' the original data frame by
stacking the two series horizontally with a created factor that holds the
variable names. Notice that I assigned the names of the two series to
Series4 and Series6 in the original data frame to correspond with the names
on the plot.
library(ggplot2)
dd - data.frame(tm = seq(as.Date('2002-1-1'), by = 'month', length = 50),
Series4 = x1, Series6 = x2)
dm - melt(dd, id = 'tm')
dd - transform(dd, ylo = pmin(Series4, Series6), yup = pmax(Series4,
Series6))
The melted data, dm, is useful for plotting the two series and associating a
color aesthetic to the different series; we use it to generate the initial
part of the plot, the two series themselves. 'variable' is a factor variable
whose levels are the variable names (which map to the legend) and 'values'
is a numeric variable containing the values of the two series.
g - ggplot(data = dm, aes(x = tm))
g + geom_line(aes(y = value, colour = variable), size = 1.2) +
scale_colour_manual(, values = c('Series4' = 'yellow',
'Series6' = 'red')) + theme_bw() + ylim(4, 12) +
geom_linerange(data = dd, aes(x = tm, ymin = ylo, ymax = yup), size =
0.8) +
opts(panel.grid.major = theme_blank()) +
opts(panel.grid.minor = theme_blank()) +
geom_hline(yintercept = c(4, 6, 8, 10, 12), size = 1, alpha = 0.1) +
scale_x_date(major = '2 months', format = '%b-%Y') +
opts(axis.text.x = theme_text(angle = 90, hjust = 1, vjust = 0.5, size =
9))
scale_colour_manual() allows specification of one's own colors rather than
ggplot2's defaults. The initial pair of double quotes gets rid of the legend
label - if you leave it off, it will use the header 'variable'. You can
always substitute in another legend header if you wish.
The vertical lines connecting the series at each date is where the work is
required. ylo and yup in data frame dd represent the min and max values of
the series at each date, respectively, produced by the pmin() and pmax()
functions. ggplot2's geom_linerange() needs these variables to represent the
lower and upper limits of the line segments at each x value (date), which is
why the data frame used in geom_linerange() is dd rather than dm. The size
parameter in geom_line() and geom_linerange() corresponds to line thickness.
The first two opts() statements get rid of the default grid lines;
geom_hline() provides the substitutes.
The scale_x_date() code basically labels every other date with some
additional options to pretty up the appearance of the labels.
HTH,
Dennis
On Tue, Aug 24, 2010 at 2:40 PM, abotaha yaseen0...@gmail.com wrote:
Thanks a lot for the nice explanation. however, you mention that it will be
nicer if there is data. yes you are right it be very kind of you if you
include a simple example. where my data are:
x1-c(11.5,9.38,9.3,9.8,9,9.06,9.42,8.8,9.05,8.14,8.2,7.59,6.92,6.47,7.12,
7.47,6.81,8.41,9.64,9.62,9.2,8.92,8,7.6,7.6,7.19,6.94,6.91,7,7.25,6.6,7.18,7.78,
7.37,7.29,6.71,7.05,6.69,6.05,6.38,6.18,7.67,7.34,7.13,6.95,6.8,6.45,6.81,6.27,6.35)
x2-c(11.5,8.959535,9.728104,9.609262,8.755494,9.221545,8.244458,7.63944,7.92052,7.690449,7.853247,
7.239616,6.609007,5.92946,6.47822,5.906936,6.966004,8.630517,9.506582,9.57479,9.236638,9.581875,8.838992,8.368731,8.608884,7.998023,7.918123,7.832322,7.930177,7.479222,6.866978,7.454062,8.206185,7.344037,7.059774,6.547646,7.005803,6.623987,5.992691,6.313481,6.194613,6.224266,7.084932,6.568976,6.43866,5.70081,6.7593,6.6929,6.46806,5.964816)
regards,
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-bar-lines-like-excel-tp2337089p2337394.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] Plot bar lines like excel
Woow, it is amazing, thank you very much. yes i forget to attach the dates, however, the dates in my case is every 16 days. so how i can use 16 day interval instead of month in by option. cheers, -- View this message in context: http://r.789695.n4.nabble.com/Plot-bar-lines-like-excel-tp2337089p2338028.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] Odp: Finding pairs
Dear Mr Petr PIKAL After reading the R code provided by you, I realized that I would have never figured out how this could have been done. I am going to re-read again and again your code to understand the logic and the commands you have provided. Thanks again from the heart for your kind advice. Regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Re: [R] Odp: Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 9:01 Hm r-help-boun...@r-project.org napsal dne 25.08.2010 09:43:26: Dear Mr Petr Pikal I am extremely sorry for the manner I have raised the query. Actually that was my first post to this R forum and in fact even I was also bit confused while drafting the query, for which I really owe sorry to all for consuming the precious time. Perhaps I will try to redraft my query in a better way as follows. I have two datasets A and B containing the names of branch offices of a particular bank say XYZ plc bank. The XYZ bank has number of main branch offices (say Parent) and some small branch offices falling under the purview of these main branch offices (say Child). The datalist A and B consists of these main branch office names as well as small branch office names. B is subset of A and these branch names are coded. Thus we have two datasets A and B as (again I am using only a portion of a large database just to have some idea) A B 144 what is here in B? Empty space?, 145 146 147 144 How do you know that 144 from B relates to 147 in A? Is it according to its positions? I.e. 4th item in B belongs to 4.th item in A? 148 145 149 147 151 148 Now the branch 144 appears in A as well as in B and in B it is mapped with 147. This means branch 147 comes under the purview of main branch 144. Again 147 is controlling the branch 149 (since 147 also has appeared in B and is mapped with 149 of A). Similarly, branch 145 is controlling branch 148 which further controls operations of bank branch 151 and like wise. Well as you did not say anything about structure of your data A-144:151 B-144:148 data.frame(A,B) A B 1 144 NA 2 145 NA 3 146 NA 4 147 144 5 148 145 6 149 146 7 150 147 8 151 148 DF-data.frame(A,B) main-DF$A[is.na(DF$B)] branch1-DF[!is.na(DF$B),] selected.branch1-branch1$A[branch1$B%in%main] branch2-branch1[!branch1$B%in%main,] selected.branch2-branch2$A[branch2$B%in%selected.branch1] and for cbinding your data which has uneven number of values see Jim Holtman's answer to this How to cbind DF:s with differing number of rows? Regards Petr So in the end I need an output something like - Main Branch Branch office1 Branch office2 144 147 149 145 148 151 146 NA NA ... .. I understand again I am not able to put forward my query properly. But I must thank all of you for giving a patient reading to my query and for reverting back earlier. Thanks once again. With warmest regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Odp: [R] Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 6:39 Hi without other details it is probably impossible to give you any reasonable advice. Do you have your data already in R? What is their form? Are they in 2 columns in data frame? How did you get them paired? So without some more information probably nobody will invest his time as it seems no trivial to me. Regards Petr r-help-boun...@r-project.org napsal dne 24.08.2010 20:28:42: Dear R Helpers, I am a newbie and recently got introduced to R. I have a large database containing the names of bank branch offices along-with other details. I am into Operational Risk as envisaged by BASEL II Accord. I am trying to express my problem and I am using only an indicative data which comes in coded format. A (branch) B (controlled by) 144 145 146 147 144 148 145 149
[R] Merging two data set in R,
Dear R Gurus, I am currently working on the two dataset ( A and B), they both have the same fields:ID , REGION, OFFICE, CSTART, CEND, NCYCLE, STATUS and CB. I want to merge the two data set by ID. The problem I have is that the in data A, the ID's are unique. However in the data set B, the ID's are not unique, thus some repeat themselves. How do I the merge or retrieve the common ones? Please advise. Kind Regards Peter South Africa +27 12 422 7357 +27 82 456 4669 Please Note: This email and its contents are subject to our email legal notice which can be viewed at http://www.sars.gov.za/Email_Disclaimer.pdf [[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] keys in score.items in package psych
Hi Ruru's, score.items in package psych is very handy for scoring test items. It has the structure score.items(keys,items). For instance: score.items(c(1,1,1),data.frame(a1=rep(1,5),a2=rep(1,5),a3=rep(1,5)))$scores correctly gives 1 on each case. But if key -1,0,1 the following happens score.items(c(0.1,0.1,0.1),data.frame(a1=rep(1,5),a2=rep(1,5),a3=rep(1,5)))$scores gives 10 on each case. It appears that the inverse of key is the relative weight of that item. My questions: is this correct? Is this an undocumented feature? Can i trust on it? Frans [[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] Cran-packages-ProfessoR- how to Automatically email a file to an address using the perl program.
Yes - please look at the function IDandEM.R (This stand for IDentification and EMail) in there it has a complete function that uses the autoemail function - I use this all the time. There is a logical option (SEND) on whether to actually mail the file to the recipient. The IDandEM function compares two lists (one from the exam the other from the list that stores the email addresses) and matches these before mailing the scores to the students. I do this to help prevent accidentally mailing students the wrong grades. In the U.S. that would be a very serious breech of protocol. I usually do not run the autoemail directly because of this. -- Jonathan M. Lees Professor THE UNIVERSITY OF NORTH CAROLINA AT CHAPEL HILL UNC-CH Geological Sciences Mitchell Hall 104 South Road CB 3315 Chapel Hill, NC 27599-3315 TEL: (919) 962-1562 FAX: (919) 966-4519 jonathan.l...@unc.edu http://www.unc.edu/~leesj -- View this message in context: http://r.789695.n4.nabble.com/Cran-packages-ProfessoR-how-to-Automatically-email-a-file-to-an-address-using-the-perl-program-tp2335184p2337984.html Sent from the R help mailing list archive at Nabble.com. [[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] using an objects contents in a text string
Hi, I am (trying) to write a script that will execute a set of functions and then to write some of the output data frames to file. I am wanting to be able to just change a single object and have this them populate through the functions etc. I have no trouble making this work when the function is calling for an argument that meets this criteria e.g. input.variable.name - DDDHHD list - function(x, input.variable = input.variable.name, ...) where DDDHHD will be then used as the input.variable in the function BUT I can not seem to get the text string used as part of the name of the output file name in the write.data.frame function. e.g. write(x, file = input.variable.name _data.txt, ...) to create an output file with the name DDDHHD_data.txt I know this syntax is incorrect but could someone guide me as to how to acieve this (I hope it is clear??). Thanks in advance. -- View this message in context: http://r.789695.n4.nabble.com/using-an-objects-contents-in-a-text-string-tp2338061p2338061.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] Repeat the first day data through all the day. Zoo
down vote favorite Hello I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a vector with just one element for each day. cheers -- View this message in context: http://r.789695.n4.nabble.com/Repeat-the-first-day-data-through-all-the-day-Zoo-tp2338069p2338069.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] lmer() causes segfault
Hi: Let's start with the data: str(test.data) 'data.frame': 100 obs. of 4 variables: $ StudentID: num 17370 17370 17370 17370 17379 ... $ GroupID : num 1 1 1 1 1 1 1 1 1 1 ... $ Time : num 1 2 3 4 1 2 3 4 1 2 ... $ Score: num 76.8 81.8 89.8 92.8 75.9 ... Both StudentID and GroupID are numeric; in the model, they would be treated as continuous covariates rather than factors, so we need to convert: test.data$StudentID - factor(test.data$StudentID) test.data$GroupID - factor(test.data$GroupID) Secondly, I believe there are some flaws in your model. After converting your variables to factors, I ran library(lme4) mlmoded1.lmer - lmer(Score ~ Time + (Time | GroupID/StudentID), data = test.data) You have two groups, so they should be treated as a fixed effect - more specifically, as a fixed blocking factor. The StudentIDs are certainly nested within GroupID, and Time is measured on each StudentID, so it is a repeated measures factor. The output of this model is mlmoded1.lmer Linear mixed model fit by REML Formula: Score ~ Time + (Time | GroupID/StudentID) Data: test.data AIC BIC logLik deviance REMLdev 393.1 416.5 -187.5376.9 375.1 Random effects: GroupsNameVariance Std.Dev. Corr StudentID:GroupID (Intercept) 0.504131 0.71002 Time 0.083406 0.28880 1.000 GroupID (Intercept) 12.809567 3.57905 Time 3.897041 1.97409 -1.000 Residual 1.444532 1.20189 Number of obs: 100, groups: StudentID:GroupID, 25; GroupID, 2 Fixed effects: Estimate Std. Error t value (Intercept) 72.803 2.552 28.530 Time 4.474 1.401 3.193 Correlation of Fixed Effects: (Intr) Time -0.994 The high correlations among the random effects and then among the fixed effects suggests that the model specification may be a bit off. The above model fits random slopes to GroupIDs and StudentIDs, along with random intercepts, but GroupID is a between-subject effect and should be at the top level. Time is a within-subject effect and StudentIDs are the observational units. I modified the model to provide fixed effects for GroupIDs, scalar random effects for StudentIDs and random slopes for StudentIDs. mod3 - lmer(Score ~ 1 + GroupID + Time + (1 | StudentID) + + (0 + Time | StudentID), data = test.data) mod3 Linear mixed model fit by REML Formula: Score ~ 1 + GroupID + Time + (1 | StudentID) + (0 + Time | StudentID) Data: test.data AIC BIC logLik deviance REMLdev 430.9 446.5 -209.4418.4 418.9 Random effects: GroupsNameVariance Std.Dev. StudentID (Intercept) 4.2186e-13 6.4951e-07 StudentID Time1.8380e+00 1.3557e+00 Residual 1.6301e+00 1.2768e+00 Number of obs: 100, groups: StudentID, 25 Fixed effects: Estimate Std. Error t value (Intercept) 70.7705 0.4204 168.33 GroupID2 4.0248 0.58546.88 Time 4.5292 0.2942 15.39 Correlation of Fixed Effects: (Intr) GrpID2 GroupID2 -0.668 Time -0.264 0.000 I didn't check the quality of the fit, but on the surface it seems to be more stable, FWIW. Perhaps one could also add a term (GroupID | StudentID), but I don't know offhand if that would make any sense. Another issue to consider is whether to fit by REML or ML, but that is secondary to getting the form of the model equation right. I don't claim this as a final model, but rather a 're-starting point'. It may well be in need of improvement, so comments are welcome. The confusion between subjects nested in time or vice versa has occurred several times this week with respect to repeated measures/longitudinal models using lmer(), so perhaps it merits a comment: subjects/experimental units are NOT nested in time. Measurements taken on an individual at several time points *entails* that time be nested within subject. Just saying... This discussion may be better continued on the R-sig-mixed list, so I've cc-ed to that group as well. HTH, Dennis On Wed, Aug 25, 2010 at 1:27 AM, Bertolt Meyer bme...@sozpsy.uzh.ch wrote: Ben Bolker bbolker at gmail.com writes: Bertolt Meyer bmeyer at sozpsy.uzh.ch writes: Hello lmer() - users, A call to the lmer() function causes my installation of R (2.11.1 on Mac OS X 10.5.8) to crash and I am trying to figure out the problem. [snip snip] detach(package:nlme) library(lme4) mod1 - lmer(performance ~ time + (time | GroupID/StudentNumber), data = dataset.long, na.action = na.omit) However, this call results in a segfault: *** caught segfault *** address 0x154c3000, cause 'memory not mapped' and a lengthy traceback. I can reproduce this error. It also occurs when I don't load nlme before lme4. Can someone tell me what I am doing wrong? Any help is greatly appreciated. This may well be a bug in lmer. There have been a number of fussy computational issues with the lme4 package on the Mac
[R] Surprising behaviour survey-package with missing values
Dear list,
I got some surprising results when using the svytotal routine from the
survey package with data containing missing values.
