[R] how to access Internet from within R?
Let's say I have some automation tasks where I have to fill a table field on a webpage, and then submit it to the server, and then obtain the returned value (a typical web query), how to do that in R? Thanks a lot! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 access Internet from within R?
On Fri, Jun 26, 2009 at 7:21 AM, Michaelcomtech@gmail.com wrote: Let's say I have some automation tasks where I have to fill a table field on a webpage, and then submit it to the server, and then obtain the returned value (a typical web query), how to do that in R? The RCurl package will let you construct POST requests to web pages - this is the mechanism that the web uses to send form data to web servers. The response comes back to the function in RCurl. That's if you know in advance the names of the required form fields. Otherwise you have to do a 'GET' request to get the form (which should be some HTML), then write some R to parse that and work out what is what, then send a POST with your response. Sounds like you don't need to do that if you know what the form is expecting. See the help for postForm in RCurl package. Basic usage is: postForm(http://example.com/form/response.html,.params=list(name=me,text=this is my response, vote=2) couldn't really be easier :) Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] what happened to the xlsReadWrite package
On Fri, 26 Jun 2009, David Scott wrote: Andrew Yee wrote: A naive question: what happened to the xlsReadWrite package? http://cran.r-project.org/web/packages/xlsReadWrite/ It says that it was removed from the CRAN repository. Note that it was archived, not removed entirely. That is done with packages that no longer install and we don't get an update. The immediate problem with xlsReadWrite was that it was Windows-only and would not install under Windows in pre-2.9.0. Are there any plans for it be available again? The second issue was binary code in the package which is at least heavily frowned on in the CRAN source repository (we host BRugs elsewhere for that reason). There was a problem with proprietary code. Only in so far as it was binary. There are a number of CRAN packages with non-open-source code. Pick up the latest version from Hans-Peter Suter's website http://treetron.googlepages.com/ Which says the problem was the binary code. David Scott _ David Scott Department of Statistics The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 Graduate Officer, Department of Statistics Director of Consulting, Department of Statistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] multiple figure windows
Hi, Is it possible to plot in multiple windows. for example Figure 1 in one window and Figure 2 in seperate window simultanously ? thanks, Nataraju -- No relationship is Static .. You either Step up or Step down [[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] Optimization and Linear Programming in R
Dear List, My student and I are looking for an optimizer for a nonlinear optimization problem we are working on. The problem we are working on is to try to pick a set of islands on which to eradicate rats for seabird conservation. We have about 50 islands, each of which has some subset of 17 seabird species. Rats are present on all islands, and will cause the seabirds to go extinct unless they are removed. Removing rats on islands costs different amounts depending on which island is chosen. The decision problem is to pick the set of islands which will save the most seabird species within a given budget. Here we use a nonlinear function to calculate the objective. The function is the proportion of individuals of each species saved divided by the total number of birds, times the log of the same number, summed up across all of the islands that are chosen to be in the set that is saved. We have a single constraint, which is a budget in our case. The contribution to this constraint is the product of the cost of each island times the decision vector. Our decision variable is a binary vector (1 for removing rats from an island, 0 for not removing them). The objective function is as I have explained above. We are looking for a solver that can deal with this nonlinear integer programming problem. We looked at a number of packages on the CRAN Task View: Optimization and Mathematical Programming, however, we have not been able to locate one that will suit our purposes. The essential features are we need to be able to write a nonlinear function for the objective (hopefully a self contained one as we need to include some data in it), we need to be able to use a binary decision (or parameter) vector, and we need to be able to use a constraint. Any suggestions as to packages or other software that will work for our problem would be much appreciated. Thanks, Chris Wilcox and Greg Thonier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Batch problem
Hi, I want to make my R program run in batch under Windows XP. To do so, I create a bat file with the command RCMD BATCH --vanilla program.R program.out and I use the bat file with the scheduled task of Windows XP. Then I log off. It works up to the log off of another user on the same computer with R-2.9.1 but this problem doesn't appear with R-1.9.1 on the same machine. Is anything wrong in the syntax of my bat file? Thanks. Regards, Alain -- Alain Guillet Statistician and Computer Scientist SMCS - Institut de statistique - Université catholique de Louvain Bureau d.126 Voie du Roman Pays, 20 B-1348 Louvain-la-Neuve Belgium tel: +32 10 47 30 50 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Calculating home ranges using mcp in adehabitat
Tom: The structure of your dataframe (mydata) differs from the structure of the example dataframe (puechabon) in that the example dataframe has the structure 'locs' in which resides the x,y data and the individual identifier. The clue to your difficulty could be found when you asked to see the object xy you created and were greeted with xy[1:19,] NULL indicating it was empty. Here is a toy example of what you want to do to successfully use mcp() library(adehabitat) mydata - data.frame(Name='XDRY', X=runif(20)-21, Y=runif(20)+64) mydata xy - mydata[,c(X,Y)] id - mydata[,Name] xy[1:20,] mcp(xy,id,percent=95) Tom Mason wrote: Hello, I've been trying to calculate home range sizes (for Icelandic geese!) using minimum convex polygons with the adehabitat package. I've tried to use the R code shown by demo(homerange) in adehabitat and when that didn't work I've fiddled around with it but to no avail...Below is the output of the demo that I've attempted to follow, followed by a subset of my data (for one individual) and the code I've been trying. Can anyone tell me where I may be going wrong??? #Demo output... data(puechabon) xy-puechabon$locs[,c(X,Y)] id-puechabon$locs$Name ## The data are: xy[1:4,] ## relocations coordinates X Y 1 699889 3161559 2 700046 3161541 3 698840 3161033 4 699809 3161496 id[1:4] ## ID [1] Brock Brock Brock Brock Levels: Brock Calou Chou Jean ### ### ### ### Home ranges ## MCP hr-mcp(xy, id) ## home range estimation # My attempts... mydata Name XY 1 XDRY -21.98389 64.06457 2 XDRY -21.99759 64.08291 3 XDRY -21.98784 64.06467 4 XDRY -21.98333 64.06058 5 XDRY -21.97889 64.06257 6 XDRY -21.98284 64.06044 7 XDRY -21.97886 64.06358 8 XDRY -21.99741 64.08124 9 XDRY -21.99715 64.08330 10 XDRY -22.00397 64.09331 11 XDRY -21.99811 64.08239 12 XDRY -22.00453 64.09337 13 XDRY -21.99713 64.08299 14 XDRY -21.99608 64.08307 15 XDRY -21.99646 64.08352 16 XDRY -21.99326 64.08361 17 XDRY -22.00770 64.09090 18 XDRY -21.98160 64.06400 19 XDRY -21.97966 64.06367 xy-mydata$locs[,c(X,Y)] id-mydata$locs$Name xy[1:19,] NULL id[1:19,] NULL mcp(xy,id,percent=95) Error in if (length(id) != nrow(xy)) stop(xy and id should be of the same length) : argument is of length zero #I've also tried things like the code below. But didn't know how to incorporate 'locs' into this... xy-cbind(X,Y) id-Name[1:19] xy[1:19,] XY [1,] -21.98389 64.06457 [2,] -21.99759 64.08291 [3,] -21.98784 64.06467 [4,] -21.98333 64.06058 [5,] -21.97889 64.06257 [6,] -21.98284 64.06044 [7,] -21.97886 64.06358 [8,] -21.99741 64.08124 [9,] -21.99715 64.08330 [10,] -22.00397 64.09331 [11,] -21.99811 64.08239 [12,] -22.00453 64.09337 [13,] -21.99713 64.08299 [14,] -21.99608 64.08307 [15,] -21.99646 64.08352 [16,] -21.99326 64.08361 [17,] -22.00770 64.09090 [18,] -21.98160 64.06400 [19,] -21.97966 64.06367 id-Name[1:19] id[1:19] [1] XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY [12] XDRY XDRY XDRY XDRY XDRY XDRY XDRY XDRY Levels: XDRY mcp(xy,id,percent=95) Error in apply(xy, 2, mean) : dim(X) must have a positive length xy1-xy[1:19,] id1-id[1:19] mcp(xy1,id1,percent=95) Error in apply(xy, 2, mean) : dim(X) must have a positive length Would be most grateful for any suggestions to where I'm going wrong! Tom Mason, University of Exeter, UK. -- View this message in context: http://www.nabble.com/Calculating-home-ranges-using-mcp-in-adehabitat-tp24207532p24216560.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] Converting S-plus project folders to R
dbosley wrote: I have many S-plus project folders that I need to convert to R workspaces. For the smaller project folders ( 200MB), using data.dump with oldStyle = T and data.restore (in the foreign package) within R seems to work fine. However, I have several project folders that are quite large (~ 4GB). When I use this procedure to try to convert these project folders, R always crashes when I perform the data.restore command. Does anyone have any suggestions? No suggestion, just to be sure: I hope R gives an error message rather than a crash? Best, Uwe Ligges Thanks, Dave Bosley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 figure windows
Nattu wrote: Hi, Is it possible to plot in multiple windows. for example Figure 1 in one window and Figure 2 in seperate window simultanously ? Sure, just start another device as in: x11() plot(1:10) x11() plot(1:10) Uwe Ligges thanks, Nataraju __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] panel.text and saving to pdf
To get reproducible results, I highly recommend to print the lattice graphics directly into the desired device, i.e. start the pdf (or jpeg or whatever) device before printing. Best, Uwe Ligges Willem Vervoort wrote: Dear all, I am not sure what I am doing wrong, but I have some unexplained behaviour when saving a lattice graph including text to a pdf file. The text seems to move around. It must have something to do with the way coordinates are set in devices other than jpg. Any suggestions would be helpful Willem Here is some example code setwd(c:/willem/research/misc) today - format(Sys.Date(),%Y%m%d) x - runif(500) y - rnorm(500) foo - data.frame(x = x, y = y, z = rep(c(a,b),250)) require(lattice) xyplot(x~y|z,data=foo) panel.text(370,470,silly graph,cex=1.2,font=2) savePlot(paste(today, jpgplot,sep=_),type=jpg) # this plots fine and text is as on the screen savePlot(paste(today, pdfplot,sep=_),type=pdf) # text has moved pdf(paste(today,pdfplot2.pdf,sep=_)) # no difference using pdf xyplot(x~y|z,data=foo) panel.text(370,470,silly graph,cex=1.2,font=2) dev.off() # There is also no difference _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 9.0 year 2009 month 04 day17 svn rev48333 language R version.string R version 2.9.0 (2009-04-17) - Dr. R.W. Vervoort McCaughey Senior Lecturer Hydrology and Catchment Management Faculty of Agriculture, Food and Natural Resources, Bldg A04, The University of Sydney, NSW 2006 http://tinyurl.com/mccaughey http://blogs.usyd.edu.au/waterhydrosu ph: +61 2 9351 8744 fax: +61 2 9351 4953 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] panel.text and saving to pdf
Willem Vervoort w.vervoort at usyd.edu.au writes: I am not sure what I am doing wrong, but I have some unexplained behaviour when saving a lattice graph including text to a pdf file. The text seems to move around. It must have something to do with the way coordinates are set in devices other than jpg. today - format(Sys.Date(),%Y%m%d) x - runif(500) y - rnorm(500) foo - data.frame(x = x, y = y, z = rep(c(a,b),250)) require(lattice) xyplot(x~y|z,data=foo) panel.text(370,470,silly graph,cex=1.2,font=2) Take this plot, resize its window: you will note that the silly graph moves relative to the graphics. The same is true for the higher-resolution pdf output. To get around this, and to have the text positioned in a relative scaled position, you could create a special panel function and call panel.text there. See the examples at the bottom of the xyplot documentation page. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] to raise in a loop more than 1
I would raise x,y and z in a loop but I won't raise of 1. I tried this but it
doesn't work
mydata=matrix(nrow=1500,ncol=3)
i=1
for(x in 0:10){
for(y in 0:20){
for(z in 0:10){
mydata[i,]=c(x,y,z)
i=i+1
z=z+2}
y=y+4}
x=x+2}
And I would have something like that
x y z
0 0 0
0 0 2
0 0 4
0 0 6
0 0 8
0 0 10
0 4 0
0 4 2
...
Could anybody help me?
I don't work on a special package to do it...
Thanks
Cordialement
Damien Landais
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] JRI - problem to access stats package
Hi Geoff,
Tks a lot for your reply.
There are many points we do not understand.
Can you clarify those one:
- At first, why do you speak about an eclipse plugin ?
We have compiled R and JRI together, so we get a jar file. We finally
include this jar file inside our java project as a referenced libraries.
So we do not understand why you are speaking about plugins and
MANIFEST.MF file.