Some example code demonstrating the behaviour is included below.
I have a stratified sampling design where I want to estimate the total
income. In some strata some of the incomes are missing. I want to
ignore these missing incomes. I would have expected that
svytotal(~income, design=mydesign, na.rm=TRUE) would do the trick.
However, when calculating the estimates 'by hand' the estimates were
different from those obtained from svytotal. The estimated mean
incomes do agree with each other. It seems that using the na.rm option
with svytotal is the same as replacing the missing values with zero's,
which is not what I would have expected, especially since this
behaviour seems to differ from that of svymean. Is there a reason for
this behaviour?
I can of course remove the missing values myself before creating the
survey object. However, with many different variables with different
missing values, this is not very practical. Is there an easy way to
get the behaviour I want?
Thanks for your help.
With regards,
Jan van der Laan
=== EXAMPLE ===
library(survey)
library(plyr)
# generate some data
data - data.frame(
id = 1:20,
stratum = rep(c(a, b), each=10),
income = rnorm(20, 100),
n = rep(c(100, 200), each=10)
)
data$income[5] - NA
# calculate mean and total income for every stratum using survey package
des - svydesign(ids=~id, strata=~stratum, data=data, fpc=~n)
svyby(~income, by=~stratum, FUN=svytotal, design=des, na.rm=TRUE)
mn - svyby(~income, by=~stratum, FUN=svymean, design=des, na.rm=TRUE)
mn
n - svyby(~n, by=~stratum, FUN=svymean, design=des)
# total does not equal mean times number of persons in stratum
mn[2] * n[2]
# calculate mean and total income 'by hand'. This does not give the same total
# as svytotal, but it does give the same mean
ddply(data, .(stratum), function(d) {
data.frame(
mean = mean(d$income, na.rm=TRUE),
n = mean(d$n),
total = mean(d$income, na.rm=TRUE) * mean(d$n)
)
})
# when we set income to 0 for missing cases and repeat the previous estimation
# we get the same answer as svytotal (but not svymean)
data2 - data
data2$income[is.na(data$income )] - 0
ddply(data2, .(stratum), function(d) {
data.frame(
mean = mean(d$income, na.rm=TRUE),
n = mean(d$n),
total = mean(d$income, na.rm=TRUE) * mean(d$n)
)
})
__
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] Merging two data set in R,
First you need to clarify what you'd like to happen when the ID in B is not unique. What do you want the resulting dataframe to look like? Some possible answers involve using different options for merge() or using unique() to remove duplicates from B before merging. But at least to me, merge or retrieve the common ones isn't clear enough to be able to say which. Sarah On Wed, Aug 25, 2010 at 5:35 AM, Mangalani Peter Makananisa pmakanan...@sars.gov.za wrote: Dear R Gurus, I am currently working on the two dataset ( A and B), they both have the same fields: ID , REGION, OFFICE, CSTART, CEND, NCYCLE, STATUS and CB. I want to merge the two data set by ID. The problem I have is that the in data A, the ID's are unique. However in the data set B, the ID's are not unique, thus some repeat themselves. How do I the merge or retrieve the common ones? Please advise. Kind Regards Peter -- 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] Repeat the first day data through all the day. Zoo
On Wed, Aug 25, 2010 at 7:43 AM, skan juanp...@gmail.com wrote: I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a vector with just one element for each day. Try ave: library(zoo) library(chron) zz - z - zoo(1:100, chron(0:9/5)) zz[] - ave(coredata(z), as.Date(time(z)), FUN = function(x) head(x, 1)) cbind(z, zz) z zz (01/01/70 00:00:00) 1 1 (01/01/70 04:48:00) 2 1 (01/01/70 09:36:00) 3 1 (01/01/70 14:24:00) 4 1 (01/01/70 19:12:00) 5 1 (01/02/70 00:00:00) 6 6 (01/02/70 04:48:00) 7 6 (01/02/70 09:36:00) 8 6 (01/02/70 14:24:00) 9 6 (01/02/70 19:12:00) 10 6 __ 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] multi day R run crashed - is there a log
Oh dear. No there isn't. I do a lot of very long runs, and have learned to write out intermediate steps. Depending on what you are doing, saving the RData file periodically may be appropriate, or writing out a csv file of results so far, or even just printing something to the output file (if you are using batch). There may well be something more elegant, but these solutions have all worked well for me in various situations. Sarah On Wed, Aug 25, 2010 at 3:46 AM, Martin Tomko martin.to...@geo.uzh.ch wrote: Dear all, I am using an R 2.10 installation on a Windows 203 server that I have no control over. After a multi-day run I found that it was terminated/crashed. Is there any log kept by R where I could see whether something/what happened? The same process has been run beofre on a smaller dataset (also at least a day of computing) without problems. Thanks Martin -- 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] Repeat the first day data through all the day. Zoo
On Wed, Aug 25, 2010 at 7:56 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Wed, Aug 25, 2010 at 7:43 AM, skan juanp...@gmail.com wrote: I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a vector with just one element for each day. Try ave: library(zoo) library(chron) zz - z - zoo(1:100, chron(0:9/5)) That should have been 10, not 100; however, it ignored 11:100 so the answer is the same. zz[] - ave(coredata(z), as.Date(time(z)), FUN = function(x) head(x, 1)) cbind(z, zz) z zz (01/01/70 00:00:00) 1 1 (01/01/70 04:48:00) 2 1 (01/01/70 09:36:00) 3 1 (01/01/70 14:24:00) 4 1 (01/01/70 19:12:00) 5 1 (01/02/70 00:00:00) 6 6 (01/02/70 04:48:00) 7 6 (01/02/70 09:36:00) 8 6 (01/02/70 14:24:00) 9 6 (01/02/70 19:12:00) 10 6 __ 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] (no subject)
Hello, http://www.rseek.org is extremely useful if you know what you want to do but not how to do it. Going there and putting quantile regression with multilevel data into the search box returns many references to functions and packages that may be of use. This list can't do much more for you without a clearer statement of your problem, as requested by the posting guide linked at the bottom of this message. Also, for future reference: I suspect many people on this list just delete messages with no subject. You will receive faster and more informed advice if you use a meaningful subject line. On Wed, Aug 25, 2010 at 5:32 AM, Mohammad Ali Vakili mavak...@yahoo.com wrote: Hi I am using repeated meaturement data for my project and I want to use quantile regression with multilevel or panel data in R. I dont find the basic version of software in R, so I have difficulty in using it. I would also appreciate if anyone more proficient in R could help me how to run this. best wishes M.A-Vakili -- 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] using an objects contents in a text string
ALSO I have had a play with cat() but have also not got this to work e.g. write(x, file = cat(input.variable.name , file = , sep = _data.txt, ), ...) but this does not seem to work and I'm sure it is not the correct use of cat() Thanks. -- View this message in context: http://r.789695.n4.nabble.com/using-an-objects-contents-in-a-text-string-tp2338061p2338106.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] Merging two data set in R,
What do you want to happen when there are duplicates? A: ID X 1 a 2 b 3 c B: ID Y 1 x 2 y 2 z What happens to ID 1? 2? 3? in your desired output? The all.x and all.y options might be of use. Sarah On Wed, Aug 25, 2010 at 8:00 AM, Mangalani Peter Makananisa pmakanan...@sars.gov.za wrote: I want to merge data set A and B, by merge(A,B, by = ID), however I am getting error massages, because the some ID's in A repeat themselves several time in data set B. Even if the ID's in B repeat themselves I want to be able to merge the two dataset and retrieve the intersection. Please help. -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: 25 August 2010 01:52 PM To: Mangalani Peter Makananisa Cc: r-help@r-project.org Subject: Re: [R] Merging two data set in R, First you need to clarify what you'd like to happen when the ID in B is not unique. What do you want the resulting dataframe to look like? Some possible answers involve using different options for merge() or using unique() to remove duplicates from B before merging. But at least to me, merge or retrieve the common ones isn't clear enough to be able to say which. Sarah On Wed, Aug 25, 2010 at 5:35 AM, Mangalani Peter Makananisa pmakanan...@sars.gov.za wrote: Dear R Gurus, I am currently working on the two dataset ( A and B), they both have the same fields: ID , REGION, OFFICE, CSTART, CEND, NCYCLE, STATUS and CB. I want to merge the two data set by ID. The problem I have is that the in data A, the ID's are unique. However in the data set B, the ID's are not unique, thus some repeat themselves. How do I the merge or retrieve the common ones? Please advise. Kind Regards Peter -- 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] using an objects contents in a text string
I'm a bit confused by your question, but you might just want paste: paste(input.variable.name, data.txt, sep=_) Sarah On Wed, Aug 25, 2010 at 8:06 AM, josquint josqu...@unimelb.edu.au wrote: ALSO I have had a play with cat() but have also not got this to work e.g. write(x, file = cat(input.variable.name , file = , sep = _data.txt, ), ...) but this does not seem to work and I'm sure it is not the correct use of cat() Thanks. -- -- 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] multi day R run crashed - is there a log
Hi Sarah, thank you very much for your answer. I have been spitting things out on screen, but unfortunately I have not run it as a batch log, but in the interactive window, so when the GUI crashed I was left without trace... I guess I should google how to run it from a batch. I should also explore using RData, I have ony been using csv files and tables so far, it seems that can also bring some added performance. As the main output of my process is a matrix, I would really need to append to a matrix after each iteration. I have identified the write.table append parameter-based solution, but that would only append rows. Is there a way to slowly grow a matrix in both directions, meaning append columns as well (it is a big distance matrix). Thanks a lot for your help, Cheers Martin On 8/25/2010 1:56 PM, Sarah Goslee wrote: Oh dear. No there isn't. I do a lot of very long runs, and have learned to write out intermediate steps. Depending on what you are doing, saving the RData file periodically may be appropriate, or writing out a csv file of results so far, or even just printing something to the output file (if you are using batch). There may well be something more elegant, but these solutions have all worked well for me in various situations. Sarah On Wed, Aug 25, 2010 at 3:46 AM, Martin Tomkomartin.to...@geo.uzh.ch wrote: Dear all, I am using an R 2.10 installation on a Windows 203 server that I have no control over. After a multi-day run I found that it was terminated/crashed. Is there any log kept by R where I could see whether something/what happened? The same process has been run beofre on a smaller dataset (also at least a day of computing) without problems. Thanks Martin -- Martin Tomko Postdoctoral Research Assistant Geographic Information Systems Division Department of Geography University of Zurich - Irchel Winterthurerstr. 190 CH-8057 Zurich, Switzerland email: martin.to...@geo.uzh.ch site: http://www.geo.uzh.ch/~mtomko mob:+41-788 629 558 tel:+41-44-6355256 fax:+41-44-6356848 __ 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] multi day R run crashed - is there a log
Hi, On Wed, Aug 25, 2010 at 8:11 AM, Martin Tomko martin.to...@geo.uzh.ch wrote: Hi Sarah, thank you very much for your answer. I have been spitting things out on screen, but unfortunately I have not run it as a batch log, but in the interactive window, so when the GUI crashed I was left without trace... I guess I should google how to run it from a batch. I have no idea how to do that in Windows, but I'm sure it's possible. :) That way the things written to the screen will be saved in a file instead. I should also explore using RData, I have ony been using csv files and tables so far, it seems that can also bring some added performance. If you need to save *everything*, RData is what you get when you use save(), or when you close a session and choose y to saving the data. It can be read in using load(). That's one way to be able to pick up where you left off. As the main output of my process is a matrix, I would really need to append to a matrix after each iteration. I have identified the write.table append parameter-based solution, but that would only append rows. Is there a way to slowly grow a matrix in both directions, meaning append columns as well (it is a big distance matrix). AFAIK, you can't append columns that way because of the way text files are written to disk. You'd need to rewrite the whole thing, or possibly write it out in lower triangular format with NA values as padding (assuming it's a symmetric distance). Or for that matter, you could just write it out as a really long vector, and turn it back into a matrix later if you need to read the saved file in after a crash. I'd recommend saving whatever variables are needed so that you can pick up exactly where you left off, if possible. Much nicer to pick up 12 hours in than to start over from the beginning. Not R, but I just finished a 5-week batch job. You can bet that I put a lot of thought into incremental save points and how to resume after an unexpected halt! 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.
[R] Estimate average standard deviation of mean of two dependent groups
Dear R-experts! I am currently running a meta-analysis with the help of the great metafor package. However I have some difficulties setting up my raw data to enter it to the meta-analysis models. I have a group of subjects that have been measured in two continuous variables (A B). I have the information about the mean of the two variables, the group size (n) and the standard deviations of the two variables. Now I would like to average both variables (A B) and get the mean and standard deviation of the merged variable (C). As for the mean this would be quiet easy: I would just take the mean of mean A and mean B to get the mean of C. However for the standard deviation this seems more tricky as it is to assume that standard deviations in A B correlate. I assume (based on further analysis) a correlation of r =0.5. I found the formula to get the standard deviation of the SUM (not the mean) of two variables: SD=SQRT(SD_A^2 + SD_B^2 + 2*r*SD_A*SD_B) with SD_B and SD_B being the standard deviation of A and B. And r*SD_A*SD_B being the covariance of A and B. Would this formula also be valid if I want to average (and not sum) my two variables? Many thanks for any help best wishes, Jokel [[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] Repeat the first day data through all the day. Zoo
On Wed, Aug 25, 2010 at 7:43 AM, skan juanp...@gmail.com wrote: down vote favorite Hello I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a vector with just one element for each day. cheers Here are a few more solutions too: library(zoo) library(chron) z - zoo(1:10, chron(0:9/5)) # aggregate / na.locf z.ag - aggregate(z, as.Date, head, 1) na.locf(z.ag, xout = time(z)) # duplicated / na.locf z.na - ifelse.zoo(!duplicated(as.Date(time(z))), z, NA) na.locf(z.na) # ave - as before zz - z zz[] - ave(coredata(z), as.Date(time(z)), FUN = function(x) head(x, 1)) zz __ 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] Merging two data set in R,
Almost. You'll need to handle the duplicate ID yourself since R has
no way of knowing which one(s) to change to NA. As I already suggested, you
can use unique() in conjunction with whatever logical rules you require
for choosing those values.
As I also already suggested, all.y and all.x are the options to merge()
that you need to consider.
A - data.frame(ID = c(1,2,3), X = c('a','b','c'))
B - data.frame(ID = c(1,2,2), Y = c('x','y','z'))
merge(A, B, all.x=FALSE, all.y=TRUE)
ID X Y
1 1 a x
2 2 b y
3 2 b z
Just think how much easier this process would have been if you had
provided a clear question with toy data and examples of what you'd
tried in your first question.
Sarah
On Wed, Aug 25, 2010 at 8:24 AM, Mangalani Peter Makananisa
pmakanan...@sars.gov.za wrote:
A:
ID X
1 a
2 b
3 c
B:
ID Y
1 x
2 y
2 z
I would like to see something like this:
Common = Merge(A,B)
Common
ID X Y
1 a x
2 b y
2 N/A z
If it is possible,
--
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] modify a nomogram
Update to the rms package which is the version now being actively
supported. New features will not be added to Design. The nomogram
function in rms separates the plotting into a plot method for easier
understanding. You can control all axes - some experimentation can
tell you if you can do what you want as I haven't tried exactly what
you are doing.
Frank
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Wed, 25 Aug 2010, david dav wrote:
Hi,
I would like to emphasize (zoom) the zone of a nomogram where the
probability are 0.01
(nomogram built with nomogram, Design).
As a consequence, I don't need to draw the part of the Total points
axis with score 60 equivalent in my case to a linear predictor 4.5
- As far as I know, this is not possible with the arguments of the function.
- Changing the code of the function is beyond my abilities
-- can not even create a nomo.f function with the same body:
body(nomo) - expression({
conf.lp - match.arg(conf.lp) . rest of the function body
this does not even work
-- I am not sure to find the piece of code responsible for
defining the axis
nomogram(logistic.fit, fun = plogis, fun.at = c(seq(0.01,1,by=.2)),
lmgp=.2, lp = T,maxscale = 100, total.sep.page = T )
Thanks
David
__
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] Plot bar lines like excel
On Wed, Aug 25, 2010 at 6:05 AM, abotaha yaseen0...@gmail.com wrote: Woow, it is amazing, thank you very much. yes i forget to attach the dates, however, the dates in my case is every 16 days. so how i can use 16 day interval instead of month in by option. Here's one way using the lubridate package: library(lubridate) today() + days(16) * seq_len(50) ymd(2008-01-01) + days(16) * seq_len(50) -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ 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] multi day R run crashed - is there a log
Hi , thanks, the lower.tri idea is I guess the best way. Will try that. Cheers Martin On 8/25/2010 2:21 PM, Sarah Goslee wrote: Hi, On Wed, Aug 25, 2010 at 8:11 AM, Martin Tomkomartin.to...@geo.uzh.ch wrote: Hi Sarah, thank you very much for your answer. I have been spitting things out on screen, but unfortunately I have not run it as a batch log, but in the interactive window, so when the GUI crashed I was left without trace... I guess I should google how to run it from a batch. I have no idea how to do that in Windows, but I'm sure it's possible. :) That way the things written to the screen will be saved in a file instead. I should also explore using RData, I have ony been using csv files and tables so far, it seems that can also bring some added performance. If you need to save *everything*, RData is what you get when you use save(), or when you close a session and choose y to saving the data. It can be read in using load(). That's one way to be able to pick up where you left off. As the main output of my process is a matrix, I would really need to append to a matrix after each iteration. I have identified the write.table append parameter-based solution, but that would only append rows. Is there a way to slowly grow a matrix in both directions, meaning append columns as well (it is a big distance matrix). AFAIK, you can't append columns that way because of the way text files are written to disk. You'd need to rewrite the whole thing, or possibly write it out in lower triangular format with NA values as padding (assuming it's a symmetric distance). Or for that matter, you could just write it out as a really long vector, and turn it back into a matrix later if you need to read the saved file in after a crash. I'd recommend saving whatever variables are needed so that you can pick up exactly where you left off, if possible. Much nicer to pick up 12 hours in than to start over from the beginning. Not R, but I just finished a 5-week batch job. You can bet that I put a lot of thought into incremental save points and how to resume after an unexpected halt! Sarah -- Martin Tomko Postdoctoral Research Assistant Geographic Information Systems Division Department of Geography University of Zurich - Irchel Winterthurerstr. 190 CH-8057 Zurich, Switzerland email: martin.to...@geo.uzh.ch site: http://www.geo.uzh.ch/~mtomko mob:+41-788 629 558 tel:+41-44-6355256 fax:+41-44-6356848 __ 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] multiple x factors using the sciplot package
Dear R community, I am a beginner using the sciplot package to graph barplots. I would like to be able to graph two x factors (Sampling.year and Period). Sampling.year: 2006, 2007, 2008, 2009 Period: First, Second, Total The parameter group is the different species I looked at. They can be seen in the legend. The parameter response is the Average percentage cover category of these species. I would like the graph so you see on the x axis the years (in a bigger font) and the period (in smaller font). It would look a bit like the barplot that can be seen at http://onertipaday.blogspot.com/2007/05/make-many-barplot-into-one-plot.html, but with the advantage of the sciplot package. Thanks a lot for your help! Jordan [[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] percentage sign in expression
On Aug 25, 2010, at 4:32 AM, e-letter wrote:
On 24/08/2010, David Winsemius dwinsem...@comcast.net wrote:
On Aug 24, 2010, at 9:37 AM, e-letter wrote:
Readers,
According to the documentation for the function 'plotmath' there
is no
apparent possibility to add the percent sign (%) to a plot function,
Where did you see an assertion made???
Within R I entered the command:
?plotmath
Also accessed using:
help.start(browser=opera)
Navigating the web browser page:
packages
packages in /usr/lib/R/library
grdevices
plotmath
In the list headed 'syntax' and 'meaning' within the section
'details'.
e.g.
plot(a[,1]~b[,2],ylab=expression(x~%),xlab=expression(z))
How to achieve this please?
Read the plotmath helo page more carefully. The section immediatedly
below the plotmath expressions points you to the use of the symbol()
expression-function and to the points help page where generation of
the available glyphs proceeds according to the advice on
help(plotmath):
In my system the paragraph immediately after the list of features
(i.e. 'syntax','meaning') describes a note to TeX users. I cannot see
reference to 'symbol()'.
It's possible that my help page is different than yours. Right after
the syntax/meaning description on mine (which is a Mac OSX system) is
a paragraph:
The symbol font uses Adobe Symbol encoding so, for example, a lower
case mu can be obtained either by the special symbol mu or by
symbol(m). This provides access to symbols that have no special
symbol name, for example, the universal, or forall, symbol is
symbol(\042). To see what symbols are available in this way
useTestChars(font=5) as given in the examples for points: some are
only available on some devices.
(In this case I would be surprised if the help pages were different
because this makes a cross-reference to the examples in points. I am
not surprised about cross-platform differences in descriptions of
graphical devices and would have included a caveat if I were
corresponding on rhelp about such. I suppose the font issues could be
platform specific so if you want to correct me on this point, I will
try to file it away. I did, however, give you the code needed to to
display Symbols and it sounds further on that it succeeded)
TestChars - function(sign=1, font=1, ...)
+ {
+if(font == 5) { sign - 1; r - c(32:126, 160:254)
+} else if (l10n_info()$MBCS) r - 32:126 else r - 32:255
+if (sign == -1) r - c(32:126, 160:255)
+par(pty=s)
+plot(c(-1,16), c(-1,16), type=n, xlab=, ylab=,
+ xaxs=i, yaxs=i)
+grid(17, 17, lty=1)
+for(i in r) try(points(i%%16, i%/%16, pch=sign*i,
font=font,...))
+ }
TestChars(font=5)
Notice that the % sign is three characters to the right (i.e.
higher) of the forall symbol that is produced by the example code
I can't see 'forall' in the code above.
Gavin has already explained why you did not. The upside-down A (==
universal or forall) was a useful reference point in the indexing,
since it is only 3 glyphs away for the % symbol.
they offer. (The numbering proceeds from bottom up which confused me
at first.)
What numbering?
Actually I did not see any numbering either, which was why I remained
confused about the location of the % symbol for several minutes.
Perhaps I should have used the term indexing.
The documentation makes reference to the command:
demo(plotmath)
I applied this command and could not see an instruction to produce the
percent (%) symbol.
I don't think I suggested it would.
--
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] Repeat the first day data through all the day. Zoo
thanks I'll try them, Why do you use the brackets in zz[] ? -- View this message in context: http://r.789695.n4.nabble.com/Repeat-the-first-day-data-through-all-the-day-Zoo-tp2338069p2338266.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 do R calculates the number of intervals between tick-marks
On Aug 24, 2010, at 7:05 PM, Antonio Olinto wrote: Hello, I want to know how do R calculates the number of intervals between tick-marks in the y axis in a plot. ?axTicks # and then look at the other citations and the code as needed I'm making a three y-axes plot and this information would help me a lot. Thanks in advance. Antonio Olinto -- 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] Comparing samples with widely different uncertainties
Hi
This is probably more of a statistics question than a specific R
question, although I will be using R and need to know how to solve the
problem in R.
I have several sets of data (ejection fraction measurements) taken in
various ways from the same set of (~400) patients (so it is paired data).
For each individual measurement I can make an estimate of the percentage
uncertainty in the measurement.
Generally the measurements in data set A are higher but they have a
large uncertainty (~20%) while the measurements in data set Bare lower
but have a small uncertainty (~4%).
I believe, from the physiology, that the true value is likely to be
nearer the value of A than of B.
I need to show that, despite the uncertainties in the measurements
(which are not themselves normally distributed), there is (or is not) a
difference between the two groups, (a straight Wilcoxon signed ranks
test shows a difference but it cannot include that uncertainty data).
Can anybody suggest what I should be looking at? Is there a language
here that I don't know? How do I do it in R?