- Nevermind, which MANIFEST.MF file are you speaking about ? the one of
the JRI plugin ? (if a JRI plugin exists...)
- Last things, you speak about a build.properties. In our project we
haven't any. Is there a work around to do the same ?
Tks for any help,
Antoine
Hi Antoine,
I have got this to work on Windows, the process should be the same for
Linux.
For an eclipse plugin you will need to add the following to your
MANIFEST.MF file:
Bundle-ClassPath: whatever you normally have,
lib/JRI.jar
Bundle-NativeCode: /lib/jri.dll; osname=winxp; processor=x86
You will also need to ensure that the *R* executable is in your path and
that *R_HOME* is set.
I did this by creating a simple batch file containing the following in a
directory called *rootfiles* within our feature, the same should work for
a .sh :
set PATH=%CD%\*R* install\exe;%PATH%
set *R_HOME*=%CD%\*R* install
eclipse.exe
I then added the following line to the build.properties file of the
feature:
*root*=*rootfiles*
This tells eclipse to copy the content of the directory to the *root* of
the application on install.
If you then *run* eclipse from the batch file, it should work.
--
Geoff Gibbs
mangosolutions
data analysis that delivers
Tel +44 (0)1249 767 700
Mob +44 (0)7791 855 620
-Original Message-
From: *r*-help-boun...@...
http://www.nabble.com/user/SendEmail.jtp?type=postpost=24199629i=0
[mailto:*r*-help-boun...@...
http://www.nabble.com/user/SendEmail.jtp?type=postpost=24199629i=1]
On Behalf Of GRAN
Sent: 25 June 2009 09:40
To: *r*-h...@...
http://www.nabble.com/user/SendEmail.jtp?type=postpost=24199629i=2
Subject: [*R*] JRI - problem to access *stats* package
Hello,
I am a new *R* user. More precisely, I am working with JRI (on a Eclipse
Java6 project under Ubuntu9).
I have difficulties to access *some* *R* packages (For example: package
*stats*, object Normal, function pnorm).
I have tried many *solutions* to set the *right* path but nothing have
succeed:
For example:
LD_LIBRARY_PATH=${*R_HOME*}/lib:${*R_HOME*}/bin:${*R_HOME*}/library
Inside the java code, I have also tried such methods:
Note that *r* is the *REngine* object coming from the connection to *R*
*r*.eval(dyn.load(\/home/gran/workspace/*R*-2.9.0/library/*stats*/libs/*stats*
.*so*\,
local = TRUE, now = TRUE));
*r*.eval(*require*(*stats*));
*r*.eval(autoload(\Normal\, \*stats*\));
*r*.eval(search());
*r*.eval(ls(\Autoloads\));
System.out.println(*r*.eval(.Autoloaded));
*r*.eval(zval = .95);
System.out.println(*r*.eval(print(zval)));
*r*.eval(p3 = pnorm(3*zval));
System.out.println(*r*.eval(print(p3)));
It gives as a *result*:
[STRING *stats*]
[*REAL** (0.95)]
null -- p3 = null because pnorm is unknown
Do you know how to use the *stats* package with JRI ?
Thanks in advance,
Antoine
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Odp: to raise in a loop more than 1
Hi
does
expand.grid(z=seq(0,6,2), y=seq(0,8,4), x=seq(0,10,2))
do what you want?
Petr
r-help-boun...@r-project.org napsal dne 26.06.2009 11:15:42:
I would raise x,y and z in a loop but I won't raise of 1. I tried this
but it
doesn't work
mydata=matrix(nrow=1500,ncol=3)
i=1
for(x in 0:10){
for(y in 0:20){
for(z in 0:10){
mydata[i,]=c(x,y,z)
i=i+1
z=z+2}
y=y+4}
x=x+2}
And I would have something like that
x y z
0 0 0
0 0 2
0 0 4
0 0 6
0 0 8
0 0 10
0 4 0
0 4 2
...
Could anybody help me?
I don't work on a special package to do it...
Thanks
Cordialement
Damien Landais
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] panel.text and saving to pdf
Hi Uwe, I should send these messages from my member e-mail number Thanks for that. As you can see from the last bit of my code, I tried that, but it gave the same results. I had a look at pdf.options() but didn't see anything there either to make the text stay on the same spot. It seems to also happen if you use png(). Willem From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: Fri 6/26/2009 6:47 PM To: Willem Vervoort Cc: r-help@r-project.org Subject: Re: [R] panel.text and saving to pdf To get reproducible results, I highly recommend to print the lattice graphics directly into the desired device, i.e. start the pdf (or jpeg or whatever) device before printing. Best, Uwe Ligges Willem Vervoort wrote: Dear all, I am not sure what I am doing wrong, but I have some unexplained behaviour when saving a lattice graph including text to a pdf file. The text seems to move around. It must have something to do with the way coordinates are set in devices other than jpg. Any suggestions would be helpful Willem Here is some example code setwd(c:/willem/research/misc) today - format(Sys.Date(),%Y%m%d) x - runif(500) y - rnorm(500) foo - data.frame(x = x, y = y, z = rep(c(a,b),250)) require(lattice) xyplot(x~y|z,data=foo) panel.text(370,470,silly graph,cex=1.2,font=2) savePlot(paste(today, jpgplot,sep=_),type=jpg) # this plots fine and text is as on the screen savePlot(paste(today, pdfplot,sep=_),type=pdf) # text has moved pdf(paste(today,pdfplot2.pdf,sep=_)) # no difference using pdf xyplot(x~y|z,data=foo) panel.text(370,470,silly graph,cex=1.2,font=2) dev.off() # There is also no difference _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 9.0 year 2009 month 04 day17 svn rev48333 language R version.string R version 2.9.0 (2009-04-17) - Dr. R.W. Vervoort McCaughey Senior Lecturer Hydrology and Catchment Management Faculty of Agriculture, Food and Natural Resources, Bldg A04, The University of Sydney, NSW 2006 http://tinyurl.com/mccaughey http://blogs.usyd.edu.au/waterhydrosu ph: +61 2 9351 8744 fax: +61 2 9351 4953 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] crr - computationally singular
Dear Sir, Thank you for your response. You were correct, I had 1 linearly dependent column. I have solved this problem and now the rank of 'covaeb' is 17 (qr(covaeb)$rank = 17). However, I still get the same error message when I use covaeb in the 'crr' function. fit=crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) 8 cases omitted due to missing values Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 3.45905e-25 Are there any other reasons why this may be happening? Thank you, Laura 2009/6/25 Ravi Varadhan rvarad...@jhmi.edu: This means that your design matrix or model matrix is rank deficient, i.e it does not have linearly independent columns. Your predictors are collinear! Just take your design matrices covaea or covaeb with 17 predcitors and compute their rank or try to invert them. You will see the problem. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Laura Bonnett Sent: Thursday, June 25, 2009 11:39 AM To: r-help@r-project.org Subject: [R] crr - computationally singular Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 1.28654e-24 This is obviously as a result of a problem with the data but apart from dataset a having 1674 rows and dataset b having 701 rows there is really no difference between them. The code I am using is as follows where covaea and covaeb are matrices of covarites, all coded as binary variables. In case a: covaea - cbind(sexa,fsha,fdra,nsigna,eega,th1a,th2a,stype1a,stype2a,stype3a,pgu 1a,pgu2a,log(agea),firstinta/1000,totsezbasea) fita - crr(snearma$with.Withtime,csaea,covaea,failcode=2,cencode=0) and in case b: covaeb - cbind(sexb,fshb,fdrb,nsignb,eegb,th1b,th2b,stype1b,stype2b,stype3b,sty pe4b,stype5b,pgu1b,pgu2b,(ageb/10)^(-1),firstintb,log(totsezbaseb)) fitb - crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) csaea and csaeb are the censoring indicators for a and b respectively which equal 1 for the event of interest, 2 for the competing risks event and 0 otherwise. Can anyone suggest a reason for the error message? I've tried running fitb with variants of covaeb and irrespective of the order of the covariates in the matrix, the code runs fine with 16 of the 17 covariates included but then produces an error message when the 17th is added. Thank you for your help, Laura __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Odp: code that will use two data sets which differ in size.
Petr Pikal
petr.pi...@precheza.cz
724008364, 581252140, 581252257
r-help-boun...@r-project.org napsal dne 26.06.2009 05:31:27:
Hi all,
This is a really basic question but I can't figure it out.
I am trying to write a piece of code that will use two datasets, z and
m..
This code is meant to assign the mean of selected values in dataset m to
a new
column z$c in datset z but the rows and columns of the two data sets
differ.
I managed to write this down but think I am missing something standard
but
basic,I will really appreciate if someone could point out where I' am
going
wrong or help me out with a piece of code for this.Google did not help
me even
and the R manuals gave me hints.
The shoulders will never grow beyond the head.
for (k in 1:nrow(z))
+ for (g in 1:nrow(m))
+ for (l in 1:3)
+ for (j in 15:43){
+ z$ccs[k]-mean(m[m$soi[g]==lm$week[g]==j,]$CCS,na.rm=TRUE)}
Maybe that I have no idea what is z I get an error too.
for (k in 1:nrow(z))
+ + for (g in 1:nrow(m))
+ + for (l in 1:3)
+ + for (j in 15:43){
+ + z$ccs[k]-mean(m[m$soi[g]==lm$week[g]==j,]$CCS,na.rm=TRUE)}
Error in nrow(z) : object 'z' not found
What kind of error do you have?
If you do not post some reproducible example it would be quite difficult
to understand what you want to get as a result. I believe some *apply
trick could help but without knowing what you want it could be just rough
guessing.
Regards
Petr
[[alternative HTML version deleted]]
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Re: [R] JRI - problem to access stats package
Sorry for the confusion, I obviously misunderstood.
I thought you were trying to use R in an eclipse application, not an
application written using eclipse.
As far as I am aware, if the path and R_HOME environment variables are
set then all should work OK.
I think from you examples that the path is set correctly, it may be that
you are using print() to get the value that you are getting null, as I
understand it print() returns invisibly after displaying the value on
screen, as there is no screen the evaluation of print does nothing but
return the result of the call. I suspect you simply want to use
System.out.println(r.eval(p3)).
--
Geoff Gibbs
mangosolutions
data analysis that delivers
Tel +44 (0)1249 767 700
Mob +44 (0)7791 855 620
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of GRAN
Sent: 26 June 2009 11:01
To: r-help@r-project.org
Cc: antoine grimault
Subject: [R] JRI - problem to access stats package
Hi Geoff,
Tks a lot for your reply.
There are many points we do not understand.
Can you clarify those one:
- At first, why do you speak about an eclipse plugin ?
We have compiled R and JRI together, so we get a jar file. We finally
include this jar file inside our java project as a referenced libraries.
So we do not understand why you are speaking about plugins and
MANIFEST.MF file.
- Nevermind, which MANIFEST.MF file are you speaking about ? the one of
the JRI plugin ? (if a JRI plugin exists...)
- Last things, you speak about a build.properties. In our project we
haven't any. Is there a work around to do the same ?
Tks for any help,
Antoine
Hi Antoine,
I have got this to work on Windows, the process should be the same for
Linux.
For an eclipse plugin you will need to add the following to your
MANIFEST.MF file:
Bundle-ClassPath: whatever you normally have, lib/JRI.jar
Bundle-NativeCode: /lib/jri.dll; osname=winxp; processor=x86
You will also need to ensure that the *R* executable is in your path and
that *R_HOME* is set.
I did this by creating a simple batch file containing the following in a
directory called *rootfiles* within our feature, the same should work
for a .sh :
set PATH=%CD%\*R* install\exe;%PATH%
set *R_HOME*=%CD%\*R* install
eclipse.exe
I then added the following line to the build.properties file of the
feature:
*root*=*rootfiles*
This tells eclipse to copy the content of the directory to the *root* of
the application on install.
If you then *run* eclipse from the batch file, it should work.
--
Geoff Gibbs
mangosolutions
data analysis that delivers
Tel +44 (0)1249 767 700
Mob +44 (0)7791 855 620
-Original Message-
From: *r*-help-boun...@...
http://www.nabble.com/user/SendEmail.jtp?type=postpost=24199629i=0
[mailto:*r*-help-boun...@...
http://www.nabble.com/user/SendEmail.jtp?type=postpost=24199629i=1]
On Behalf Of GRAN
Sent: 25 June 2009 09:40
To: *r*-h...@...
http://www.nabble.com/user/SendEmail.jtp?type=postpost=24199629i=2
Subject: [*R*] JRI - problem to access *stats* package
Hello,
I am a new *R* user. More precisely, I am working with JRI (on a Eclipse
Java6 project under Ubuntu9).
I have difficulties to access *some* *R* packages (For example: package
*stats*, object Normal, function pnorm).