Many thanks for your help
Sandy
--
Sandy Small
Clinical Physicist
NHS Greater Glasgow and Clyde
and
NHS Forth Valley
Phone: 01412114592
E-mail: sandy.sm...@nhs.net
This message may contain confidential information. If yo...{{dropped:21}}
__
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] Repeat the first day data through all the day. Zoo
On Wed, Aug 25, 2010 at 9:48 AM, skan juanp...@gmail.com wrote: thanks I'll try them, Why do you use the brackets in zz[] ? So it stays a zoo object with the same index. We are only replacing the data part. __ 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 do R calculates the number of intervals between tick-marks
Take a look at pretty function. On Tue, Aug 24, 2010 at 8:05 PM, Antonio Olinto aolint...@bignet.com.brwrote: Hello, I want to know how do R calculates the number of intervals between tick-marks in the y axis in a plot. I'm making a three y-axes plot and this information would help me a lot. Thanks in advance. Antonio Olinto Webmail - iBCMG Internet http://www.ibcmg.com.br __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] Odp: Finding pairs
Hi well, I will add some explanation r-help-boun...@r-project.org napsal dne 25.08.2010 11:24:38: Dear Mr Petr PIKAL After reading the R code provided by you, I realized that I would have never figured out how this could have been done. I am going to re-read again and again your code to understand the logic and the commands you have provided. Thanks again from the heart for your kind advice. Regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Re: [R] Odp: Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 9:01 Hm r-help-boun...@r-project.org napsal dne 25.08.2010 09:43:26: Dear Mr Petr Pikal I am extremely sorry for the manner I have raised the query. Actually that was my first post to this R forum and in fact even I was also bit confused while drafting the query, for which I really owe sorry to all for consuming the precious time. Perhaps I will try to redraft my query in a better way as follows. I have two datasets A and B containing the names of branch offices of a particular bank say XYZ plc bank. The XYZ bank has number of main branch offices (say Parent) and some small branch offices falling under the purview of these main branch offices (say Child). The datalist A and B consists of these main branch office names as well as small branch office names. B is subset of A and these branch names are coded. Thus we have two datasets A and B as (again I am using only a portion of a large database just to have some idea) A B 144 what is here in B? Empty space?, 145 146 147 144 How do you know that 144 from B relates to 147 in A? Is it according to its positions? I.e. 4th item in B belongs to 4.th item in A? 148 145 149 147 151 148 Now the branch 144 appears in A as well as in B and in B it is mapped with 147. This means branch 147 comes under the purview of main branch 144. Again 147 is controlling the branch 149 (since 147 also has appeared in B and is mapped with 149 of A). Similarly, branch 145 is controlling branch 148 which further controls operations of bank branch 151 and like wise. Well as you did not say anything about structure of your data A-144:151 B-144:148 data.frame(A,B) A B 1 144 NA 2 145 NA 3 146 NA 4 147 144 5 148 145 6 149 146 7 150 147 8 151 148 DF-data.frame(A,B) This was just making a data frame with 2 columns to have some data to play with main-DF$A[is.na(DF$B)] Above are codes from A which are NA in B branch1-DF[!is.na(DF$B),] Above is data frame of remaining codes (other than main) selected.branch1-branch1$A[branch1$B%in%main] Above is codes from column A for which B column and main are the same branch2-branch1[!branch1$B%in%main,] This is the rest of yet not selected rows selected.branch2-branch2$A[branch2$B%in%selected.branch1] and this is selection of values from column A for which B column and selected.branch1 values are same. But it works for this particular data, I am not sure how it behaves with duplicates and further issues. It also depends on how your data is organised. And if you are in reading you could also go through setdiff, merge and maybe sqldf package and Rdata Import/export manual Regards Petr and for cbinding your data which has uneven number of values see Jim Holtman's answer to this How to cbind DF:s with differing number of rows? Regards Petr So in the end I need an output something like - Main Branch Branch office1 Branch office2 144 147 149 145 148 151 146 NA NA ... .. I understand again I am not able to put forward my query properly. But I must thank all of you for giving a patient reading to my query and for reverting back earlier. Thanks once again. With warmest regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Odp: [R] Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 6:39 Hi without other details it is probably impossible to give you any reasonable advice. Do you
[R] approxfun-problems (yleft and yright ignored)
Dear all,
I have run into a problem when running some code implemented in the
Bioconductor panp-package (applied to my own expression data), whereby gene
expression values of known true negative probesets (x) are interpolated onto
present/absent p-values (y) between 0 and 1 using the *approxfun -
function*{stats}; when I have used R version 2.8, everything had
worked fine,
however, after updating to R 2.11.1., I got unexpected output (explained
below).
Please correct me here, but as far as I understand, the yleft and yright
arguments set the extreme values of the interpolated y-values in case the
input x-values (on whose approxfun is applied) fall outside range(x). So if
I run approxfun with yleft=1 and yright=0 with y-values between 0 and 1,
then I should never get any values higher than 1. However, this is not the
case, as this code-example illustrates:
### define the x-values used to construct the approxfun, basically these
are 2000 expression values ranging from ~ 3 to 7:
xNeg - NegExprs[, 1]
xNeg - sort(xNeg, decreasing = TRUE)
### generate 2000 y-values between 0 and 1:
yNeg - seq(0, 1, 1/(length(xNeg) - 1))
### define yleft and yright as well as the rule to clarify what should
happen if input x-values lie outside range(x):
interp - approxfun(xNeg, yNeg, yleft = 1, yright = 0, rule=2)
Warning message:
In approxfun(xNeg, yNeg, yleft = 1, yright = 0, rule = 2) :
collapsing to unique 'x' values
### apply the approxfun to expression data that range from ~2.9 to 13.9
and can therefore lie outside range(xNeg):
PV - sapply(AllExprs[, 1], interp)
range(PV)
[1]0.000 6208.932
summary(PV)
Min. 1st Qu.Median Mean 3rd Qu. Max.
0.000e+00 0.000e+00 2.774e-03 1.299e+00 3.164e-01 6.209e+03
So the resulting output PV object contains data ranging from 0 to 6208, the
latter of which lies outside yleft and is not anywhere close to extreme
y-values that were used to set up the interp-function. This seems wrong to
me, and from what I understand, yleft and yright are simply ignored?
I have attached a few histograms that visualize the data distributions of
the objects I xNeg, yNeg, AllExprs[,1] (== input x-values) and PV (the
output), so that it is easier to make sense of the data structures...
Does anyone have an explanation for this or can tell me how to fix the
problem?
Thanks a million for any help, best, Sam
sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
locale:
[1] en_IE.UTF-8/en_IE.UTF-8/C/C/en_IE.UTF-8/en_IE.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] panp_1.18.0 affy_1.26.1 Biobase_2.8.0
loaded via a namespace (and not attached):
[1] affyio_1.16.0 preprocessCore_1.10.0
--
-
Samuel Wuest
Smurfit Institute of Genetics
Trinity College Dublin
Dublin 2, Ireland
Phone: +353-1-896 2444
Web: http://www.tcd.ie/Genetics/wellmer-2/index.html
Email: wue...@tcd.ie
--
__
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] Comparing samples with widely different uncertainties
On Aug 25, 2010, at 3:57 PM, Sandy Small wrote: Hi This is probably more of a statistics question than a specific R question, although I will be using R and need to know how to solve the problem in R. I have several sets of data (ejection fraction measurements) taken in various ways from the same set of (~400) patients (so it is paired data). For each individual measurement I can make an estimate of the percentage uncertainty in the measurement. Generally the measurements in data set A are higher but they have a large uncertainty (~20%) while the measurements in data set Bare lower but have a small uncertainty (~4%). I believe, from the physiology, that the true value is likely to be nearer the value of A than of B. I need to show that, despite the uncertainties in the measurements (which are not themselves normally distributed), there is (or is not) a difference between the two groups, (a straight Wilcoxon signed ranks test shows a difference but it cannot include that uncertainty data). Can anybody suggest what I should be looking at? Is there a language here that I don't know? How do I do it in R? Many thanks for your help Sandy Hm, well... I don't think the issue is entirely well-defined, but let me try and give you some pointers: For bivariate normal data (X,Y), the situation is that even if V(X) != V(Y) you still end up looking at X-Y if the question is whether the means are the same. It's sort of the only thing you _can_ do... For non-normal data, it is not clear what the null hypothesis really is. The signed-rank test assumes that X-Y has a symmetric distribution, which is dubious if X is not symmetric and its variation dominates that of Y. You could also do a sign test and see if the differences has a median of zero (bear in mind that the median of a difference is different from the difference of the medians, but it could actually suffice.) I'd probably start off with a simple plot of Y vs X and look for fan-out effects indicating that the variance depends on the mean. If it does, perhaps take logs or square roots and see if it makes the variances appear more stable and perhaps improve on the normality. Then maybe just do a paired t-test. -- Peter Dalgaard 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.
Re: [R] Comparing samples with widely different uncertainties
On Aug 25, 2010, at 8:57 AM, Sandy Small wrote: Hi This is probably more of a statistics question than a specific R question, although I will be using R and need to know how to solve the problem in R. I have several sets of data (ejection fraction measurements) taken in various ways from the same set of (~400) patients (so it is paired data). For each individual measurement I can make an estimate of the percentage uncertainty in the measurement. Generally the measurements in data set A are higher but they have a large uncertainty (~20%) while the measurements in data set Bare lower but have a small uncertainty (~4%). I believe, from the physiology, that the true value is likely to be nearer the value of A than of B. I need to show that, despite the uncertainties in the measurements (which are not themselves normally distributed), there is (or is not) a difference between the two groups, (a straight Wilcoxon signed ranks test shows a difference but it cannot include that uncertainty data). Can anybody suggest what I should be looking at? Is there a language here that I don't know? How do I do it in R? Many thanks for your help Sandy The first place that I would start is at Martin Bland's page pertaining to the design and analysis of measurement studies: http://www-users.york.ac.uk/~mb55/meas/meas.htm The papers he co-authored with Doug Altman are the go to resource for this domain. 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] SEM : Warning : Could not compute QR decomposition of Hessian
Hi useRs,
I'm trying for the first time to use a sem. The model finally runs,
but gives a warning saying :
In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names
= vars, : Could not compute QR decomposition of Hessian.
Optimization probably did not converge.
I found in R-help some posts on this warning, but my attemps to modify
the code didn't change the warning message (i tried to give an error
of 1 to the latente variables). I can't figure what the problem is.
Here is the code :
tab-read.table(F:/Mes documents/stats/sem/donnees_corr.txt,
header=T, sep=,na.strings = NA)
tab[,46]-as.factor(tab[,46])
tab[,24]-as.factor(tab[,24])
tab[,40]-as.factor(tab[,40])
fct_cor-hetcor(tab, ML=T)
cor_tab- fct_cor$correlations
moment_tab-read.moments(diag=F, names=c('c1','c2', 'c3','c4','c5',
'c6','c7', 'c8', 'c9', 'ind_plando', 'long_sup15', 'long_inf15',
'pente', 'est', 'sud','ouest', 'nord' ,'reg_hydriq', 'prof_sol',
'pierro', 'efferv', 'struct','drainage','texture', 'route1_pond',
'route2_pond',
'pourcactif', 'tx_chomage', 'pourcagric', 'pourc_jeunes', 'pop99',
'rev_imp_foyer','eq_CONC', 'eq_sante', 'eq_edu', 'sold_nat',
'sold_mig', 'tps_dom_emp','TXEMPLOI','ORIECO','dist_paris','axe1',
'axe2', 'axe3', 'dist_protect','urbanisation','pays_incli','pays_alti'))
# after comes the moment matrix (triangular)
ram_tab-specify.model()
type_paysage-pays_alti,NA,1
type_paysage-pays_incli, pays2, NA
pedo-reg_hydriq, NA, 1
pedo-prof_sol, ped8, NA
pedo-pierro, ped9, NA
pedo-efferv, ped10, NA
pedo-struct, ped11, NA
pedo-drainage, ped12, NA
pedo-texture, ped13, NA
adj_99-c1, NA,1
adj_99-c2, adj2,NA
adj_99-c3, adj3,NA
adj_99-c4, adj4,NA
adj_99-c5, adj5,NA
adj_99-c6, adj6,NA
adj_99-c7, adj7,NA
adj_99-c8, adj8,NA
adj_99-c9, adj9,NA
etat_hexa-axe1, NA, 1
etat_hexa-axe2, et2, NA
etat_hexa-axe3, et3, NA
socioBV-sold_mig, BV1, NA
socioBV-sold_nat, BV2, NA
socioBV-TXEMPLOI, BV3, NA
socioBV-ORIECO, BV4, NA
socioBV-tps_dom_emp, NA, 1
eqBV-eq_CONC, NA, 1
eqBV-eq_sante, eq2, NA
eqBV-eq_edu, eq3, NA
socio_com-pourcactif , NA, 1
socio_com-tx_chomage, com2, NA
socio_com-pourcagric, com3, NA
socio_com-pourc_jeunes, com4, NA
socio_com-pop99, com5, NA
socio_com-rev_imp_foyer, com7, NA
access_hexa-route1_pond, NA, 1
access_hexa-route2_pond, acc2, NA
hydro-ind_plando, NA, 1
hydro-long_sup15, eau2, NA
hydro-long_inf15, eau3, NA
topog-pente, NA, 1
topog-est, top2, NA
topog-sud, top3, NA
topog-nord, top4, NA
topog-ouest, top5, NA
dist_protect- urbanisation, cor1,NA
dist_protect- adj_99, cor2, NA
dist_protect- etat_hexa, cor3, NA
topog- urbanisation, cor4, NA
topog- adj_99, cor5, NA
topog- etat_hexa, cor6, NA
topog- access_hexa, cor7, NA
topog-hydro, cor8, NA
topog-pedo, cor9, NA
pedo- urbanisation, cor10, NA
pedo- adj_99, cor11, NA
pedo- etat_hexa, cor12, NA
pedo-hydro, cor1, NA
hydro- urbanisation, cor13, NA
hydro- adj_99, cor14, NA
hydro- etat_hexa, cor15, NA
access_hexa- urbanisation, cor16, NA
access_hexa- etat_hexa, cor17, NA
socio_com- etat_hexa, cor18, NA
socio_com- adj_99, cor19, NA
socio_com- urbanisation, cor20, NA
dist_paris- socio_com, cor21, NA
dist_paris- access_hexa, cor22, NA
dist_paris- adj_99, cor23, NA
dist_paris- etat_hexa, cor24, NA
dist_paris- urbanisation, cor25, NA
dist_paris- socioBV, cor26, NA
socioBV- eqBV, cor27, NA
socioBV- urbanisation, cor28, NA
socioBV- adj_99, cor29, NA
socioBV- etat_hexa, cor30, NA
eqBV- etat_hexa, cor31, NA
eqBV- adj_99, cor32, NA
eqBV- urbanisation, cor33, NA
etat_hexa- urbanisation, cor34, NA
etat_hexa- adj_99, cor35, NA
adj_99- urbanisation, cor36, NA
type_paysage- urbanisation, cor37, NA
type_paysage- adj_99, cor38, NA
type_paysage- etat_hexa, cor39, NA
dist_paris-dist_paris, auto1, NA
dist_protect-dist_protect, auto2, NA
c1 - c1, auto4, NA
c2 - c2 , auto5, NA
c3 - c3 , auto6, NA
c4 - c4 , auto7, NA
c5 - c5 , auto8, NA
c6 - c6 , auto9, NA
c7 - c7 , auto10, NA
c8 - c8 , auto11, NA
c9 - c9 , auto12, NA
ind_plando - ind_plando, auto13, NA
long_sup15 - long_sup15 , auto14, NA
long_inf15 - long_inf15 , auto15, NA
pente - pente , auto16, NA
est- est , auto17, NA
sud - sud , auto18, NA
ouest- ouest , auto19, NA
nord - nord , auto20, NA
reg_hydriq - reg_hydriq , auto21, NA
prof_sol- prof_sol , auto22, NA
pierro - pierro , auto23, NA
efferv - efferv , auto24, NA
struct - struct , auto25, NA
drainage - drainage, auto26, NA
texture - texture , auto27, NA
route1_pond -route1_pond , auto30, NA
route2_pond - route2_pond , auto31, NA
pourcactif - pourcactif , auto32, NA
tx_chomage - tx_chomage, auto33, NA
pourcagric - pourcagric , auto34, NA
pourc_jeunes- pourc_jeunes , auto36, NA
pop99- pop99 , auto36, NA
rev_imp_foyer - rev_imp_foyer , auto38, NA
eq_CONC - eq_CONC, auto39, NA
eq_sante -eq_sante , auto40, NA
eq_edu-eq_edu , auto41, NA
sold_nat- sold_nat , auto42, NA
sold_mig - sold_mig , auto43, NA
tps_dom_emp - tps_dom_emp , auto44, NA
TXEMPLOI - TXEMPLOI , auto45, NA
ORIECO - ORIECO , auto46, NA
axe1 - axe1 , auto47, NA
[R] frequency, count rows, data for heat map
Hi all, I have read posts of heat map creation but I am one step prior -- Here is what I am trying to do and wonder if you have any tips? We are trying to map sequence reads from tumors to viral genomes. Example input file : 111 abc 111 sdf 111 xyz 1079 abc 1079 xyz 1079 xyz 5576 abc 5576 sdf 5576 sdf How may xyz's are there for 1079 and 111? How many abc's, etc? How many times did reads from sample (1079) align to virus xyz. In some cases there are thousands per virus in a give sample, sometimes one. The original file (two columns by tens of thousands of rows; 20 MB) is text file (tab delimited). Output file: abc sdf xyz 111 1 1 1 1079 1 0 2 5576 1 2 0 Or, other ways to generate this data so I can then use it for heat map creation? Thanks for any help you may have, rtsweeney palo alto, ca -- View this message in context: http://r.789695.n4.nabble.com/frequency-count-rows-data-for-heat-map-tp2338363p2338363.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] lattice help required
hello, i want to stack two lattice plots beneath each other using one x-axis and sharing the same text-panels, like: # library(lattice) y1 - rnorm(100,100,10) y2 - rnorm(100,10,1) facs-expand.grid(Sites=rep(c(Site I,Site II),25),Treatment=c(A,B)) pl1-dotplot(y1 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 pl2-dotplot(y2 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 print(pl1, split=c(1,2,1,2), more=TRUE) print(pl2, split=c(1,1,1,2)) # but as said, ideally the plots should be stacked with only the lower plot giving the x-axis annotation and only the upper plot with text-panels. thanks a lot, kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338382.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] lmer() causes segfault
Dennis, just wow. Thank you so much. I knew it was something trivial - in this case the variable type of the of the grouping variables. However, something as trivial as this should not throw a segfault IMHO. I tried subscribing to R-sig-mixed this morning, but the corresponding mail server at the ETH's stats department seems to be down. And thank you so much for changing the model, that is a great new starting point. Can you recommend a good book that deals with multilevel models in lmer() that include longitudinal data? I was not aware of the difference between scalar random effects and random slopes and would like to read up on that. Again, thanks a lot. Regards, Bertolt Am 25.08.2010 um 13:47 schrieb Dennis Murphy: Hi: Let's start with the data: str(test.data) 'data.frame': 100 obs. of 4 variables: $ StudentID: num 17370 17370 17370 17370 17379 ... $ GroupID : num 1 1 1 1 1 1 1 1 1 1 ... $ Time : num 1 2 3 4 1 2 3 4 1 2 ... $ Score: num 76.8 81.8 89.8 92.8 75.9 ... Both StudentID and GroupID are numeric; in the model, they would be treated as continuous covariates rather than factors, so we need to convert: test.data$StudentID - factor(test.data$StudentID) test.data$GroupID - factor(test.data$GroupID) Secondly, I believe there are some flaws in your model. After converting your variables to factors, I ran library(lme4) mlmoded1.lmer - lmer(Score ~ Time + (Time | GroupID/StudentID), data = test.data) You have two groups, so they should be treated as a fixed effect - more specifically, as a fixed blocking factor. The StudentIDs are certainly nested within GroupID, and Time is measured on each StudentID, so it is a repeated measures factor. The output of this model is mlmoded1.lmer Linear mixed model fit by REML Formula: Score ~ Time + (Time | GroupID/StudentID) Data: test.data AIC BIC logLik deviance REMLdev 393.1 416.5 -187.5376.9 375.1 Random effects: GroupsNameVariance Std.Dev. Corr StudentID:GroupID (Intercept) 0.504131 0.71002 Time 0.083406 0.28880 1.000 GroupID (Intercept) 12.809567 3.57905 Time 3.897041 1.97409 -1.000 Residual 1.444532 1.20189 Number of obs: 100, groups: StudentID:GroupID, 25; GroupID, 2 Fixed effects: Estimate Std. Error t value (Intercept) 72.803 2.552 28.530 Time 4.474 1.401 3.193 Correlation of Fixed Effects: (Intr) Time -0.994 The high correlations among the random effects and then among the fixed effects suggests that the model specification may be a bit off. The above model fits random slopes to GroupIDs and StudentIDs, along with random intercepts, but GroupID is a between-subject effect and should be at the top level. Time is a within-subject effect and StudentIDs are the observational units. I modified the model to provide fixed effects for GroupIDs, scalar random effects for StudentIDs and random slopes for StudentIDs. mod3 - lmer(Score ~ 1 + GroupID + Time + (1 | StudentID) + + (0 + Time | StudentID), data = test.data) mod3 Linear mixed model fit by REML Formula: Score ~ 1 + GroupID + Time + (1 | StudentID) + (0 + Time | StudentID) Data: test.data AIC BIC logLik deviance REMLdev 430.9 446.5 -209.4418.4 418.9 Random effects: GroupsNameVariance Std.Dev. StudentID (Intercept) 4.2186e-13 6.4951e-07 StudentID Time1.8380e+00 1.3557e+00 Residual 1.6301e+00 1.2768e+00 Number of obs: 100, groups: StudentID, 25 Fixed effects: Estimate Std. Error t value (Intercept) 70.7705 0.4204 168.33 GroupID2 4.0248 0.58546.88 Time 4.5292 0.2942 15.39 Correlation of Fixed Effects: (Intr) GrpID2 GroupID2 -0.668 Time -0.264 0.000 I didn't check the quality of the fit, but on the surface it seems to be more stable, FWIW. Perhaps one could also add a term (GroupID | StudentID), but I don't know offhand if that would make any sense. Another issue to consider is whether to fit by REML or ML, but that is secondary to getting the form of the model equation right. I don't claim this as a final model, but rather a 're-starting point'. It may well be in need of improvement, so comments are welcome. The confusion between subjects nested in time or vice versa has occurred several times this week with respect to repeated measures/ longitudinal models using lmer(), so perhaps it merits a comment: subjects/experimental units are NOT nested in time. Measurements taken on an individual at several time points *entails* that time be nested within subject. Just saying... This discussion may be better continued on the R-sig-mixed list, so I've cc-ed to that group as well. HTH, Dennis On Wed, Aug 25, 2010 at 1:27 AM, Bertolt Meyer bme...@sozpsy.uzh.ch wrote: Ben Bolker bbolker at gmail.com
Re: [R] frequency, count rows, data for heat map
Your problem is not completely clear to me, but perhaps something like
data - data.frame(
a = rep(c(1,2), each=10),
b = rep(c('a', 'b', 'c', 'd'), 5))
library(plyr)
daply(data, a ~ b, nrow)
does what you need.
Regards,
Jan
On Wed, Aug 25, 2010 at 4:53 PM, rtsweeney tripswee...@gmail.com wrote:
Hi all,
I have read posts of heat map creation but I am one step prior --
Here is what I am trying to do and wonder if you have any tips?
We are trying to map sequence reads from tumors to viral genomes.
Example input file :
111 abc
111 sdf
111 xyz
1079 abc
1079 xyz
1079 xyz
5576 abc
5576 sdf
5576 sdf
How may xyz's are there for 1079 and 111? How many abc's, etc?
How many times did reads from sample (1079) align to virus xyz.
In some cases there are thousands per virus in a give sample, sometimes one.
The original file (two columns by tens of thousands of rows; 20 MB) is
text file (tab delimited).
Output file:
abc sdf xyz
111 1 1 1
1079 1 0 2
5576 1 2 0
Or, other ways to generate this data so I can then use it for heat map
creation?
Thanks for any help you may have,
rtsweeney
palo alto, ca
--
View this message in context:
http://r.789695.n4.nabble.com/frequency-count-rows-data-for-heat-map-tp2338363p2338363.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] Repeat the first day data through all the day. Zoo
# duplicated / na.locf doesn't work it says Error in fix.by(by.x, x) : 'by' must specify valid column(s) if I use ifelse instead of ifelse.zoo it works but it gives me a non zoo vector. Myabe is because my zoo version is older. cheers -- View this message in context: http://r.789695.n4.nabble.com/Repeat-the-first-day-data-through-all-the-day-Zoo-tp2338069p2338409.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] Estimate average standard deviation of mean of two dependent groups
Hi Jokel,
If I remember correctly
1) The variance of the sum of two variables X and Y is:
var(X) + var(Y) + (2 * cov(X, Y))
If you create some sample data, you can verify this by:
var((X + Y))
2) The variance of a random variable (X) multiplied by some constant (K)
is equal to the the variance of X multiplied by the square of K. So:
var(X * K) = var(X) * K^2
I have never seen these combined, but I imagine they could be.
Let Z = X + Y
The mean of X and Y is (X + Y)/2 = Z/2
From the formula above (1), the variance of Z is:
var(X) + var(Y) + (2 * cov(X, Y))
Because Z/2 = Z * 0.5, the variance of Z * 0.5 is given by:
var(Z) * (0.5^2)
and substituting, the variance of the mean of X and Y is:
(var(X) + var(Y) + (2 * cov(X, Y))) * (0.5^2)
This held up under several empirical tests I did where I had actual
data for X and Y.
***Discalimer
I am piecing this together, and I am far from an expert on the
subject. I do not know if it is actually true, and there may be
additional assumptions.
I tested this using:
myfun - function() {
x - rnorm(100)
y - rnorm(100)
var.x - var(x)
var.y - var(y)
cov.xy - cov(x, y)
calc.var.xplusy - var.x + var.y + 2*cov.xy
var.meanxy - calc.var.xplusy * (.5^2)
empirical.var.meanxy - var( (x + y)/2 )
output - all.equal(var.meanxy, empirical.var.meanxy)
return(output)
}
temp - vector(logical, 1000)
for(i in 1:1000) {temp[i] - myfun()}
all(temp)
HTH,
Josh
On Wed, Aug 25, 2010 at 5:29 AM, Jokel Meyer jokel.me...@googlemail.com wrote:
Dear R-experts!
I am currently running a meta-analysis with the help of the great metafor
package. However I have some difficulties setting up my raw data to enter it
to the meta-analysis models.
I have a group of subjects that have been measured in two continuous
variables (A B). I have the information about the mean of the two
variables, the group size (n) and the standard deviations of the two
variables.
Now I would like to average both variables (A B) and get the mean and
standard deviation of the merged variable (C).
As for the mean this would be quiet easy: I would just take the mean of mean
A and mean B to get the mean of C.
However for the standard deviation this seems more tricky as it is to assume
that standard deviations in A B correlate. I assume (based on further
analysis) a correlation of r =0.5.
I found the formula to get the standard deviation of the SUM (not the mean)
of two variables:
SD=SQRT(SD_A^2 + SD_B^2 + 2*r*SD_A*SD_B)
with SD_B and SD_B being the standard deviation of A and B. And r*SD_A*SD_B
being the covariance of A and B.
Would this formula also be valid if I want to average (and not sum) my two
variables?