I have tried many *solutions* to set the *right* path but nothing have
succeed:
For example:
LD_LIBRARY_PATH=${*R_HOME*}/lib:${*R_HOME*}/bin:${*R_HOME*}/library
Inside the java code, I have also tried such methods:
Note that *r* is the *REngine* object coming from the connection to
*R*
*r*.eval(dyn.load(\/home/gran/workspace/*R*-2.9.0/library/*stats*/libs
/*stats*
.*so*\,
local = TRUE, now = TRUE));
*r*.eval(*require*(*stats*));
*r*.eval(autoload(\Normal\, \*stats*\));
*r*.eval(search());
*r*.eval(ls(\Autoloads\));
System.out.println(*r*.eval(.Autoloaded));
*r*.eval(zval = .95);
System.out.println(*r*.eval(print(zval)));
*r*.eval(p3 = pnorm(3*zval));
System.out.println(*r*.eval(print(p3)));
It gives as a *result*:
[STRING *stats*]
[*REAL** (0.95)]
null -- p3 = null because pnorm is unknown
Do you know how to use the *stats* package with JRI ?
Thanks in advance,
Antoine
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Re: [R] GeoXp package
The problem is that the binary Ubuntu install of gdal has not made sure that its dependencies are properly satisfied, so you'll need to do this manually. You can also make sure that the directory the GRASS libraries are found in is seen by ldconfig, either that or making sure that your running R sees LD_LIBRARY_PATH. This is a very detailed question, more suited to GRASS or GDAL lists, or R-sig-geo. You'll see when you have the solution because the GDAL utilities will work at the shell, for example gdalinfo --formats - I think that it doesn't at the moment. Roger Bivand epoizot wrote: Zeljko Vrba wrote: On Wed, Jun 10, 2009 at 08:21:06AM +0200, Poizot Emmanuel wrote: Error in fun(...) : GDAL Error 1: libgrass_I.so: Ne peut ouvrir le fichier d'objet partagé: Aucun fichier ou dossier de ce type (sorry for the french :) ) It would have been far more useful had you translated the error message to english than to have apologized. Try doing export LD_LIBRARY_PATH=/path/to/dir/where/so/is/located before starting R from the shell. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Hi, I tried this but with no success. Still enable to load the library rgdal and so GeoXp. This problem of for me now of high importance as I need to export interpolated data sets to grids. and rgdal allows that kind of operations. Regards -- View this message in context: http://www.nabble.com/GeoXp-package-tp23964536p24218092.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 avoid ifelse statement converting factor to character
Stavros Macrakis wrote: On Thu, Jun 25, 2009 at 12:47 PM, Craig P. Pyramecrap...@gmail.com wrote: The man page Stavros quotes states that the class attribute of the result is taken from 'test', which clearly is not the case: Actually, the behavior is documented pretty clearly: The mode of the answer will be coerced from logical to accommodate first any values taken from 'yes' and then any values taken from 'no'. Whether this is a good design or not is another issue Perhaps the justification is that it avoids evaluating the yes or no arguments (to determine their class) in cases where their value is not needed. Thank you for pointing me to this. Now I get a headache from trying to figure out what does mode have to do with class - I thought that the class of the result should be that of test, and that the mode is something entirely different. Why does coercing the mode also affect the class? If the man page said The class attribute is taken from test, and it will be coerced ... or The mode of the result is taken from test, and it will be coreced ..., would this be wrong? What is the class-mode mixture about? Why does this fail: r = as.raw(TRUE) ifelse(TRUE, r, r) = error This gives an error which I take for saying that raw cannot be coerced to logical, but yes it can: as.logical(r) = TRUE and raw can even be used as the condition vector in ifelse: ifelse(r, 1, 2) = 1 Best regards, Craig Example: ifelse(c(T,F),1,a) = c(1,a) This has the same effect as res - c(T,F) res[1] - 1 res[2] - a which is in fact pretty much the way it is implemented. And also, I find myself incapable of making sense of the may in the mode of the result may depend on the value of 'test' - may in what sense? See the examples at the end of ? ifelse -s __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] The Claw Density and LOCFIT
I am trying to reproduce Figure 10.5 of Loader's book: Local Regression and Likelihood. The code provided in the book does not seem to work. I have managed (a while ago) to get the accompanied R-code for the figures in the book (file called lffigs.R) from somewhere - cannot find it on the web anymore. The code in the .R script file does not work either. Could anybody please direct me in finding an updated version of this document, or help me correct the code given in the file. The (out-of-date) code is as follows: data(claw54) fit1 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.315), ev = grid(100, ll = -3.5, ur = 2.7)) fit2 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.985), ev = grid(100, ll = -3.5, ur = 2.7)) x - seq(-3.5, 2.7, length.out = 200) y - dnorm(x, -1., 0.1) + dnorm(x, -0.5, 0.1) + dnorm(x, 0, 0.1) + dnorm(x, 0.5, 0.1) + dnorm(x, 1., 0.1) y - (y + 5 * dnorm(x))/10 plot(fit1, get.data = T, main = h=0.315, ylim = c(0, max(y))) lines(x, y, lty = 2) plot(fit2, get.data = T, main = h=0.985, ylim = c(0, max(y))) lines(x, y, lty = 2) THANKS FOR ANY ASSISTANCE. ps: This code differs from that in the book. I have tried both, without success. Even if I just use, for example, fit1 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.315)) I do not get the same result. Jacob L van Wyk, Dept. of Statistics, University of Johannesburg (APK), Box 524, Auckland Park, 2006. Office: +27 11 559 3080, Fax: +27 11 559 2499 This email and all contents are subject to the following disclaimer: http://www.uj.ac.za/UJ_email_legal_disclaimer.htm [[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] Matching data to a new column
Hi all, I wonder if anyone can help, I have a dataframe with columns for... 'I.D' 'age' 'mothers I.D' 01 5 03 02 6 06 03 16 NA 04 8 06 05 3 NA 06 17 NA I need to create a new column for 'mothers age' which puts the age of the individual with 'mothers i.d' into the row for her offspring (so individual 01 would have 16 in the mothers age column, as thats the age of individual 03) Hope that makes sense, any help appreciated, Thanks Laura [[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] Matching data to a new column
Kubasiewicz, Laura laura.kubasiewicz08 at imperial.ac.uk writes: I have a dataframe with columns for... 'I.D' 'age' 'mothers I.D' 01 5 03 02 6 06 03 16 NA 04 8 06 05 3 NA 06 17 NA I need to create a new column for 'mothers age' which puts the age of the individual with 'mothers i.d' into the row for her offspring (so individual 01 would have 16 in the mothers age column, as thats the age of individual 03) That's at typical join in SQL language, and it is handled in R with function merge. In your special case, you would merge the dataframe with itself, so something like merge(fr,fr, by... and check the NA cases) Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Optimization and Linear Programming in R
Hello, On 6/26/09, chris.wil...@csiro.au chris.wil...@csiro.au wrote: We are looking for a solver that can deal with this nonlinear integer programming problem. We looked at a number of packages on the CRAN Task View: Optimization and Mathematical Programming, however, we have not been able to locate one that will suit our purposes. The essential features are we need to be able to write a nonlinear function for the objective (hopefully a self contained one as we need to include some data in it), we need to be able to use a binary decision (or parameter) vector, and we need to be able to use a constraint. Any suggestions as to packages or other software that will work for our problem would be much appreciated. There is `solnp', part of the Integer and Nonlinear Optimization in R (RINO) project [1]. It is still in early beta, though. [1] http://r-forge.r-project.org/projects/rino/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 I get just the two last tokens of each string in a vector?
Dear list, Sorry for asking this very silly question on the list, but I seem to have made my life complicated by going into string manipulation in vectors. What I need is to get the last part of a sting (the two last tokens, separated by a space), and of course, this should be done for all strings in a vector, creating a new vector of exual size. So, a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) should be made into a vector c( L*H H%, H* H%, L*H %, L*H % ) I have tried strsplit, but it seems to produce a structure I cannot get to work in this context. Any ideas on how to solve this? Thankful for all the help I can get. /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I get just the two last tokens of each string in a vector?
This should do it a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) # split the strings x - strsplit(a, ' ') # get last two tokens sapply(x, function(.tokens) paste(tail(.tokens, 2), collapse=' ')) [1] L*H H% H* H% L*H % L*H % On Fri, Jun 26, 2009 at 7:21 AM, Fredrik Karlsson dargo...@gmail.comwrote: Dear list, Sorry for asking this very silly question on the list, but I seem to have made my life complicated by going into string manipulation in vectors. What I need is to get the last part of a sting (the two last tokens, separated by a space), and of course, this should be done for all strings in a vector, creating a new vector of exual size. So, a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) should be made into a vector c( L*H H%, H* H%, L*H %, L*H % ) I have tried strsplit, but it seems to produce a structure I cannot get to work in this context. Any ideas on how to solve this? Thankful for all the help I can get. /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] Optimization and Linear Programming in R
Chris.Wilcox at csiro.au writes: Dear List, [...] We are looking for a solver that can deal with this nonlinear integer programming problem. We looked at a number of packages on the CRAN Task View: Optimization and Mathematical Programming, however, we have not been able to locate one that will suit our purposes. The essential features are we need to be able to write a nonlinear function for the objective (hopefully a self contained one as we need to include some data in it), we need to be able to use a binary decision (or parameter) vector, and we need to be able to use a constraint. Any suggestions as to packages or other software that will work for our problem would be much appreciated. What you desribe appears to be a 'Mixed Integer NonLinear Programming' (MINLP) problem. R may not be the right place to look for such kind of solvers. You can have a look into the NEOS Optimization Software Guide or into the Projects page of the COmputational INfrastructure for Operations Research (COIN-OR). The 'Bonmin' COIN-OR project could perhaps provide an appropriate solver for this problem. The RINO project on 'R-forge' plans to provide Rbonmin and Rlago packages, but this may take some time. Perhaps by reconsidering your problem approach you can (at least partly) linearize your target function or see if it can be made convex, etc.; it did not sound to be too complex. --Hans Werner P.S.: As far as I know, 'solnp' does not solve mixed integer problems. Thanks, Chris Wilcox and Greg Thonier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] to raise in a loop more than 1
Dear Damien,
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of damien landais
Sent: Friday, June 26, 2009 11:16 AM
To: R-help@r-project.org
Subject: [R] to raise in a loop more than 1
I would raise x,y and z in a loop but I won't raise of 1. I
tried this but it doesn't work
mydata=matrix(nrow=1500,ncol=3)
i=1
for(x in 0:10){
for(y in 0:20){
for(z in 0:10){
mydata[i,]=c(x,y,z)
i=i+1
z=z+2}
y=y+4}
x=x+2}
And I would have something like that
x y z
0 0 0
0 0 2
0 0 4
0 0 6
0 0 8
0 0 10
0 4 0
0 4 2
...
I am not exactly sure what you want. If my guess is correct, this will
reproduce the desired output:
x - seq(from=0, to=10, by=2)
y - seq(from=0, to=20, by=4)
z - seq(from=0, to=10, by=2)
mydata2 - expand.grid(x=as.factor(x), y=as.factor(y), z=as.factor(z))
mydata3 - with(mydata2,
mydata2[order(x,y,z),]
)
head(mydata3, n=8)
Hope this helps,
Roland
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Re: [R] How to avoid ifelse statement converting factor to character
It gives me a headache, too! I think you'll have to wait for a more expert user than me to supply explanations of these behaviors and their rationales. -s On 6/26/09, Craig P. Pyrame crap...@gmail.com wrote: Stavros Macrakis wrote: On Thu, Jun 25, 2009 at 12:47 PM, Craig P. Pyramecrap...@gmail.com wrote: The man page Stavros quotes states that the class attribute of the result is taken from 'test', which clearly is not the case: Actually, the behavior is documented pretty clearly: The mode of the answer will be coerced from logical to accommodate first any values taken from 'yes' and then any values taken from 'no'. Whether this is a good design or not is another issue Perhaps the justification is that it avoids evaluating the yes or no arguments (to determine their class) in cases where their value is not needed. Thank you for pointing me to this. Now I get a headache from trying to figure out what does mode have to do with class - I thought that the class of the result should be that of test, and that the mode is something entirely different. Why does coercing the mode also affect the class? If the man page said The class attribute is taken from test, and it will be coerced ... or The mode of the result is taken from test, and it will be coreced ..., would this be wrong? What is the class-mode mixture about? Why does this fail: r = as.raw(TRUE) ifelse(TRUE, r, r) = error This gives an error which I take for saying that raw cannot be coerced to logical, but yes it can: as.logical(r) = TRUE and raw can even be used as the condition vector in ifelse: ifelse(r, 1, 2) = 1 Best regards, Craig Example: ifelse(c(T,F),1,a) = c(1,a) This has the same effect as res - c(T,F) res[1] - 1 res[2] - a which is in fact pretty much the way it is implemented. And also, I find myself incapable of making sense of the may in the mode of the result may depend on the value of 'test' - may in what sense? See the examples at the end of ? ifelse -s __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] to raise in a loop more than 1
I assume that you are getting this error:
mydata=matrix(nrow=1500,ncol=3)
i=1
for(x in 0:10){
+ for(y in 0:20){
+ for(z in 0:10){
+ mydata[i,]=c(x,y,z)
+ i=i+1
+ z=z+2}
+ y=y+4}
+ x=x+2}
Error in mydata[i, ] = c(x, y, z) : subscript out of bounds
No suitable frames for recover()
i
[1] 1501
You have indexed outside the limits. For what you are doing, you need the
matrix to be 11*21*11=2451
mydata=matrix(nrow=2541,ncol=3)
i=1
for(x in 0:10){
+ for(y in 0:20){
+ for(z in 0:10){
+ mydata[i,]=c(x,y,z)
+ i=i+1
+ z=z+2}
+ y=y+4}
+ x=x+2}
head(mydata)
[,1] [,2] [,3]
[1,]000
[2,]001
[3,]002
[4,]003
[5,]004
[6,]005
You still don't get what you are expecting because you are trying to change
the loop variable within the loop which is not allowed (or really ignored).
On Fri, Jun 26, 2009 at 5:15 AM, damien landais damien.land...@tdf.frwrote:
I would raise x,y and z in a loop but I won't raise of 1. I tried this but
it doesn't work
mydata=matrix(nrow=1500,ncol=3)
i=1
for(x in 0:10){
for(y in 0:20){
for(z in 0:10){
mydata[i,]=c(x,y,z)
i=i+1
z=z+2}
y=y+4}
x=x+2}
And I would have something like that
x y z
0 0 0
0 0 2
0 0 4
0 0 6
0 0 8
0 0 10
0 4 0
0 4 2
...
Could anybody help me?
I don't work on a special package to do it...
Thanks
Cordialement
Damien Landais
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What is the problem that you are trying to solve?
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Re: [R] apply on xts
The way to get the best answers on r-help (and the method advised in
the Posting Guide) is to offer the audience a specification of the
libraries being used, to create examples that illustrate the sort of
input expected, to offer the code that is being used, and to clearly
specify what output is needed. I only saw one of those aspects in your
posting.
The lack of example input was the feature that held me off from
further efforts, although others may have had their own reasons. You
may think ticker objects are a well defined data type, but they are
not so for me.
(You should also send plain text, again as the Posting Guide specifies.)
--
DW
On Jun 25, 2009, at 11:27 PM, R_help Help wrote:
Thanks David for reading for me. Anyhow, is there better way to do
looping on xts objects than using apply? Thanks again.
adschai
On Thu, Jun 25, 2009 at 12:39 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Jun 24, 2009, at 9:26 PM, R_help Help wrote:
Hi,
I do not understand why after I called apply on a function that
returns an
xts (getIdvAdjSeries) it returns a matrix whose columns are just
numeric
value of time series in xts instead of a list of xts objects.
Basically, I called the following:
apply(matrix(tickers,ncol=1),1,FUN=getDivAdjSeries)
getDivAdjSeries - function(ticker) {
seriesName - paste(ticker,Adjusted,sep=.);
command - parse(text=paste(ticker,[,',seriesName,'],sep=));
s - eval(command);
dimnames(s)[[2]] - ticker;
command - parse(text=paste(ticker,@index,sep=));
s - xts(s,index=eval(command));
return(s);
}
This doesn't seem to work. Can anyone shed some light please? Thank
you.
That would appear to be the expected behavior after reading the Value
section of the help page for apply.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] crr - computationally singular
Still the same reasons. It is possible to have collinearity without having any one column be a multiple of another. xyz - data.frame(x=sample(1:1000, 5), y=sample(1:1000, 5) , xx=sample(1:1000, 5) ,yy=sample(1:1000, 5) ) xyz$z - xyz$x + xyz$y + xyz$xx solve(xyz) Error in solve.default(xyz) : system is computationally singular: reciprocal condition number = 6.39164e-20 On Jun 26, 2009, at 6:22 AM, Laura Bonnett wrote: Dear Sir, Thank you for your response. You were correct, I had 1 linearly dependent column. I have solved this problem and now the rank of 'covaeb' is 17 (qr(covaeb)$rank = 17). However, I still get the same error message when I use covaeb in the 'crr' function. fit=crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) 8 cases omitted due to missing values Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 3.45905e-25 Are there any other reasons why this may be happening? Thank you, Laura 2009/6/25 Ravi Varadhan rvarad...@jhmi.edu: This means that your design matrix or model matrix is rank deficient, i.e it does not have linearly independent columns. Your predictors are collinear! Just take your design matrices covaea or covaeb with 17 predcitors and compute their rank or try to invert them. You will see the problem. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org ] On Behalf Of Laura Bonnett Sent: Thursday, June 25, 2009 11:39 AM To: r-help@r-project.org Subject: [R] crr - computationally singular Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 1.28654e-24 This is obviously as a result of a problem with the data but apart from dataset a having 1674 rows and dataset b having 701 rows there is really no difference between them. The code I am using is as follows where covaea and covaeb are matrices of covarites, all coded as binary variables. In case a: covaea - cbind (sexa,fsha,fdra,nsigna,eega,th1a,th2a,stype1a,stype2a,stype3a,pgu 1a,pgu2a,log(agea),firstinta/1000,totsezbasea) fita - crr(snearma$with.Withtime,csaea,covaea,failcode=2,cencode=0) and in case b: covaeb - cbind (sexb,fshb,fdrb,nsignb,eegb,th1b,th2b,stype1b,stype2b,stype3b,sty pe4b,stype5b,pgu1b,pgu2b,(ageb/10)^(-1),firstintb,log(totsezbaseb)) fitb - crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) csaea and csaeb are the censoring indicators for a and b respectively which equal 1 for the event of interest, 2 for the competing risks event and 0 otherwise. Can anyone suggest a reason for the error message? I've tried running fitb with variants of covaeb and irrespective of the order of the covariates in the matrix, the code runs fine with 16 of the 17 covariates included but then produces an error message when the 17th is added. Thank you for your help, Laura __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories 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] The Claw Density and LOCFIT
Hi Jaap, Could anybody please direct me in finding an updated version of this document, or help me correct the code given in the file. The (out-of-date) code is as follows: You are not helping yourself, or anyone else, by not including the error messages you get when trying to execute your code. The matter is in fact quite straightfowards: ev does not accept an evaluation structure called grid. The current documentation for locfit does tell you this: ?locfit.raw (sub ev). When I executed your code I got fit1 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.315), ev = + grid(100, ll = -3.5, ur = 2.7)) Error in grid(100, ll = -3.5, ur = 2.7) : unused argument(s) (ll = -3.5, ur = 2.7) This explicitly tells you that the problem lies with the call to ev = grid(...), which should in fact be ev = lfgrid(...). The following works for me and should do so for you. fit1 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.315), ev = lfgrid(100, ll = -3.5, ur = 2.7)) HTH, Mark. R version 2.10.0 Under development (unstable) (2009-06-20 r48806) i386-pc-mingw32 locale: [1] LC_COLLATE=English_South Africa.1252 LC_CTYPE=English_South Africa.1252 [3] LC_MONETARY=English_South Africa.1252 LC_NUMERIC=C [5] LC_TIME=English_South Africa.1252 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] locfit_1.5-4lattice_0.17-25 akima_0.5-2 DPpackage_1.0-7 ade4_1.4-11 Design_2.2-0 [7] survival_2.35-4 Hmisc_3.6-0 loaded via a namespace (and not attached): [1] cluster_1.12.0 grid_2.10.0tools_2.10.0 Van Wyk, Jaap wrote: I am trying to reproduce Figure 10.5 of Loader's book: Local Regression and Likelihood. The code provided in the book does not seem to work. I have managed (a while ago) to get the accompanied R-code for the figures in the book (file called lffigs.R) from somewhere - cannot find it on the web anymore. The code in the .R script file does not work either. Could anybody please direct me in finding an updated version of this document, or help me correct the code given in the file. The (out-of-date) code is as follows: data(claw54) fit1 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.315), ev = grid(100, ll = -3.5, ur = 2.7)) fit2 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.985), ev = grid(100, ll = -3.5, ur = 2.7)) x - seq(-3.5, 2.7, length.out = 200) y - dnorm(x, -1., 0.1) + dnorm(x, -0.5, 0.1) + dnorm(x, 0, 0.1) + dnorm(x, 0.5, 0.1) + dnorm(x, 1., 0.1) y - (y + 5 * dnorm(x))/10 plot(fit1, get.data = T, main = h=0.315, ylim = c(0, max(y))) lines(x, y, lty = 2) plot(fit2, get.data = T, main = h=0.985, ylim = c(0, max(y))) lines(x, y, lty = 2) THANKS FOR ANY ASSISTANCE. ps: This code differs from that in the book. I have tried both, without success. Even if I just use, for example, fit1 - locfit( ~ claw54, deg = 0, kern = gauss, alpha = c(0, 0.315)) I do not get the same result. Jacob L van Wyk, Dept. of Statistics, University of Johannesburg (APK), Box 524, Auckland Park, 2006. Office: +27 11 559 3080, Fax: +27 11 559 2499 This email and all contents are subject to the following disclaimer: http://www.uj.ac.za/UJ_email_legal_disclaimer.htm [[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. -- View this message in context: http://www.nabble.com/The-Claw-Density-and-LOCFIT-tp24218274p24220007.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] crr - computationally singular
But I have centred all the dummy variables for the covariates... 2009/6/26 David Winsemius dwinsem...@comcast.net: Still the same reasons. It is possible to have collinearity without having any one column be a multiple of another. xyz - data.frame(x=sample(1:1000, 5), y=sample(1:1000, 5) , xx=sample(1:1000, 5) ,yy=sample(1:1000, 5) ) xyz$z - xyz$x + xyz$y + xyz$xx solve(xyz) Error in solve.default(xyz) : system is computationally singular: reciprocal condition number = 6.39164e-20 On Jun 26, 2009, at 6:22 AM, Laura Bonnett wrote: Dear Sir, Thank you for your response. You were correct, I had 1 linearly dependent column. I have solved this problem and now the rank of 'covaeb' is 17 (qr(covaeb)$rank = 17). However, I still get the same error message when I use covaeb in the 'crr' function. fit=crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) 8 cases omitted due to missing values Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 3.45905e-25 Are there any other reasons why this may be happening? Thank you, Laura 2009/6/25 Ravi Varadhan rvarad...@jhmi.edu: This means that your design matrix or model matrix is rank deficient, i.e it does not have linearly independent columns. Your predictors are collinear! Just take your design matrices covaea or covaeb with 17 predcitors and compute their rank or try to invert them. You will see the problem. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Laura Bonnett Sent: Thursday, June 25, 2009 11:39 AM To: r-help@r-project.org Subject: [R] crr - computationally singular Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 1.28654e-24 This is obviously as a result of a problem with the data but apart from dataset a having 1674 rows and dataset b having 701 rows there is really no difference between them. The code I am using is as follows where covaea and covaeb are matrices of covarites, all coded as binary variables. In case a: covaea - cbind(sexa,fsha,fdra,nsigna,eega,th1a,th2a,stype1a,stype2a,stype3a,pgu 1a,pgu2a,log(agea),firstinta/1000,totsezbasea) fita - crr(snearma$with.Withtime,csaea,covaea,failcode=2,cencode=0) and in case b: covaeb - cbind(sexb,fshb,fdrb,nsignb,eegb,th1b,th2b,stype1b,stype2b,stype3b,sty pe4b,stype5b,pgu1b,pgu2b,(ageb/10)^(-1),firstintb,log(totsezbaseb)) fitb - crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) csaea and csaeb are the censoring indicators for a and b respectively which equal 1 for the event of interest, 2 for the competing risks event and 0 otherwise. Can anyone suggest a reason for the error message? I've tried running fitb with variants of covaeb and irrespective of the order of the covariates in the matrix, the code runs fine with 16 of the 17 covariates included but then produces an error message when the 17th is added. Thank you for your help, Laura __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories 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] Optimization and Linear Programming in R
On Fri, 26 Jun 2009, Liviu Andronic wrote: There is `solnp', part of the Integer and Nonlinear Optimization in R (RINO) project [1]. It is still in early beta, though. [1] http://r-forge.r-project.org/projects/rino/ I do not find it at the indicated location on: http://r-forge.r-project.org/R/?group_id=162 [herr...@centos-5 R2spec]$ wget \ http://r-forge.r-project.org/src/contrib/Rsolnp_0.1.tar.gz HTTP request sent, awaiting response... 404 Not Found 09:37:55 ERROR 404: Not Found. [herr...@centos-5 R2spec]$ -- Russ herrold __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 avoid ifelse statement converting factor to character
Stavros Macrakis wrote: It gives me a headache, too! I think you'll have to wait for a more expert user than me to supply explanations of these behaviors and their rationales. Thanks, Stavros. I hope someone with expertise will shed more light on this, and in the meantime I'll try to learn more from the manuals. Best regards, Craig -s On 6/26/09, Craig P. Pyrame crap...@gmail.com wrote: Stavros Macrakis wrote: On Thu, Jun 25, 2009 at 12:47 PM, Craig P. Pyramecrap...@gmail.com wrote: The man page Stavros quotes states that the class attribute of the result is taken from 'test', which clearly is not the case: Actually, the behavior is documented pretty clearly: The mode of the answer will be coerced from logical to accommodate first any values taken from 'yes' and then any values taken from 'no'. Whether this is a good design or not is another issue Perhaps the justification is that it avoids evaluating the yes or no arguments (to determine their class) in cases where their value is not needed. Thank you for pointing me to this. Now I get a headache from trying to figure out what does mode have to do with class - I thought that the class of the result should be that of test, and that the mode is something entirely different. Why does coercing the mode also affect the class? If the man page said The class attribute is taken from test, and it will be coerced ... or The mode of the result is taken from test, and it will be coreced ..., would this be wrong? What is the class-mode mixture about? Why does this fail: r = as.raw(TRUE) ifelse(TRUE, r, r) = error This gives an error which I take for saying that raw cannot be coerced to logical, but yes it can: as.logical(r) = TRUE and raw can even be used as the condition vector in ifelse: ifelse(r, 1, 2) = 1 Best regards, Craig Example: ifelse(c(T,F),1,a) = c(1,a) This has the same effect as res - c(T,F) res[1] - 1 res[2] - a which is in fact pretty much the way it is implemented. And also, I find myself incapable of making sense of the may in the mode of the result may depend on the value of 'test' - may in what sense? See the examples at the end of ? ifelse -s __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Deleting repeated rows
dreamworx wrote: Appologies if this is a simple problem. I have a matrix which is 86x3 and each row contains three integers. The problem I have is that some of the rows of integers are repeated in other rows and I only wish to have one copy of these rows. Is there a function that I can use which will identify these repeated rows and delete them from the matrix? I have searched for help with this problem but the only function I have found is 'grep' which I'm not sure is of any use. ?unique # generate data set.seed(1001) m - matrix(sample(1:3,20*3,TRUE),ncol=3);m # get unique values unique(m) As you can see, row 7 and 20 are 1,1,1... and you get only one row with 1,1,1 in the second matrix. -- View this message in context: http://www.nabble.com/Deleting-repeated-rows-tp24219989p24220573.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 I get just the two last tokens of each string in a vector?