Many thanks for any help best wishes,
Jokel
[[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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.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] lattice help required
Kay, doe this do what you want? dotplot(y1+y2 ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list(x = list(rot = 90, tck=c(1,0))), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE) On Wed, Aug 25, 2010 at 11:04 AM, Kay Cichini kay.cich...@uibk.ac.atwrote: hello, i want to stack two lattice plots beneath each other using one x-axis and sharing the same text-panels, like: # library(lattice) y1 - rnorm(100,100,10) y2 - rnorm(100,10,1) facs-expand.grid(Sites=rep(c(Site I,Site II),25),Treatment=c(A,B)) pl1-dotplot(y1 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 pl2-dotplot(y2 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 print(pl1, split=c(1,2,1,2), more=TRUE) print(pl2, split=c(1,1,1,2)) # but as said, ideally the plots should be stacked with only the lower plot giving the x-axis annotation and only the upper plot with text-panels. thanks a lot, kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338382.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.htmlhttp://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] Draw a perpendicular line?
At 3:04 PM -0700 8/23/10, CZ wrote: Hi, I am trying to draw a perpendicular line from a point to two points. Mathematically I know how to do it, but to program it, I encounter some problem and hope can get help. Thanks. I have points, A, B and C. I calculate the slope and intercept for line drawn between A and B. I am trying to check whether I can draw a perpendicular line from C to line AB and get the x,y value for the point D at the intersection. Assume I get the slope of the perpendicular line, I will have my point (D) using variable x and y which is potentially on line AB. My idea was using |AC|*|AC| = |AD|*|AD|+ |CD|*|CD|. I don't know what function I may need to call to calculate the values for point D (uniroot?). This is easier than you think. Think of the x,y coordinates of each point : Then, the slope is slope = rise/run = (By- Ay)/(Bx- Ax) The Dx coordinate = Cx and the Dy = (Dx - Ax) * slope Then, to draw the line segment from C to D lines(C,D) In R: A - c(2,4) B - c(4,1) C - c(8,10) slope -( C[2]- A[2])/(C[1]-A[1]) #rise/run D - c(B[1],(B[1]-A[1])*slope + A[2]) # find D my.data - rbind(A,B,C,D) colnames(my.data) - c(X,Y) my.data#show it plot(my.data,type=n) #graph it without the points text(my.data,rownames(my.data)) #put the points in segments(A[1],A[2],C[1],C[2]) #draw the line segments segments(B[1],B[2],D[1],D[2]) Bill Thank you. -- View this message in context: http://r.789695.n4.nabble.com/Draw-a-perpendicular-line-tp2335882p2335882.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Change value of a slot of an S4 object within a method.
Dear all,
I have an S4 class with a slot extra which is a list. I want to be
able to add an element called name to that list, containing the
object value (which can be a vector, a dataframe or another S4
object)
Obviously
setMethod(add.extra,signature=c(PM10Data,character,vector),
function(object,name,value){
obj...@extra[[name]] - value
}
)
Contrary to what I would expect, the line :
eval(eval(substitute(expression(obj...@extra[[name]] - value
gives the error :
Error in obj...@extra[[name]] - value : object 'object' not found
Substitute apparently doesn't work any more in S4 methods...
I found a work-around by calling the initializer every time, but this
influences the performance. Returning the object is also not an
option, as I'd have to remember to assign that one each time to the
original name.
Basically I'm trying to do some call by reference with S4, but don't
see how I should do that. How would I be able to tackle this problem
in an efficient and elegant way?
Thank you in advance
Cheers
Joris
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
__
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] multiple x factors using the sciplot package
Hi: You can probably do what you want in either ggplot2 or lattice, but I would recommend at least a couple different approaches: (1) Plot individual bar charts by combinations of year and period. This is easy to do in both ggplot2 and lattice: in ggplot2, one would use geom_bar(x) + facet_grid(year ~ period), where x is the count variable. The same thing in lattice would be barchart( ~ x | year * period, data = mydata). Suitable options can be added to either plot for better presentation. All the plots are on the same horizontal and vertical scales so that visual comparisons across the rows and columns are easy to make. (2) Use a Cleveland dot chart. This is much less wasteful of ink than bar charts, and multiple groups can be compared without too much difficulty. See dotplot() in lattice for details. The VADeaths example is likely the best one to emulate for your problem. Both the input data structure and plot code matter. (3) If your purpose is to compare the two groups with respect to a count variable, a third option is to consider a mosaic plot, which can be thought of as a 'visual chi-square test'. These can be found in package vcd, which contains a nice 40+ page vignette to show you in detail how to create a suitable mosaic plot (and how to interpret it). I prefer to see R used to promote good graphics and sound statistical principles. In that spirit, performing a two-factor comparison with count data is easier, and quite possibly more instructive, with a 2D visual metaphor (which all three of the suggestions above have in common) than with the 1D metaphor of multiple dodged/side-by-side bar charts. Of course, this may be against the standard practice in your field, in which case you have a decision to make. HTH, Dennis On Wed, Aug 25, 2010 at 6:16 AM, Jordan Ouellette-Plante jordan_opla...@hotmail.com wrote: Dear R community, I am a beginner using the sciplot package to graph barplots. I would like to be able to graph two x factors (Sampling.year and Period). Sampling.year: 2006, 2007, 2008, 2009 Period: First, Second, Total The parameter group is the different species I looked at. They can be seen in the legend. The parameter response is the Average percentage cover category of these species. I would like the graph so you see on the x axis the years (in a bigger font) and the period (in smaller font). It would look a bit like the barplot that can be seen at http://onertipaday.blogspot.com/2007/05/make-many-barplot-into-one-plot.html, but with the advantage of the sciplot package. Thanks a lot for your help! Jordan [[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] Compiling Fortran for R : .so: mach-o, but wrong architecture
Hi Marie, this link may be helpful if you want to build it for both architectures, 32-bit and 64-bit. I struggled with the same problem not too long ago. http://cran.rakanu.com/bin/macosx/RMacOSX-FAQ.html#Building-universal-package I ended up writing a very simple Makefile to accompany [my version of] the bar.f code. There is almost certainly a better way to do it. Maybe with Makevars? David -- View this message in context: http://r.789695.n4.nabble.com/Compiling-Fortran-for-R-so-mach-o-but-wrong-architecture-tp2337198p2338507.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] Change value of a slot of an S4 object within a method.
Hi Joris,
On Wed, Aug 25, 2010 at 11:56 AM, Joris Meys jorism...@gmail.com wrote:
Dear all,
I have an S4 class with a slot extra which is a list. I want to be
able to add an element called name to that list, containing the
object value (which can be a vector, a dataframe or another S4
object)
Obviously
setMethod(add.extra,signature=c(PM10Data,character,vector),
function(object,name,value){
obj...@extra[[name]] - value
}
)
Contrary to what I would expect, the line :
eval(eval(substitute(expression(obj...@extra[[name]] - value
gives the error :
Error in obj...@extra[[name]] - value : object 'object' not found
Substitute apparently doesn't work any more in S4 methods...
I found a work-around by calling the initializer every time, but this
influences the performance. Returning the object is also not an
option, as I'd have to remember to assign that one each time to the
original name.
Basically I'm trying to do some call by reference with S4, but don't
see how I should do that. How would I be able to tackle this problem
in an efficient and elegant way?
In lots of my own S4 classes I define a slot called .cache which is
an environment for this exact purpose.
Using this solution for your scenario might look something like this:
setMethod(add.extra,signature=c(PM10Data,character,vector),
function(object, name, value) {
obj...@.cache$extra[[name]] - value
})
I'm not sure what your entire problem looks like, but to get your
extra list, or a value form it, you could:
setMethod(extra, signature=PM10Data,
function(object, name=NULL) {
if (!is.null(name)) {
obj...@.cache$extra[[name]]
} else {
obj...@.cache$extra
})
... or something like that.
The last thing you have to be careful of is that you nave to make sure
that each new(PM10Data) object you have initializes its *own* cache:
setClass(PM10Data, representation(..., .cache='environment'))
setMethod(initialize, PM10Data,
function(.Object, ..., .cache=new.env()) {
callNextMethod(.Object, .cache=.cache, ...)
})
Does that make sense?
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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] Odp: Finding pairs
Hi: I'm just ideating here (think IBM commercial...) but perhaps a graphical model approach might be worth looking into. It seems to me that Mr. Rhodes is looking for clusters of banks that are under the same ownership umbrella. That information is not directly available in a single variable, but can evidently be inferred from the matches between the two variables: B[i] controls A[i] if B[i] is nonempty. In the bank 144 - 147 - 149 example, 149 controls 147 and 147 controls 144, so it appears that some transitive relation holds among the set of matches as well. (Why is PacMan going through my head? :) I know next to nothing about graphical models, but I'm thinking about igraph and some of the tools in the statnet bundle to tackle this problem. Does that make sense to anyone? Alternatives? FWIW, Dennis On Wed, Aug 25, 2010 at 2:24 AM, Mike Rhodes mike_simpso...@yahoo.co.ukwrote: Dear Mr Petr PIKAL After reading the R code provided by you, I realized that I would have never figured out how this could have been done. I am going to re-read again and again your code to understand the logic and the commands you have provided. Thanks again from the heart for your kind advice. Regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Re: [R] Odp: Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 9:01 Hm r-help-boun...@r-project.org napsal dne 25.08.2010 09:43:26: Dear Mr Petr Pikal I am extremely sorry for the manner I have raised the query. Actually that was my first post to this R forum and in fact even I was also bit confused while drafting the query, for which I really owe sorry to all for consuming the precious time. Perhaps I will try to redraft my query in a better way as follows. I have two datasets A and B containing the names of branch offices of a particular bank say XYZ plc bank. The XYZ bank has number of main branch offices (say Parent) and some small branch offices falling under the purview of these main branch offices (say Child). The datalist A and B consists of these main branch office names as well as small branch office names. B is subset of A and these branch names are coded. Thus we have two datasets A and B as (again I am using only a portion of a large database just to have some idea) A B 144 what is here in B? Empty space?, 145 146 147 144 How do you know that 144 from B relates to 147 in A? Is it according to its positions? I.e. 4th item in B belongs to 4.th item in A? 148 145 149 147 151 148 Now the branch 144 appears in A as well as in B and in B it is mapped with 147. This means branch 147 comes under the purview of main branch 144. Again 147 is controlling the branch 149 (since 147 also has appeared in B and is mapped with 149 of A). Similarly, branch 145 is controlling branch 148 which further controls operations of bank branch 151 and like wise. Well as you did not say anything about structure of your data A-144:151 B-144:148 data.frame(A,B) A B 1 144 NA 2 145 NA 3 146 NA 4 147 144 5 148 145 6 149 146 7 150 147 8 151 148 DF-data.frame(A,B) main-DF$A[is.na(DF$B)] branch1-DF[!is.na(DF$B),] selected.branch1-branch1$A[branch1$B%in%main] branch2-branch1[!branch1$B%in%main,] selected.branch2-branch2$A[branch2$B%in%selected.branch1] and for cbinding your data which has uneven number of values see Jim Holtman's answer to this How to cbind DF:s with differing number of rows? Regards Petr So in the end I need an output something like - Main Branch Branch office1 Branch office2 144 147 149 145 148 151 146 NA NA ... .. I understand again I am not able to put forward my query properly. But I must thank all of you for giving a patient reading to my query and for reverting back earlier. Thanks once again. With warmest regards Mike --- On Wed, 25/8/10, Petr PIKAL petr.pi...@precheza.cz wrote: From: Petr PIKAL petr.pi...@precheza.cz Subject: Odp: [R] Finding pairs To: Mike Rhodes mike_simpso...@yahoo.co.uk Cc: r-help@r-project.org Date: Wednesday, 25 August, 2010, 6:39 Hi without other details it is probably impossible to give you any reasonable advice. Do you have your data already in R? What is their form? Are they in 2 columns in data frame? How did you get
[R] package MuMI
Hello, I am using the package MuMI to run all the possible combinations of variables in my full model, and select my best models. When I enter my variables in the original model I write them like this lm(y~ a +b +c +a:b) However, MuMI will also use the variable b:a, which I do not want in my model. How do I stop that from happening? Thank you, Agnese __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice help required
exactly - thanks a lot, richard! kay Zitat von RICHARD M. HEIBERGER r...@temple.edu: Kay, doe this do what you want? dotplot(y1+y2 ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list(x = list(rot = 90, tck=c(1,0))), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE) On Wed, Aug 25, 2010 at 11:04 AM, Kay Cichini kay.cich...@uibk.ac.atwrote: hello, i want to stack two lattice plots beneath each other using one x-axis and sharing the same text-panels, like: # library(lattice) y1 - rnorm(100,100,10) y2 - rnorm(100,10,1) facs-expand.grid(Sites=rep(c(Site I,Site II),25),Treatment=c(A,B)) pl1-dotplot(y1 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 pl2-dotplot(y2 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 print(pl1, split=c(1,2,1,2), more=TRUE) print(pl2, split=c(1,1,1,2)) # but as said, ideally the plots should be stacked with only the lower plot giving the x-axis annotation and only the upper plot with text-panels. thanks a lot, kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338382.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.htmlhttp://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] Repeat the first day data through all the day. Zoo
On Wed, Aug 25, 2010 at 11:18 AM, skan juanp...@gmail.com wrote: # duplicated / na.locf doesn't work it says Error in fix.by(by.x, x) : 'by' must specify valid column(s) if I use ifelse instead of ifelse.zoo it works but it gives me a non zoo vector. Myabe is because my zoo version is older. They all work: library(zoo) library(chron) z - zoo(1:10, chron(0:9/5)) # aggregate / na.locf z.ag - aggregate(z, as.Date, head, 1) na.locf(z.ag, xout = time(z)) (01/01/70 00:00:00) (01/01/70 04:48:00) (01/01/70 09:36:00) (01/01/70 14:24:00) 1 1 1 1 (01/01/70 19:12:00) (01/02/70 00:00:00) (01/02/70 04:48:00) (01/02/70 09:36:00) 1 6 6 6 (01/02/70 14:24:00) (01/02/70 19:12:00) 6 6 # duplicated / na.locf z.na - ifelse.zoo(!duplicated(as.Date(time(z))), z, NA) na.locf(z.na) (01/01/70 00:00:00) (01/01/70 04:48:00) (01/01/70 09:36:00) (01/01/70 14:24:00) 1 1 1 1 (01/01/70 19:12:00) (01/02/70 00:00:00) (01/02/70 04:48:00) (01/02/70 09:36:00) 1 6 6 6 (01/02/70 14:24:00) (01/02/70 19:12:00) 6 6 # ave - as before zz - z zz[] - ave(coredata(z), as.Date(time(z)), FUN = function(x) head(x, 1)) zz (01/01/70 00:00:00) (01/01/70 04:48:00) (01/01/70 09:36:00) (01/01/70 14:24:00) 1 1 1 1 (01/01/70 19:12:00) (01/02/70 00:00:00) (01/02/70 04:48:00) (01/02/70 09:36:00) 1 6 6 6 (01/02/70 14:24:00) (01/02/70 19:12:00) 6 6 packageDescription(zoo)$Version [1] 1.6-4 R.version.string [1] R version 2.11.1 Patched (2010-05-31 r52167) __ 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] several odfWeave questions
[Sending both to the maintainer and to R-help, in case anyone else has
answers ...]