I have the feeling that this is a very clumsy way to do it but I think it does
what you want.
=
a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % )
mylist - strsplit(a, )
pick - function(x) {tail(x,2) }
unlist(pick(mylist)
=
--- On Fri, 6/26/09, Fredrik Karlsson dargo...@gmail.com wrote:
From: Fredrik Karlsson dargo...@gmail.com
Subject: [R] How do I get just the two last tokens of each string in a vector?
To: R-Help List r-h...@stat.math.ethz.ch
Received: Friday, June 26, 2009, 7:21 AM
Dear list,
Sorry for asking this very silly question on the list, but
I seem to
have made my life complicated by going into string
manipulation in
vectors.
What I need is to get the last part of a sting (the two
last tokens,
separated by a space), and of course, this should be done
for all
strings in a vector, creating a new vector of exual size.
So,
a - c( %L H*L L*H H%, %L H* H%, %L L*H
%, %L L*H % )
should be made into a vector
c( L*H H%, H* H%, L*H %, L*H
% )
I have tried strsplit, but it seems to produce a structure
I cannot
get to work in this context. Any ideas on how to solve
this?
Thankful for all the help I can get.
/Fredrik
--
Life is like a trumpet - if you don't put anything into
it, you don't
get anything out of it.
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mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] what happened to the xlsReadWrite package
Thanks for the detailed clarification. Perhaps the page here http://cran.r-project.org/web/packages/xlsReadWrite/ could include these explanations as well. I imagine I'm not the only user with these questions! Thanks, Andrew On Fri, Jun 26, 2009 at 2:48 AM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote: On Fri, 26 Jun 2009, David Scott wrote: Andrew Yee wrote: A naive question: what happened to the xlsReadWrite package? http://cran.r-project.org/web/packages/xlsReadWrite/ It says that it was removed from the CRAN repository. Note that it was archived, not removed entirely. That is done with packages that no longer install and we don't get an update. The immediate problem with xlsReadWrite was that it was Windows-only and would not install under Windows in pre-2.9.0. Are there any plans for it be available again? The second issue was binary code in the package which is at least heavily frowned on in the CRAN source repository (we host BRugs elsewhere for that reason). There was a problem with proprietary code. Only in so far as it was binary. There are a number of CRAN packages with non-open-source code. Pick up the latest version from Hans-Peter Suter's website http://treetron.googlepages.com/ Which says the problem was the binary code. David Scott _ David Scott Department of Statistics The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 Graduate Officer, Department of Statistics Director of Consulting, Department of Statistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[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] How to create separate plots for all combinations of some factors
Hi, All I have a data frame as follows: data.class(tapes) [1] data.frame names(tapes) [1] dateloc classdrp datascratch reclaim total Date is a date; loc, class and drp are factors; the rest are numerics. I want to generate separate plots by date for the numeric variables for all combinations of the factors. I tried to do this as follows: by(tapes,list(loc,class,drp),plot(date,scratch)) Error in FUN(X[[1L]], ...) : could not find function FUN As well as the error I get a single plot for all values of scratch and date which is meaningless. What am I going wrong here? For clarification what I want would be done is SAS as something like: proc gplot; plot scratch*date; by loc class drp; quit; run; Jim Lane Capacity Planner RBC Financial Group 315 Front St W 6th Floor - H14 Toronto, Ontario CANADA M5V 3A4 416-348-6024 Display of superior knowledge is as great a vulgarity as display of superior wealth - greater indeed, inasmuch as knowledge should tend more definitely than wealth towards discretion and good manners. - H. W. Fowler, Modern English Usage ___ This e-mail may be privileged and/or confidential, and the sender does not waive any related rights and obligations. Any distribution, use or copying of this e-mail or the information it contains by other than an intended recipient is unauthorized. If you received this e-mail in error, please advise me (by return e-mail or otherwise) immediately. Ce courrier électronique est confidentiel et protégé. L'expéditeur ne renonce pas aux droits et obligations qui s'y rapportent. Toute diffusion, utilisation ou copie de ce message ou des renseignements qu'il contient par une personne autre que le (les) destinataire(s) désigné(s) est interdite. Si vous recevez ce courrier électronique par erreur, veuillez m'en aviser immédiatement, par retour de courrier électronique ou par un autre moyen. [[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] Matching data to a new column
x - I.D age 'MID' 01 5 03 02 6 06 03 16 NA 04 8 06 05 3 NA 06 17 NA xx - read.table(textConnection(x), header=TRUE); xx closeAllConnections() ag1 - xx[, c(1,2)] ; ag1 ag2 - xx[, c(1,3)] ; ag2 names(ag2[2]) - I.D merge(ag1,ag2, by=I.D) --- On Fri, 6/26/09, Kubasiewicz, Laura laura.kubasiewic...@imperial.ac.uk wrote: From: Kubasiewicz, Laura laura.kubasiewic...@imperial.ac.uk Subject: [R] Matching data to a new column To: r-help@r-project.org r-help@r-project.org Received: Friday, June 26, 2009, 6:55 AM Hi all, I wonder if anyone can help, I have a dataframe with columns for... 'I.D' 'age' 'mothers I.D' 01 5 03 02 6 06 03 16 NA 04 8 06 05 3 NA 06 17 NA I need to create a new column for 'mothers age' which puts the age of the individual with 'mothers i.d' into the row for her offspring (so individual 01 would have 16 in the mothers age column, as thats the age of individual 03) Hope that makes sense, any help appreciated, Thanks Laura [[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. __ Be smarter than spam. See how smart SpamGuard i ions in Mail and switch to New Mail today or register for free at http://mail.yahoo.ca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 I get just the two last tokens of each string in a vector?
HI Jim, Thank you! Works perfectly. /Fredrik On Fri, Jun 26, 2009 at 1:00 PM, jim holtmanjholt...@gmail.com wrote: This should do it a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) # split the strings x - strsplit(a, ' ') # get last two tokens sapply(x, function(.tokens) paste(tail(.tokens, 2), collapse=' ')) [1] L*H H% H* H% L*H % L*H % On Fri, Jun 26, 2009 at 7:21 AM, Fredrik Karlsson dargo...@gmail.com wrote: Dear list, Sorry for asking this very silly question on the list, but I seem to have made my life complicated by going into string manipulation in vectors. What I need is to get the last part of a sting (the two last tokens, separated by a space), and of course, this should be done for all strings in a vector, creating a new vector of exual size. So, a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) should be made into a vector c( L*H H%, H* H%, L*H %, L*H % ) I have tried strsplit, but it seems to produce a structure I cannot get to work in this context. Any ideas on how to solve this? Thankful for all the help I can get. /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] predicted values after fitting gamma2 function
On Jun 25, 2009, at 11:30 PM, Steven Matthew Anderson wrote: Question: after fitting a gamma function to some data, how do I get predicted values? I'm a SAS programmer, I new R, and am having problems getting my brain to function with the concept of object as class The following is specifics of what I am doing: I'm trying to determine the pdf from data I have created in a simulation. I have generated frequency counts using the following: Max.brks - pretty(range(Max.Spread$Distance), 100) Max.f-hist(x=Max.Spread$Distance, breaks=Max.brks,plot=FALSE ) Max.cnt-as.data.frame(cbind(sim,Max.f$mids,Max.f$counts)) colnames(Max.cnt)-c(Simulation,MidPoint,Count) then I fit this to a gamma distribution function: Using a non-base function without including the appropriate require() or library() call is a bit like asking a SAS programmer to debug code but not telling what PROC it's from; modl- vglm (Count ~ MidPoint ,gamma2 ,data = subset(Max.cnt,select=(simulation,MidPoint,Count),trace=TRUE,crit=c) print(coef(modl2,matrix=TRUE)) print(summary(modl2)) This produces the output: snipped output Now - how do I get this information to give me predicted values given the same x-values I used in the experimental model (i.e. from Max.brks - pretty(range(Max.Spread$Distance), 100)). Most regression functions in R, and vglm is no exception, have predict methods. The default is to give back predictions for the data from which the parameters were estimated, but if you want predictions on specific new values there is a newdata option. ?predict.vglm David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crr - computationally singular
How did you determine that you have full rank model matrix comprising 17 predictors? Are you able to invert the model matrix using `solve'? If not, you still have collinearity problem. If you are, then the problem might be in the Newton's method used by `crr' to solve the partial-likelihood optimization. The hessian matrix of the parameters might be singular during the iterations. If this is the case, your best bet would be to just simplify the model, i.e. use fewer predictors. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h tml -Original Message- From: Laura Bonnett [mailto:l.j.bonn...@googlemail.com] Sent: Friday, June 26, 2009 6:22 AM To: Ravi Varadhan Cc: r-help@r-project.org Subject: Re: [R] crr - computationally singular Dear Sir, Thank you for your response. You were correct, I had 1 linearly dependent column. I have solved this problem and now the rank of 'covaeb' is 17 (qr(covaeb)$rank = 17). However, I still get the same error message when I use covaeb in the 'crr' function. fit=crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) 8 cases omitted due to missing values Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 3.45905e-25 Are there any other reasons why this may be happening? Thank you, Laura 2009/6/25 Ravi Varadhan rvarad...@jhmi.edu: This means that your design matrix or model matrix is rank deficient, i.e it does not have linearly independent columns. Your predictors are collinear! Just take your design matrices covaea or covaeb with 17 predcitors and compute their rank or try to invert them. You will see the problem. Ravi. -- -- --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Vara dhan.h tml -- -- -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Laura Bonnett Sent: Thursday, June 25, 2009 11:39 AM To: r-help@r-project.org Subject: [R] crr - computationally singular Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) : system is computationally singular: reciprocal condition number = 1.28654e-24 This is obviously as a result of a problem with the data but apart from dataset a having 1674 rows and dataset b having 701 rows there is really no difference between them. The code I am using is as follows where covaea and covaeb are matrices of covarites, all coded as binary variables. In case a: covaea - cbind(sexa,fsha,fdra,nsigna,eega,th1a,th2a,stype1a,stype2a,stype3a,pg u 1a,pgu2a,log(agea),firstinta/1000,totsezbasea) fita - crr(snearma$with.Withtime,csaea,covaea,failcode=2,cencode=0) and in case b: covaeb - cbind(sexb,fshb,fdrb,nsignb,eegb,th1b,th2b,stype1b,stype2b,stype3b,st y pe4b,stype5b,pgu1b,pgu2b,(ageb/10)^(-1),firstintb,log(totsezbaseb)) fitb - crr(snearmb$with.Withtime,csaeb,covaeb,failcode=2,cencode=0) csaea and csaeb are the censoring indicators for a and b respectively which equal 1 for the event of interest, 2 for the competing risks event and 0 otherwise. Can anyone suggest a reason for the error message? I've tried running fitb with variants of covaeb and irrespective of the order of the covariates in the matrix, the code runs fine with 16 of the 17 covariates included but then produces an error message when the 17th is added. Thank you for your help, Laura __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
Re: [R] Calculating distance between spatial points
If your goal is to get the distance between geographic points, you might try computing the distance along the great circle arc and forget about projection. Several packages have functions that do this. See https://stat.ethz.ch/pipermail/r-help/2007-October/144546.html. hope that helps, Ian Tim Clark wrote: Dear List, I am trying to determine the speed an animal is traveling on each leg of a track. My data is in longitude and latitude, so I am using the package rgdal to convert it into a spatial points data frame and transform it to UTM. I would then like to find the difference between successive longitudes and latitudes, find the euclidean distance between points, and compute the speed of the animal on each leg. [ removed from quote ] Aloha, Tim Tim Clark Department of Zoology University of Hawaii -- View this message in context: http://www.nabble.com/Calculating-distance-between-spatial-points-tp24212875p24220923.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] changing the loss function in the logistic regression?
Hi all, Is there a way to change the loss function in the logistic regression? Or we could provide a customized loss function in the logistic regression so we could use that loss function in the Cross Validation in logistic regression? Thanks a lot! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] parallel R?
Hi all, Lots of big IT companies are renting out their computing facilities. Amazon has one such service. In my understanding, this will dramatically improve the speed of my R program -- currently the cross validation and model selection part is the bottle neck. It take a few days to just finish optimal parameter tuning via CV. Could anybody shed some lights on this? Is it completely transparent to move my R program to a rented parallel computing facility? Moreover, at my PC level, I have a 4-core PC, is there anything we could do in R to speed up my CV programs? Thanks a lot! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 I get just the two last tokens of each string in a vector?
One way is: a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) sub(^.*(^| )([^ ]+ [^ ]+$),\\2,a) [1] L*H H% H* H% L*H % L*H % Just be aware that this is not terribly efficient for very large strings. -s On Fri, Jun 26, 2009 at 7:21 AM, Fredrik Karlssondargo...@gmail.com wrote: Dear list, Sorry for asking this very silly question on the list, but I seem to have made my life complicated by going into string manipulation in vectors. What I need is to get the last part of a sting (the two last tokens, separated by a space), and of course, this should be done for all strings in a vector, creating a new vector of exual size. So, a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) should be made into a vector c( L*H H%, H* H%, L*H %, L*H % ) I have tried strsplit, but it seems to produce a structure I cannot get to work in this context. Any ideas on how to solve this? Thankful for all the help I can get. /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R 2.9.1 is released
I've rolled up R-2.9.1.tar.gz a few hours ago. This is a maintenance release and fixes a number of mostly minor issues. See the full list of changes below. You can get it from http://cran.r-project.org/src/base/R-2/R-2.9.1.tar.gz or wait for it to be mirrored at a CRAN site nearer to you. Binaries for various platforms will appear in due course. For the R Core Team Peter Dalgaard These are the md5sums for the freshly created files, in case you wish to check that they are uncorrupted: 70447ae7f2c35233d3065b004aa4f331 INSTALL 433182754c05c2cf7a04ad0da474a1d0 README 4f004de59e24a52d0f500063b4603bcb OONEWS ff4bd9073ef440b1eb43b1428ce96872 ONEWS 11a746c387c5a94f172a27939b96dd4c NEWS 7abcbbc7480df75a11a00bb09783db90 THANKS 070cca21d9f8a6af15f992edb47a24d5 AUTHORS a6f89e2100d9b6cdffcea4f398e37343 COPYING.LIB eb723b61539feef013de476e68b5c50a COPYING 020479f381d5f9038dcb18708997f5da RESOURCES 8ea34a9306ef5cf5dc03b45883586129 FAQ 54a79eebdf0cec3fd2c489fc94d99b00 R-2.9.1.tar.gz 54a79eebdf0cec3fd2c489fc94d99b00 R-latest.tar.gz Here is the relevant part of the NEWS file: CHANGES IN R VERSION 2.9.1 NEW FEATURES o New function anyDuplicated(x) returns 0 (= FALSE) or the index of the first duplicated entry of x. o matplot(), matlines() and matpoints() now also obey a 'lend' argument, determining line end styles. (Wish of PR#13619). o bw.SJ(), bw.bcv() and bw.ucv() now gain an optional 'tol' argument allowing more accurate estimates. o new.packages() no longer regards packages with the same name as a member of an installed bundle as 'new' (this is now consistent with the dependency checks in install.packages()). It no longer reports on partially installed bundles (since members can be updated individually if a bundle is unbundled). o old.packages() and hence updates.packages() will look for updates to members of package bundles before updates to the whole bundle: this allow bundles to be split and installations updated. o nlminb() gives better non-convergence messages in some cases. o S3 method dispatch will support S4 class inheritance for S3 methods, for primitives and via UseMethod(), if the argument S3methods=TRUE is given to setClass(). S4 method dispatch will use S3 per-object inheritance if S3Class() is set on the object. See ?Methods and the paper referenced there. o R CMD INSTALL is more tolerant of (malformed) packages with a 'man' directory but no validly named .Rd files. o R CMD check now reports where options are used that cause some of the checks to be skipped. o RSiteSearch has been updated to be consistent with the new layout of the search site itself, which now includes separate options for vignettes, views, and r-sig-mixed-models, as well as changed names for r-help. (Contributed by Jonathan Baron.) o That R CMD check makes use of a pkg/tests/Examples/pkg-Ex.Rout.save file as a reference result is now documented in 'Writing R Extensions'. DEPRECATED DEFUNCT o print.atomic() (defunct since 1.9.0) has been removed since it caused confusion for an S4 class union atomic. o png(type=cairo1) is deprecated -- it was only needed for platforms with 1.0 = cairo 1.2. BUG FIXES o The ... argument was not handled properly when ... was found in the enclosure of the current function, rather than in the function header itself. (This caused integrate() to fail in certain cases.) o col2rgb(#0080, TRUE) would return the background colour. (Reported by Hadley Wickham.) o interaction() now ensures that the levels of the result are unique. o packageDescription() and hence sessionInfo() now report the correct package version also for a non-attached loaded namespace of a version different from the default lib.loc. o smoothScatter() now also works when e.g. xlim[2] xlim[1]. o Invalid use of sprintf() such as sprintf(%S%) now give an error instead of a segmentation fault, as do very unusual cases such as sprintf(%s, tryCatch(stop(), error=identity)). (It was always documented that misuse could crash R in platform-dependent ways.) o parse_Rd() would mishandle braces when they occurred at the start of a line within an R string in an Rd file (reported by Alex Couture-Beil) or when they occurred in an R comment (reported by Mark Bravington). o readNEWS() missed version numbers with more than one digit. o building R --without-x no longer fails (PR#13665) o printCoefmat(cbind(0,1)) now works too (PR#13677) o bw.SJ(c(1:99, 1e6)) now works too. o Rd2txt() could not handle empty descriptions of items in an Rd
[R] Anova with multiple responses without loop
Hello, I have one question about Anova in car package. Here is my situation: I have a response matrix (Y), consisting of 7 different responses for a group of subjects, 2 categorical factors (A, B), 3 covariates (C,D,E). I would like to know the main effects of 2 factors and the interaction between them on each of 7 responses. So, I used Linear regression (lm) and Anova with a model, Y[,i]~A+B+A:B+C+D+E in a loop. This worked as I expected it to work. However, I want to know how to perform the same task without using any loop to increase the performance. Basically, my question is How to perform multiples of LM and Anova without using a loop. If anyone know how to do this, would you advise me? Thank you in advance for your any help. -- Sungeun Kim [[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] changing the loss function in the logistic regression?
Michael wrote: Hi all, Is there a way to change the loss function in the logistic regression? Or we could provide a customized loss function in the logistic regression so we could use that loss function in the Cross Validation in logistic regression? Thanks a lot! The goal is to use a loss function that yields optimality, with a sensible definition of optimality. For many purposes, maximum likelihood or penalized maximum likelihood is optimum. So don't change the optimality criteria just because you are cross-validating a different measure. By the way, it's often not a good idea to cross-validate a different measure. At least the accuracy index should be information-preserving. Deviance, log-likelihood, and AIC are your friends. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matching data to a new column
Hi Laura, The function merge() works well, but another elegant to do it would involve match(). (Assuming your data.frame object is called x) x$motherAge - with(x, age[match(mothers.I.D,I.D)]) Cheers, -- *Luc Villandré* /Biostatistician McGill University Health Center - Montreal Children's Hospital Research Institute/ John Kane wrote: x - I.D age 'MID' 01 5 03 02 6 06 03 16 NA 04 8 06 05 3 NA 06 17 NA xx - read.table(textConnection(x), header=TRUE); xx closeAllConnections() ag1 - xx[, c(1,2)] ; ag1 ag2 - xx[, c(1,3)] ; ag2 names(ag2[2]) - I.D merge(ag1,ag2, by=I.D) --- On Fri, 6/26/09, Kubasiewicz, Laura laura.kubasiewic...@imperial.ac.uk wrote: From: Kubasiewicz, Laura laura.kubasiewic...@imperial.ac.uk Subject: [R] Matching data to a new column To: r-help@r-project.org r-help@r-project.org Received: Friday, June 26, 2009, 6:55 AM Hi all, I wonder if anyone can help, I have a dataframe with columns for... 'I.D' 'age' 'mothers I.D' 01 5 03 02 6 06 03 16 NA 04 8 06 05 3 NA 06 17 NA I need to create a new column for 'mothers age' which puts the age of the individual with 'mothers i.d' into the row for her offspring (so individual 01 would have 16 in the mothers age column, as thats the age of individual 03) Hope that makes sense, any help appreciated, Thanks Laura [[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. __ Be smarter than spam. See how smart SpamGuard i ions in Mail and switch to New Mail today or register for free at http://mail.yahoo.ca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] parallel R?
On 26 June 2009 at 07:40, Michael wrote: | Hi all, | | Lots of big IT companies are renting out their computing facilities. | Amazon has one such service. In my understanding, this will | dramatically improve the speed of my R program -- currently the cross | validation and model selection part is the bottle neck. It take a few | days to just finish optimal parameter tuning via CV. | | Could anybody shed some lights on this? Is it completely transparent | to move my R program to a rented parallel computing facility? | | Moreover, at my PC level, I have a 4-core PC, is there anything we | could do in R to speed up my CV programs? Try this link: http://lmgtfy.com/?q=parallel+computing+with+R A survey paper on parallel computing with R by Schmidberger et al will be fortcoming shortly in Journal of Statistical Software -- that is at http://jstatsoft.org and a Google query will lead to preprints. Also note that R has a support site network called 'CRAN' which contains so called 'Task Views'. Amnong these you may find this one of interest: http://cran.r-project.org/web/views/HighPerformanceComputing.html As to your initial question: yes, people do use R on the Amazon service. Dirk -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 I get just the two last tokens of each string in a vector?
Here is a similar solution using strapply in gsubfn. The pattern matches non-spaces, [^ ]+, followed by a space followed by non-spaces followed by end-of-string, $. library(gsubfn) strapply(a, [^ ]+ [^ ]+$, simplify = c) [1] L*H H% H* H% L*H % L*H % On Fri, Jun 26, 2009 at 7:21 AM, Fredrik Karlssondargo...@gmail.com wrote: Dear list, Sorry for asking this very silly question on the list, but I seem to have made my life complicated by going into string manipulation in vectors. What I need is to get the last part of a sting (the two last tokens, separated by a space), and of course, this should be done for all strings in a vector, creating a new vector of exual size. So, a - c( %L H*L L*H H%, %L H* H%, %L L*H %, %L L*H % ) should be made into a vector c( L*H H%, H* H%, L*H %, L*H % ) I have tried strsplit, but it seems to produce a structure I cannot get to work in this context. Any ideas on how to solve this? Thankful for all the help I can get. /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Converting S-plus project folders to R
Thanks for the reply. Sometimes R does crash (quits completely). Other times, it just hangs after giving an error box with some sort of exception code, even though it looks like it's still trying to execute the data.restore. -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: Friday, June 26, 2009 2:21 AM To: Bosley, David Cc: r-help@r-project.org Subject: Re: [R] Converting S-plus project folders to R dbosley wrote: I have many S-plus project folders that I need to convert to R workspaces. For the smaller project folders ( 200MB), using data.dump with oldStyle = T and data.restore (in the foreign package) within R seems to work fine. However, I have several project folders that are quite large (~ 4GB). When I use this procedure to try to convert these project folders, R always crashes when I perform the data.restore command. Does anyone have any suggestions? No suggestion, just to be sure: I hope R gives an error message rather than a crash? Best, Uwe Ligges Thanks, Dave Bosley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] parallel R?
I guess when we move to Amazon AWS, we have to rewrite the whole R programs? On Fri, Jun 26, 2009 at 8:05 AM, Dirk Eddelbuettele...@debian.org wrote: On 26 June 2009 at 07:40, Michael wrote: | Hi all, | | Lots of big IT companies are renting out their computing facilities. | Amazon has one such service. In my understanding, this will | dramatically improve the speed of my R program -- currently the cross | validation and model selection part is the bottle neck. It take a few | days to just finish optimal parameter tuning via CV. | | Could anybody shed some lights on this? Is it completely transparent | to move my R program to a rented parallel computing facility? | | Moreover, at my PC level, I have a 4-core PC, is there anything we | could do in R to speed up my CV programs? Try this link: http://lmgtfy.com/?q=parallel+computing+with+R A survey paper on parallel computing with R by Schmidberger et al will be fortcoming shortly in Journal of Statistical Software -- that is at http://jstatsoft.org and a Google query will lead to preprints. Also note that R has a support site network called 'CRAN' which contains so called 'Task Views'. Amnong these you may find this one of interest: http://cran.r-project.org/web/views/HighPerformanceComputing.html As to your initial question: yes, people do use R on the Amazon service. Dirk -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Modifying Sweave.sty to allow escapes with fancyvrb package in LaTeX
Dear Colleagues:
In an attempt to have things like # See page \pageref{this} inside
comments in R code chunks I have modified Sweave.sty as below. I have
followed fancyvrb's manual with regard to the use of the commandchars
argument. But when compiling with LaTeX (using attached test file) I
get a LaTeX error
(/usr/share/texmf-texlive/tex/latex/ae/t1aett.fd)
! Missing \endcsname inserted.
to be read again
\global
l.25 } else y - 4
The line in question is
\DefineVerbatimEnvironment{Sinput}{Verbatim}{formatcom=\lstset{fancyvrb=true},commandchars=\\\{\}}
Can anyone give me a pointer to how to properly change Sweave.sty to
allow it to use fancyvrb in this way? I am following the model in
SweaveListingUtils for having Sweave use the listings package to do
pretty-printing.
Thanks very much
Frank
Sweavel.sty:
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{Sweavel}{}
\RequirePackage{listings,ifthen,graphicx,fancyvrb,relsize}
\RequirePackage{ifthen}
\newboolean{swe...@gin}
\setboolean{swe...@gin}{true}
\newboolean{swe...@ae}
\setboolean{swe...@ae}{true}
\declareoption{nogin}{\setboolean{swe...@gin}{false}}
\declareoption{noae}{\setboolean{swe...@ae}{false}}
\ProcessOptions
\IfFileExists{upquote.sty}{\RequirePackage{upquote}}{}
\ifthenelse{\boolean{swe...@gin}}{\setkeys{gin}{width=0.8\textwidth}}{}%
\ifthenelse{\boolean{swe...@ae}}{%
\RequirePackage[T1]{fontenc}
\RequirePackage{ae}
}{}%
%\DefineVerbatimEnvironment{Sinput}{Verbatim}{fontshape=sl}
\DefineVerbatimEnvironment{Sinput}{Verbatim}{formatcom=\lstset{fancyvrb=true},commandchars=\\\{\}}
\DefineVerbatimEnvironment{Soutput}{Verbatim}{}
\DefineVerbatimEnvironment{Scode}{Verbatim}{fontshape=sl}
\newenvironment{Schunk}{}{}
\newcommand{\Sconcordance}[1]{%
\ifx\pdfoutput\undefined%
\csname newcount\endcsname\pdfoutput\fi%
\ifcase\pdfoutput\special{#1}%
\else\immediate\pdfobj{#1}\fi}
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
\documentclass{article}
\usepackage[nogin]{Sweavel}
\begin{document}
\lstloadlanguages{R}
\lstset{language=R,fancyvrb=true,escapechar=`,%
basicstyle=\smaller,commentstyle=\ttfamily\itshape\smaller,%
keywordstyle=\bf,%
showstringspaces=false,%
xleftmargin=4ex,literate={-}{{$\leftarrow$}}2{-}{{$\twoheadleftarrow$}}2 {~}{{$\sim$}}1 {=}{{$\leq$}}2 {=}{{$\geq$}}2 {^}{{$^\wedge$}}1,%
alsoother={$},alsoletter={.-},%
otherkeywords={!,!=,~,$,*,\,\%/\%,\%*\%,\%\%,-,-,/}}%
This is page \pageref{this}.\label{this}
\begin{Schunk}
\begin{Sinput}
b - 4
a - b # this is a test line
i - 2
if(i==3) { # another line, for y^2 \pageref{this}
y - 3^3
z - 'this string'
qqcat - y ~ pol(x,2)
} else y - 4
\end{Sinput}
\end{Schunk}
\end{document}
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and provide commented, minimal, self-contained, reproducible code.
[R] Determining if swap memory is turned off
In order to run some performance tests on optimization tools, I want to
be able
to avoid the use of swap memory. In *nix systems, at least Linux ones, I
can issue
a 'sudo swapoff -a' command and use just the RAM available. If I don't
do this, at
some point swap will be used, the disk goes ballistic and the machine is
unresponsive
because it is thrashing data to the swap. I suspect others have
encountered this
too. Clearly I'd rather fail out than get into the thrashing situation.
Generally I
prefer no swap -- memory is cheap enough that it is easier not to worry
about these
sorts of woes. However, I'd like to make packages I'm developing
bullet-proof.
What I need to know is if there are similar facilities on other
platforms and how to
use them. I see memory.size and memory.limit for Windows -- making R
less platform
independent -- and they seem to work on a WinXP virtual machine I have
available.
However, I suspect that the memory.limit includes the swap. When I was
running
Windoze, I tried valiantly to get the PageFile.sys killed and found it
kept being set
up again despite following all the recommended steps. (I needed to avoid
the
plaintext of an encrypted file being saved for me!) I know nothing
about Mac
swap and how to control it.
Advice welcome. Below is a crude but self-contained example.
John Nash
## memcrash.R -- try to hit the wall with memory
cat(memcrash.R -- try to hit the wall\n)
cat(You should have swap turned off\n)
temp-readline(Do you?)
if (temp != y) stop(Fix swap)
for (j in 1:10) {
n - 5^j
cat(building matrix of order ,n,\n)
A-matrix(nrow=n, ncol=n)
} # end loop
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to create separate plots for all combinations of some factors
You did not supply any executable examples for testing, but perhaps
you could benefit by looking at:
?expand.grid
?tapply
Those to split or index your data for all combinations of factors.
?mapply # to do the plotting.
On Jun 26, 2009, at 9:54 AM, Lane, Jim wrote:
Hi, All
I have a data frame as follows:
data.class(tapes)
[1] data.frame
names(tapes)
[1] dateloc classdrp datascratch
reclaim total
Date is a date; loc, class and drp are factors; the rest are numerics.
I want to generate separate plots by date for the numeric variables
for
all combinations of the factors. I tried to do this as follows:
by(tapes,list(loc,class,drp),plot(date,scratch))
Error in FUN(X[[1L]], ...) : could not find function FUN
Welcome to functional programming. Generally with R when you pass
arguments, you need to create a function that receives those
arguments, unless you happen to know that the proper objects would be
coming to plot (and then you would not use (date, scratch). Perhaps
(ObWAG, untested, complete guesswork) :
by(tapes, list(loc,class,drp), function(x) { with(x,
plot(date,scratch))} )
(I generally try tapply() first rather than using by().)
# What gets passed to the function is not named date and scratch
in any case, but is named whatever you use in the function argument
list. You need to pull that object apart properly within the function.
As well as the error I get a single plot for all values of scratch and
date which is meaningless. What am I going wrong here?
For clarification what I want would be done is SAS as something like:
An executable example would have been a better way of getting an
accurate answer.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting S-plus project folders to R
Bosley, David wrote: Thanks for the reply. Sometimes R does crash (quits completely). Other times, it just hangs after giving an error box with some sort of exception code, even though it looks like it's still trying to execute the data.restore. - Hmmm, that should not happen. Can you generate some reproducible example (e.g. by generating some random numbers in some file?)? - I guess we are talking about R-2.9.0? Today's release of R-2.9.1 won't fix it probably, but it is worth to try at least. Best wishes, Uwe Ligges -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: Friday, June 26, 2009 2:21 AM To: Bosley, David Cc: r-help@r-project.org Subject: Re: [R] Converting S-plus project folders to R dbosley wrote: I have many S-plus project folders that I need to convert to R workspaces. For the smaller project folders ( 200MB), using data.dump with oldStyle = T and data.restore (in the foreign package) within R seems to work fine. However, I have several project folders that are quite large (~ 4GB). When I use this procedure to try to convert these project folders, R always crashes when I perform the data.restore command. Does anyone have any suggestions? No suggestion, just to be sure: I hope R gives an error message rather than a crash? Best, Uwe Ligges Thanks, Dave Bosley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Determining if swap memory is turned off
On windows XP | start | Control Panel | System | Advanced |
Performance | Settings | Advanced | Virtual Memory | and select No
Pagefile. I suspect Vista is similar.
John
2009/6/26 Prof. John C Nash nas...@uottawa.ca:
In order to run some performance tests on optimization tools, I want to be
able
to avoid the use of swap memory. In *nix systems, at least Linux ones, I can
issue
a 'sudo swapoff -a' command and use just the RAM available. If I don't do
this, at
some point swap will be used, the disk goes ballistic and the machine is
unresponsive
because it is thrashing data to the swap. I suspect others have encountered
this
too. Clearly I'd rather fail out than get into the thrashing situation.
Generally I
prefer no swap -- memory is cheap enough that it is easier not to worry
about these
sorts of woes. However, I'd like to make packages I'm developing
bullet-proof.
What I need to know is if there are similar facilities on other platforms
and how to
use them. I see memory.size and memory.limit for Windows -- making R less
platform
independent -- and they seem to work on a WinXP virtual machine I have
available.
However, I suspect that the memory.limit includes the swap. When I was
running
Windoze, I tried valiantly to get the PageFile.sys killed and found it kept
being set
up again despite following all the recommended steps. (I needed to avoid the
plaintext of an encrypted file being saved for me!) I know nothing about
Mac
swap and how to control it.
Advice welcome. Below is a crude but self-contained example.