I've looked in odfWeave documentation, vignette, and poked around on
the web some, and haven't found answers yet.
1a. am I right in believing that odfWeave does not respect the
'keep.source' option? Am I missing something obvious?
1b. is there a way to set global options analogous to \SweaveOpts{}
directives in Sweave? (I looked at odfWeaveControl, it doesn't seem to
do it.)
2. I tried to write a Makefile directive to process files from the
command line:
%.odt: %_in.odt
$(RSCRIPT) -e library(odfWeave);
odfWeave(\$*_in.odt\,\$*.odt\);
This works, *but* the resulting output file gives a warning (The file
'odftest2.odt' is corrupt and therefore cannot be opened.
OpenOffice.org can try to repair the file ...). Based on looking at
the contents, it seems that a spurious/unnecessary 'Rplots.pdf' file is getting
created and zipped in with the rest of the archive; when I unzip, delete
the Rplots.pdf file and re-zip, the ODT file opens without a warning.
Obviously I could post-process but it would be nicer to find a
workaround within R ...
3. I find the requirement that all file paths be specified as absolute
rather than relative paths somewhat annoying -- I understand the reason,
but it goes against one practice that I try to encourage for
reproducibility, which is *not* to use absolute file paths -- when
moving a same set of data and analysis files across computers, it's hard
to enforce them all ending up in the same absolute location, which then
means that the recipient has to edit the ODT file. It would be nice if
there were hooks for read.table() and load() as there are for plotting
and package/namespace loading -- then one could just copy them into the
working directory on the fly.
has anyone experienced this/thought of any workarounds?
(I guess one solution is to zip any necessary source files into the archive
beforehand,
as illustrated in the vignette.)
My test files are posted at
http://www.math.mcmaster.ca/~bolker/junk/odftest_in.odt and
http://www.math.mcmaster.ca/~bolker/junk/odftest.odt
thanks,
Ben Bolker
sessionInfo()
R version 2.11.1 (2010-05-31)
i486-pc-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] odfWeave_0.7.16 XML_3.1-1 lattice_0.19-9
loaded via a namespace (and not attached):
[1] grid_2.11.1
__
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] frequency, count rows, data for heat map
Hi:
Here are a couple of ways to render a basic 2D table. Let's call your input
data frame dat:
names(dat) - c('samp', 'sequen')
ssTab - as.data.frame(with(dat, table(samp, sequen)))
ssTab # data frame version
samp sequen Freq
1 111abc1
2 1079abc1
3 5576abc1
4 111sdf1
5 1079sdf0
6 5576sdf2
7 111xyz1
8 1079xyz2
9 5576xyz0
with(dat, table(samp, sequen)) # table version
sequen
samp abc sdf xyz
1111 1 1
1079 1 0 2
5576 1 2 0
HTH,
Dennis
On Wed, Aug 25, 2010 at 7:53 AM, rtsweeney tripswee...@gmail.com wrote:
Hi all,
I have read posts of heat map creation but I am one step prior --
Here is what I am trying to do and wonder if you have any tips?
We are trying to map sequence reads from tumors to viral genomes.
Example input file :
111 abc
111 sdf
111 xyz
1079 abc
1079 xyz
1079 xyz
5576 abc
5576 sdf
5576 sdf
How may xyz's are there for 1079 and 111? How many abc's, etc?
How many times did reads from sample (1079) align to virus xyz.
In some cases there are thousands per virus in a give sample, sometimes
one.
The original file (two columns by tens of thousands of rows; 20 MB) is
text file (tab delimited).
Output file:
abc sdf xyz
111 1 1 1
1079 1 0 2
5576 1 2 0
Or, other ways to generate this data so I can then use it for heat map
creation?
Thanks for any help you may have,
rtsweeney
palo alto, ca
--
View this message in context:
http://r.789695.n4.nabble.com/frequency-count-rows-data-for-heat-map-tp2338363p2338363.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] Change value of a slot of an S4 object within a method.
Hi Steve,
thanks for the tip. I'll definitely take a closer look at your
solution for implementation for future use. But right now I don't
have the time to start rewriting my class definitions.
Luckily, I found where exactly things were going wrong. After reading
into the documentation about the evaluation in R, I figured out I have
to specify the environment where substitute should look explicitly as
parent.frame(1). I still don't understand completely why exactly, but
it does the job.
Thus :
eval(eval(substitute(expression(obj...@extra[[name]] - value
should become :
eval(
eval(
substitute(
expression(obj...@extra[[name]] - value)
,env=parent.frame(1) )
)
)
Tried it out and it works.
Cheers
Joris
On Wed, Aug 25, 2010 at 6:21 PM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
Hi Joris,
On Wed, Aug 25, 2010 at 11:56 AM, Joris Meys jorism...@gmail.com wrote:
Dear all,
I have an S4 class with a slot extra which is a list. I want to be
able to add an element called name to that list, containing the
object value (which can be a vector, a dataframe or another S4
object)
Obviously
setMethod(add.extra,signature=c(PM10Data,character,vector),
function(object,name,value){
obj...@extra[[name]] - value
}
)
Contrary to what I would expect, the line :
eval(eval(substitute(expression(obj...@extra[[name]] - value
gives the error :
Error in obj...@extra[[name]] - value : object 'object' not found
Substitute apparently doesn't work any more in S4 methods...
I found a work-around by calling the initializer every time, but this
influences the performance. Returning the object is also not an
option, as I'd have to remember to assign that one each time to the
original name.
Basically I'm trying to do some call by reference with S4, but don't
see how I should do that. How would I be able to tackle this problem
in an efficient and elegant way?
In lots of my own S4 classes I define a slot called .cache which is
an environment for this exact purpose.
Using this solution for your scenario might look something like this:
setMethod(add.extra,signature=c(PM10Data,character,vector),
function(object, name, value) {
obj...@.cache$extra[[name]] - value
})
I'm not sure what your entire problem looks like, but to get your
extra list, or a value form it, you could:
setMethod(extra, signature=PM10Data,
function(object, name=NULL) {
if (!is.null(name)) {
obj...@.cache$extra[[name]]
} else {
obj...@.cache$extra
})
... or something like that.
The last thing you have to be careful of is that you nave to make sure
that each new(PM10Data) object you have initializes its *own* cache:
setClass(PM10Data, representation(..., .cache='environment'))
setMethod(initialize, PM10Data,
function(.Object, ..., .cache=new.env()) {
callNextMethod(.Object, .cache=.cache, ...)
})
Does that make sense?
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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 obtain seed after generating random number?
If you find yourself doing things like this often, but don't want to explicitly
set the seed, or save seeds before simulating, then you can run the following
code (or put it into .Rprofile or similar):
.Last.Random.seed - .Random.seed
addTaskCallback( function(expr, val, ok, visible){
if(!isTRUE( all.equal(.Last.Random.seed, .Random.seed)) ) {
.Last.Random.seed - .Random.seed
}
TRUE
})
Then the previous seed will be stored in .Last.Random.seed and you can restore
the seed to be able to rerun the same values, e.g.:
rnorm(10)
[1] -0.28361138 0.86951931 -0.54435528 0.62880324 -1.42233446 -1.22751263
[7] -1.67410552 0.08439848 -0.20612566 1.44187164
.Random.seed - .Last.Random.seed
rnorm(10)
[1] -0.28361138 0.86951931 -0.54435528 0.62880324 -1.42233446 -1.22751263
[7] -1.67410552 0.08439848 -0.20612566 1.44187164
rnorm(10)
[1] -0.0417821 1.3537545 1.9452253 -0.4909382 0.3884391 -0.8448933
[7] 0.7379904 -1.0797603 -1.0264739 0.2887934
The above code only keeps the most recent seed, but the code could be modified
to store a longer history if that were desired.
If you want to set your own seeds (a little easier to save, pass to others),
but find integers to unimaginative, then look at the char2seed function in the
TeachingDemos package.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Bogaso Christofer
Sent: Tuesday, August 24, 2010 11:12 AM
To: r-help@r-project.org
Subject: [R] How to obtain seed after generating random number?
Dear all, I was doing an experiment to disprove some theory therefore
performing lot of random simulation. Goal is to show the audience that
although something has very rare chance to occur but it doesn't mean
that
event would be impossible.
In this case after getting that rare event I need to show that same
scenario
for multiple times to explain other audience. Hence I need to somehow
save
that seed which generates that random numbers after doing the
experiment.
However as it is very rare event it is not very practical to start with
a
fixed seed and then generate random numbers. Hence I am looking for
some way
which will tell me about that corresponding seed which was responsible
to
generate that particular series of random numbers responsible for
occurrence
of that rare event.
In short, I need to know the seed ***after*** generating the random
numbers.
Is there any possibility to know this?
Thanks and regards,
[[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] nls self starting function
I need the simple function for the following set of data. This is a toy version of my data, but the error is persistent in both. To compare with excel, I would just 1) format trendline, 2) display equation and R-squared on chart. I obviously tried to use a nls and wanted to use the self startup function SSasymp rm(list=ls()) t - c( 0,.1,.2,.7,.4,.5,.6,1,.8,.4,1.2,1.3,1.4,1.5,1.6, 1.6,1.7,1.8,1.9,2.0, 3,6,11) y-c(2,3.5,3.01,3.09,3,3.25,3.36,3.49,3.64, 3.81, 4.44, 4.69, 4.96, 5.25, 6, 6.1, 5.89, 6.24, 6.61, 7.00, 12.00, 39, 124.00) plot(y~t ,xlab=t, ylab=y) model-nls(y~SSasymp(t,a,b)) summary(model) Thanks in advance, keith -- M. Keith Cox, Ph.D. Alaska NOAA Fisheries, National Marine Fisheries Service Auke Bay Laboratories 17109 Pt. Lena Loop Rd. Juneau, AK 99801 keith@noaa.gov marlink...@gmail.com U.S. (907) 789-6603 [[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] lattice help required
... i added relation=free to account for diffferent ranges of y1 and y2: dotplot(y1+y2 ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list(y = list(relation=free), x = list(rot = 90, tck=c(1,0))), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE) but then i get four different y-lims. more suited there should be only two - one for each response. can someone tell me how i have to change the code? best, kay Kay, doe this do what you want? dotplot(y1+y2 ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list(x = list(rot = 90, tck=c(1,0))), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE) On Wed, Aug 25, 2010 at 11:04 AM, Kay Cichini kay.cich...@uibk.ac.atwrote: hello, i want to stack two lattice plots beneath each other using one x-axis and sharing the same text-panels, like: # library(lattice) y1 - rnorm(100,100,10) y2 - rnorm(100,10,1) facs-expand.grid(Sites=rep(c(Site I,Site II),25),Treatment=c(A,B)) pl1-dotplot(y1 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 pl2-dotplot(y2 ~ facs$Treatment|facs$Sites, scales = list(x = list(rot = 90, tck=c(1,0 print(pl1, split=c(1,2,1,2), more=TRUE) print(pl2, split=c(1,1,1,2)) # but as said, ideally the plots should be stacked with only the lower plot giving the x-axis annotation and only the upper plot with text-panels. thanks a lot, kay - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338382.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Kay Cichini Postgraduate student Institute of Botany Univ. of Innsbruck -- View this message in context: http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338641.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice help required
The multiple y axes are protecting you in this situation. z - cbind(rnorm(100,c(1,10),1), rnorm(100,c(20,30),1)) dotplot(z[,1]+z[,2] ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list( y = list( relation=free)), ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE, main=problem) dotplot(z[,1]+z[,2] ~ facs$Treatment|facs$Sites, outer=TRUE, scales = list( y = list( relation=free, limits=list(c(-5,13),c(-5,13),c(18,32),c(18,32, ylab=c(y1, y2), xlab=c(Site 1, Site 2), strip=FALSE, main=protected) For more control (such as suppressing the y-tick labels in the right-hand column, I recommend Deepayan Sarkar's book. Rich [[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] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
Hi,
I am new to R, and as a first exercise, I decided to try to implement an XIRR
function using the secant method. I did a quick search and saw another posting
that used the Bisection method but wanted to see if it was possible using the
secant method.