John Nash
## memcrash.R -- try to hit the wall with memory
cat(memcrash.R -- try to hit the wall\n)
cat(You should have swap turned off\n)
temp-readline(Do you?)
if (temp != y) stop(Fix swap)
for (j in 1:10) {
n - 5^j
cat(building matrix of order ,n,\n)
A-matrix(nrow=n, ncol=n)
} # end loop
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
John C Frain
Economics Department
Trinity College Dublin
Dublin 2
Ireland
www.tcd.ie/Economics/staff/frainj/home.html
mailto:fra...@tcd.ie
mailto:fra...@gmail.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Placing text outside graph boundary
If you are using base graphics: Create an outer margin using something like: par(oma=c(0,0,3,0)) Then use mtext with the outer argument to place the text in the outer margin. 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 Payam Minoofar Sent: Thursday, June 25, 2009 5:51 PM To: r-help@r-project.org Subject: [R] Placing text outside graph boundary Hello, Is possible to place a label on a multipanel figure outside the boundaries of any single one of the existing graphs in a panel? Specifically, I have a function that creates a panel with quartz() and then fills it with 6 graphs consisting of exactly the same plot for 6 different groups. I would like to add one title for the entire figure at the top of the figure, but I cannot figure out how to do that without anchoring the title one of the graphs. Is it possible to do that? Thank you very much. Payam -- Payam Minoofar, Ph.D. Scientist Meissner Filtration Products 4181 Calle Tesoro Camarillo, CA 93012 USA +1 805 388 9911 +1 805 388 5948 fax payam.minoo...@meissner.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] GeoXp package
Hi, I resolved partially my problem, by adding the path of the grass library where is located libgrass_I.so. So I add the following line to my .bashrc file: export LD_LIBRARY_PATH=$LD_LIBRARY_PATH:/usr.lib/grass/lib Now, I can install and load rgdal and GeoXp libraries from R. Thanks for your help. Emmanuel Poizot Roger Bivand wrote: The problem is that the binary Ubuntu install of gdal has not made sure that its dependencies are properly satisfied, so you'll need to do this manually. You can also make sure that the directory the GRASS libraries are found in is seen by ldconfig, either that or making sure that your running R sees LD_LIBRARY_PATH. This is a very detailed question, more suited to GRASS or GDAL lists, or R-sig-geo. You'll see when you have the solution because the GDAL utilities will work at the shell, for example gdalinfo --formats - I think that it doesn't at the moment. Roger Bivand epoizot wrote: Zeljko Vrba wrote: On Wed, Jun 10, 2009 at 08:21:06AM +0200, Poizot Emmanuel wrote: Error in fun(...) : GDAL Error 1: libgrass_I.so: Ne peut ouvrir le fichier d'objet partagé: Aucun fichier ou dossier de ce type (sorry for the french :) ) It would have been far more useful had you translated the error message to english than to have apologized. Try doing export LD_LIBRARY_PATH=/path/to/dir/where/so/is/located before starting R from the shell. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Hi, I tried this but with no success. Still enable to load the library rgdal and so GeoXp. This problem of for me now of high importance as I need to export interpolated data sets to grids. and rgdal allows that kind of operations. Regards -- View this message in context: http://www.nabble.com/GeoXp-package-tp23964536p24219649.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] calculate AIC
Dear all,  I want to calculate AIC values of PLSR models. But I find that AIC and extractAIC functions in R could not be used to calculate AIC values of PLSR models. Now I write a section of code(below) to calculate it. But I don't known whether the result is right or not. If I am wrong, please give me some suggestions. Thanks a lot.  Rong Huang  data-data.frame( y=c(-1,-1,-0.9,-0.88,-0.85,-0.41,-0.38,-0.04,0,0.1,0.23,0.3,0.32, 0.42,0.43,0.48,0.77,0.82,0.82,0.89,0.92,1.02,1.03,1.07,1.13,1.36,1.36,1.4,1.41, 1.55,1.84), X1=c(-0.101,-0.1629,-0.1004,-0.4688,-0.1005,-0.1006,-0.098,-0.1478,-0.5146, -0.1611,-0.1619,-0.1442,-0.2566,-0.1308,-0.16,-0.1634,-0.1623,-0.1469,-0.148, -0.1628,-0.161,-0.1618,-0.1619,-0.1612,-0.1603,-0.161,-0.0938,-0.1616,-0.1628, -0.1611,-0.076), X2=c(-1.6953,-0.9866,-1.7833,-1.1233,-1.7022,-1.7001,-1.6017,-0.8415,-0.9508, -0.9389,-1.0148,-0.861,-1.3785,-1.6241,-1.0883,-2.0914,-1.5157,-0.7912,-0.8419, -0.8149,-0.7935,-0.8606,-0.8699,-0.7949,-0.7832,-0.824,-1.0875,-0.8313,-0.8149, -0.7949,-1.0338), X3=c(250.4,307.8,396.2,342.3,380.0,282.8,328.0,405.0,353.6,304.8,280.4, 353.6,328.8,328.8,328.3,216.2,328.5,320.6,372.6,340.2,332.0,321.6,321.6, 336.1,333.8,332.3,387.1,320.6,340.2,354.0,313.8), X4=c(3.007,3.686,7.23,5.73,7.239,4.06 5,5.96,9.52,5.67,4.888,5.354,5.681, 7.372,7.372,6.811,4.654,6.18,6.994,8.47,8.466,6.113,6.695,6.695,7.269,6.205, 6.838,9.3,6.994,7.408,7.869,6.755), X5=c(0.25,0.67,0.28,0.52,0.25,0.25,0.25,0.67,0.54,0.67,0.67,0.54,0.2,0.2, 0.67,0.67,0.67,0.67,0.67,0.67,0.67,0.67,0.67,0.67,0.67,0.67,1.08,0.67,0.67, 0.67,0.51) ) ncomp-5 model-plsr(y~ .,data=data,ncomp=ncomp,validation=LOO,x=TRUE,y=TRUE,model=TRUE) y-model$y y.hat-model$fitted.value[,,ncomp] residuals-model$residuals[,,ncomp] sigmaML-sqrt(mean(residuals^2)) loglik-sum(dnorm(y,y.hat,sigmaML,log=T)) AIC--2*loglik+2*(ncomp+1) ___ 好ç©è´ºå¡çä½ åï¼é®ç®±è´ºå¡å ¨æ°ä¸çº¿ï¼ [[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] Deleting repeated rows
Appologies if this is a simple problem. I have a matrix which is 86x3 and each row contains three integers. The problem I have is that some of the rows of integers are repeated in other rows and I only wish to have one copy of these rows. Is there a function that I can use which will identify these repeated rows and delete them from the matrix? I have searched for help with this problem but the only function I have found is 'grep' which I'm not sure is of any use. -- View this message in context: http://www.nabble.com/Deleting-repeated-rows-tp24219989p24219989.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] parallel R?
losemind wrote: Moreover, at my PC level, I have a 4-core PC, is there anything we could do in R to speed up my CV programs? I have seen one very nice paper that compared parallelization options for R: http://epub.ub.uni-muenchen.de/8991/ losemind wrote: we have to rewrite the whole R programs? I would like to disclaim that I haven't had the time to parallelize any of my R scripts. That being said, I noticed that many of the packages above implement parallelized versions of the apply() functions. Maybe taking advantage of apply wherever possible is a way to ensure you can get some benefit from parallelization without rewriting large portions of code. Hope this helps! -Charlie - Charlie Sharpsteen Undergraduate Environmental Resources Engineering Humboldt State University -- View this message in context: http://www.nabble.com/parallel-R--tp24221547p24223277.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] Optimization and Linear Programming in R
One option is to use optim with method='SANN' which is simulate annealing. The parameter vector is your vector of 0's and 1's, the fn argument is the function that you are trying to maximize with respect to the par vector (sounds like you have that pretty much worked out), then the gr argument is a function that will randomly change the par vector, here is where you put in the budget constraint, only do changes that meet the budget. There are a couple of control parameters that you need to set, see the last set of examples to optim. 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 chris.wil...@csiro.au Sent: Friday, June 26, 2009 1:24 AM To: r-help@r-project.org Subject: [R] Optimization and Linear Programming in R Dear List, My student and I are looking for an optimizer for a nonlinear optimization problem we are working on. The problem we are working on is to try to pick a set of islands on which to eradicate rats for seabird conservation. We have about 50 islands, each of which has some subset of 17 seabird species. Rats are present on all islands, and will cause the seabirds to go extinct unless they are removed. Removing rats on islands costs different amounts depending on which island is chosen. The decision problem is to pick the set of islands which will save the most seabird species within a given budget. Here we use a nonlinear function to calculate the objective. The function is the proportion of individuals of each species saved divided by the total number of birds, times the log of the same number, summed up across all of the islands that are chosen to be in the set that is saved. We have a single constraint, which is a budget in our case. The contribution to this constraint is the product of the cost of each island times the decision vector. Our decision variable is a binary vector (1 for removing rats from an island, 0 for not removing them). The objective function is as I have explained above. We are looking for a solver that can deal with this nonlinear integer programming problem. We looked at a number of packages on the CRAN Task View: Optimization and Mathematical Programming, however, we have not been able to locate one that will suit our purposes. The essential features are we need to be able to write a nonlinear function for the objective (hopefully a self contained one as we need to include some data in it), we need to be able to use a binary decision (or parameter) vector, and we need to be able to use a constraint. Any suggestions as to packages or other software that will work for our problem would be much appreciated. Thanks, Chris Wilcox and Greg Thonier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: Deleting repeated rows
?unique as an example mat-matrix(c(1,2,3,1,1,2,1,2,3,4,7,5), ncol=3, byrow=T) mat #rows 1 and 3 are identical [,1] [,2] [,3] [1,]123 [2,]112 [3,]123 [4,]475 unique(mat) [,1] [,2] [,3] [1,]123 [2,]112 [3,]475 Stefano -Messaggio originale- Da: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]per conto di dreamworx Inviato: venerdì 26 giugno 2009 15.10 A: r-help@r-project.org Oggetto: [R] Deleting repeated rows Appologies if this is a simple problem. I have a matrix which is 86x3 and each row contains three integers. The problem I have is that some of the rows of integers are repeated in other rows and I only wish to have one copy of these rows. Is there a function that I can use which will identify these repeated rows and delete them from the matrix? I have searched for help with this problem but the only function I have found is 'grep' which I'm not sure is of any use. -- View this message in context: http://www.nabble.com/Deleting-repeated-rows-tp24219989p24219989.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. Rispetta l'ambiente: Se non ti è necessario, non stampare questa mail. Le informazioni contenute nel presente messaggio di posta elettronica e in ogni suo allegato sono da considerarsi riservate e il destinatario della email è l'unico autorizzato ad usarle, copiarle e, sotto la propria responsabilità, divulgarle. Chiunque riceva questo messaggio per errore senza esserne il destinatario deve immediatamente rinviarlo al mittente cancellando l'originale. Eventuali dati personali e sensibili contenuti nel presente messaggio e/o suoi allegati vanno trattati nel rispetto della normativa in materia di privacy ( DLGS n.196/'03). __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] panel.text and saving to pdf
On Fri, Jun 26, 2009 at 1:53 AM, Dieter
Mennedieter.me...@menne-biomed.de wrote:
Willem Vervoort w.vervoort at usyd.edu.au writes:
I am not sure what I am doing wrong, but I have some unexplained behaviour
when saving a lattice graph
including text to a pdf file. The text seems to move around. It must have
something to do with the way
coordinates are set in devices other than jpg.
today - format(Sys.Date(),%Y%m%d)
x - runif(500)
y - rnorm(500)
foo - data.frame(x = x, y = y, z = rep(c(a,b),250))
require(lattice)
xyplot(x~y|z,data=foo)
panel.text(370,470,silly graph,cex=1.2,font=2)
Take this plot, resize its window: you will note that the silly graph moves
relative to the graphics. The same is true for the higher-resolution pdf
output.
Yes, and that this even works at all is an accident of implementation.
Once a lattice graph is drawn, the only officially supported way to
add to it is to use trellis.focus() (or it's underlying grid
equivalent downViewport()), e.g.;
xyplot(1:10 ~ 1:10)
trellis.focus(toplevel) ## has coordinate system [0,1] x [0,1]
panel.text(0.5, 0.2, silly graph, cex = 1.2, font = 2)
trellis.unfocus()
If you know beforhand what you want to add, another option is to use
the page argument:
xyplot(1:10 ~ 1:10,
page = function(n) {
panel.text(0.5, 0.2, silly graph, cex = 1.2, font = 2)
})
-Deepayan
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[R] 95% confidence interval coverage
Hi R-users, I would like to compute 95%CI coverage for each bootstrapped sample. But, I don't know how to get 95% CI of estimate for each bootstrapped sample. I could only see the estimate of each bootstrapped sample. Thanks, Becky __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] parallel R?
On Fri, Jun 26, 2009 at 8:28 AM, Michaelcomtech@gmail.com wrote: I guess when we move to Amazon AWS, we have to rewrite the whole R programs? Not necessarily. I use foreach (currently available in our REvolution R Enterprise distribution and coming very soon to CRAN), and test out the parallel code on my dual-core laptop. When I'm ready to run the big job, I just launch (say) 10 instances on Amazon EC2, use joinSleigh() to hook them together, and then just run the same foreach code. Now the same code is running on 10 distributed machines, no rewrite necessary. There's an extended example using foreach here: http://blog.revolution-computing.com/2009/05/parallelized-backtesting-with-foreach.html # David Smith -- David M Smith da...@revolution-computing.com Director of Community, REvolution Computing www.revolution-computing.com Tel: +1 (206) 577-4778 x3203 (San