I would input a Cash Flow and Date vector as well as an initial guess. I
hardcoded today's initial date so I could do checks in Excel. This code seems
to only converge when my initial guess is very close to the correct IRR.
Maybe I have some basic errors in my coding/logic? Any help would be greatly
appreciated.
The Wikipedia article to secant method and IRR:
http://en.wikipedia.org/wiki/Internal_rate_of_return#Numerical_solution
Thanks!
ANXIRR - function (cashFlow, cfDate, guess){
cfDate-as.Date(cfDate,format=%m/%d/%Y)
irrprev - c(0); irr- guess
pvPrev- sum(cashFlow)
pv-
sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=days))/360)))
print(pv)
print(Hi)
while (abs(pv) = 0.001) {
t-irrprev; irrprev- irr;
irr-irr-((irr-t)*pv/(pv-pvPrev));
pvPrev-pv;
pv-sum(cashFlow/((1+irr)^(as.numeric(difftime(cfDate,2010-08-24,units=days))/365)))
print(irr);print(pv)
}
}
Please consider the environment before printing this e-mail.
[[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] Merging two data set in R,
try the following merge command. merge(A,B, by = intersect(names(A), names(B)), all.x=FALSE, all.y=FALSE) or merge(A,B, by = ID, all.x=FALSE, all.y=FALSE) Dannemora On Wed, Aug 25, 2010 at 5:35 AM, Mangalani Peter Makananisa pmakanan...@sars.gov.za wrote: Dear R Gurus, I am currently working on the two dataset ( A and B), they both have the same fields:ID , REGION, OFFICE, CSTART, CEND, NCYCLE, STATUS and CB. I want to merge the two data set by ID. The problem I have is that the in data A, the ID's are unique. However in the data set B, the ID's are not unique, thus some repeat themselves. How do I the merge or retrieve the common ones? Please advise. Kind Regards Peter South Africa +27 12 422 7357 +27 82 456 4669 Please Note: This email and its contents are subject to our email legal notice which can be viewed at http://www.sars.gov.za/Email_Disclaimer.pdf [[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] Showing a line function on the ploat area
I have a collection of results. I use R to get the linearization presented. how can I get R to show the equation and R^2 value on the plot area next to the graph? -- View this message in context: http://r.789695.n4.nabble.com/Showing-a-line-function-on-the-ploat-area-tp2338389p2338389.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] Plot bar lines like excel
Great .. thanks for the to much help and i too appreciate hel p and explanation Cheers -- View this message in context: http://r.789695.n4.nabble.com/Plot-bar-lines-like-excel-tp2337089p2338158.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] GLM outputs in condensed versus expanded table
Hi I'm having different outputs from GLM when using a condensed table V1 V2 V3 Present Absent 0 0 0 3 12 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 7 20 1 0 1 0 0 1 1 0 3 0 1 1 1 6 0 resp=cbind(Present, Absent) glm(resp~V1+V2+V3+I(V1*V2*V3),family=binomial) Deviance Residuals: [1] 0 0 0 0 0 0 0 0 etc and also coefficients... And when using the same but expanded table V1 V2 V3 condition (1 present 0 abscent) Id1 1 0 0 1 id2 1 1 1 1 ... etc glm(condition~V1+V2+V3+I(V1*V2*V3),family=binomial) Deviance Residuals: Min 1Q Median 3Q Max -0.7747317 -0.7747317 -0.6680472 0.0001315 1.7941226 and also coefficients are different from above. What could I be doing wrong? -- View this message in context: http://r.789695.n4.nabble.com/GLM-outputs-in-condensed-versus-expanded-table-tp2338407p2338407.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] NewJerseyR Meeting
Mango Solutions are proud to announce the first NewJerseyR Meeting to be held
on the 16th September 2010.
Please see details below:
NewJerseyR
Date: Tuesday 16th September
Time: 6.30pm - 9.30pm
Venue: Renaissance Woodbridge Hotel - Iselin, New Jersey
The agenda has yet to be finalised, I shall send details out as soon as they
have been confirmed.
Please contact: newjers...@mango-solutions.com to be added to the NewJerseyR
mailing list to receive meeting updates and R news.
To register for this event please contact: newjers...@mango-solutions.com
We currently hold 2 very successful R meetings - LondonR and BaselR. NewJerseyR
will be our first R meeting in America. NewJerseyR will be a free informal
event, held quarterly and intended to serve as a platform for all local (and
regional) R users to present and exchange their experiences and ideas around
the usage of R. A typical meeting will consist of 3-4 talks of about 20-25 min
to give plenty of room for sharing your R experiences, discussions and exchange
of ideas.
For presentations from our other R meetings can be found at:
www.londonr.org and www.baselr.org
The NewJerseyR website will be up and running shortly.
For further information, please contact newjers...@mango-solutions.com
Sarah Lewis
Hadley Wickham, Creator of ggplot2 - first time teaching in the UK. 1st - 2nd
November 2010.
To book your seat please go to http://mango-solutions.com/news.html
T: +44 (0)1249 767700 Ext: 200
F: +44 (0)1249 767707
M: +44 (0)7746 224226
www.mango-solutions.com
Unit 2 Greenways Business Park
Bellinger Close
Chippenham
Wilts
SN15 1BN
UK
LEGAL NOTICE
This message is intended for the use o...{{dropped:9}}
__
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 plot y-axis on the right of x-axis
Dear Elaine I'm developing a code to make a 3 y-axes plot. It may can help you. Also your question leads to mine, posted yesterday: How does R calculate the interval between tick-marks. Below follows the code I'm developing. Data: ANO CKG NUP NDE 200526352158150025014 200632514789210035024 200727458126180030254 200828568971150043254 200932564789300060320 Code: # I use comma as decimal symbol so I use dec=, dat.caupde - read.delim(clipboard,dec=,,header=T) attach(dat.caupde) summary(dat.caupde) ANOCKGNUPNDE Min. :2005 Min. :26352158 Min. :1500 Min. :25014 1st Qu.:2006 1st Qu.:27458126 1st Qu.:1500 1st Qu.:30254 Median :2007 Median :28568971 Median :1800 Median :35024 Mean :2007 Mean :29491767 Mean :1980 Mean :38773 3rd Qu.:2008 3rd Qu.:32514789 3rd Qu.:2100 3rd Qu.:43254 Max. :2009 Max. :32564789 Max. :3000 Max. :60320 # here I indicate the limits of each axes and calculate their ranges y1min - 26 y1max - 33 y1range - y1max-y1min y2min - 1500 y2max - 3000 y2range - y2max-y2min y3min - 25 y3max - 61 y3range - y3max-y3min # making the plot # y1range*((NUP-y2min)/y2range)+y1min calculates the proportion to between axes par(mar=c(6, 6, 2,12)) plot(CKG/100, ylim=c(y1min,y1max), ylab=landings (1000 t), type=l,col=red,las=0, cex.axis=1.2,cex.lab=1.4,xaxt=n,xlab=) points( y1range*((NUP-y2min)/y2range)+y1min,type=l,col=blue) points( y1range*((NDE/1000-y3min)/y3range)+y1min,type=l,col=darkgreen) # the number 1 in axis(4,at=seq(y1min,y1max,1 ... is the interval between each tick-mark axis(1,at=1:5,labels=as.character(ANO),las=2,cex.axis=1.2) axis(4,at=seq(y1min,y1max,1),labels=as.integer(seq(y2min,y2max,y2range/y1range)),las=0,cex.axis=1.2) axis(4,at=seq(y1min,y1max,1), labels=as.integer(seq(y3min,y3max,y3range/y1range)),cex.axis=1.2, las=0,line=5) mtext(nº of fishing boats,4,3,cex=1.4) mtext(nº of fishing trips (x1000),4,8, cex=1.4) legend(4,33,c(L,Nº FB,Nº FT),bty=n,lty=c(1,1,1),col=c(red,blue,darkgreen),cex=1.4) Now I want to replace the number 1 by the formula used to calculate the interval between tick-marks. Why, with the range 26-33, R choose unitary intervals for the y axis (26, 27, 28 ...)? All the best, Antonio Olinto Citando elaine kuo elaine.kuo...@gmail.com: Dear List, I have a richness data distributing across 20 N to 20 S latitude. (120 E-140 E longitude). I would like to draw the richness in the north hemisphere and a regression line in the plot (x-axis: latitude, y-axis: richness in the north hemisphere). The above demand is done using plot. Then, south hemisphere richness and regression are required to be generated using the same y-axis above but an x-axis on the left side of the y-axis. (The higher latitude in the south hemisphere, the left it would move) Please kindly share how to design the south plot and regression line for richness. Also, please advise if any more info is in need. Elaine [[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. Webmail - iBCMG Internet http://www.ibcmg.com.br __ 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 all objects except a few specified objects?
Thanks to De-Jian and Peter. Peter's way is neat and cool! -- View this message in context: http://r.789695.n4.nabble.com/How-to-remove-all-objects-except-a-few-specified-objects-tp2335651p2338221.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] Problem with clusterCall, Error in checkForRemoteErrors(lapply(cl, recvResult)) :
Hi all,
I am trying to use snow package to do a parallel MCMC. I have read a few
guides and articles, the following is that I came up with.
When I run it I got the error message:
Error in checkForRemoteErrors(lapply(cl, recvResult)) :
4 nodes produced errors; first error: could not find function ui.Next
The data is a longitudinal data with few repeated readings on a number of
individuals. However, the response is organised in a vector rather than a
matrix. E.g. (y_11 , y_12 , y_13 , y_14 , y_21 , y_22 , ... , y_n1 , ... ,
y_nTn )^T
X , Z are covariates.
beta is a matrix of coefficients associated with X
C is the latent class membership indicator
sigma.sq is the diagonal elements of the covariance matrix of u, which is a
random effect parameter and the one I am trying to sample with the function.
sd is the diagonal elements of the covariance matrix of the proposal
distribution.
The following is the particular function:
ui.Full.Sample.Cluster = function( levels , Y , X , beta , Z , C , sigma.sq
, sd , burnin , iteration )
{
cluster = makeCluster( 4 , type = MPI);
clusterSetupRNG( cluster );
patients = unique( levels );
q = length( X[1,] );
u = ui.Ini( q , length( patients ) );
n = levels[ length(patients) ];
marker = levels == patients[1];
y = Y[ marker ];
x = X[ marker, ];
z = Z[ marker, ];
u[1,] = ui.1.Sample( u[1,] , y , x , beta , z , C[1] , sigma.sq , sd ,
burnin , iteration )$uFinal;
for( i in 2:n)
{
marker = levels == patients[i];
y = Y[ marker ];
x = X[ marker, ];
z = Z[ marker, ];
print( i );
u[i, ] = clusterCall( cluster , ui.1.Sample, u[i,] , y , x ,
beta , z ,
C[i] , sigma.sq , sd , burnin , iteration )$uFinal;
print( i );
}
stopCluster( cluster );
u;
}
If anyone could help that would be much appreciated! But big thanks for
taking your time to read the post!
TelM8
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-clusterCall-Error-in-checkForRemoteErrors-lapply-cl-recvResult-tp2338375p2338375.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] What does this warning message (from optim function) mean?
Hi R users,
I am trying to use the optim function to maximize a likelihood funciton, and
I got the following warning messages.
Could anyone explain to me what messege 31 means exactly? Is it a cause for
concern?
Since the value of convergence turns out to be zero, it means that the
converging is successful, right?
So can I assume that the parameter estimates generated thereafter are
reliable MLE estimates?
Thanks a lot for your help.
Maomao
p-optim(c(0,0,0), f, method =BFGS, hessian =T, y=y,X=X,W=W)
There were 31 warnings (use warnings() to see them)
warnings()
Warning messages:
1: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
2: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
3: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
4: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
5: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
6: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
7: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
8: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
9: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
10: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
11: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
12: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
13: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
14: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
15: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
16: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
17: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
18: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
19: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
20: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
21: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
22: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
23: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
24: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
25: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
26: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
27: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
28: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
29: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
30: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
31: In if (hessian) { ... :
the condition has length 1 and only the first element will be used
p$counts
function gradient
148 17
p$convergence
[1] 0
[[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] pairwise correlation of many samples
Dear list, I have a csv file like below, and I have similar 150 files (responses from 150 samples) in a folder. Id like to get the mean score of pairwise correlation score among 150 respondents by hour(0~20hour). All I can think of is bring up two files and get the pairwise correlation score and repeating this (e.g. (p1,p2),(p1,P3) (P4,P5)..)and get the total mean of correlation score. However, since therere 150 samples I cant think of doing this manually.. Hour Response 0 -0.11703 0.016667 -0.10427 0.03 -0.091347 0.05 -0.078313 0.07 -0.065226 0.08 -0.052147 0.1 -0.039136 . . Can anyone please point out the best way to do? .. Its been only a month since I started to use R Thank you! [[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] Problem with clusterCall, Error in checkForRemoteErrors(lapply(cl, recvResult)) :
Oh, I forgot to say, ui.Next() is a function I defined to sample from the proposal distribution given the current state. -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-clusterCall-Error-in-checkForRemoteErrors-lapply-cl-recvResult-tp2338375p2338378.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] multivariate analysis of variance
Hi, can anyone tell me how i may find out the between-group variance and within-group variance for multivariate case? I have 6 Groups and 73 Variables. (with MANOVA ? wie) dim(data) [1] 2034 76 Thanks Celal __ 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.
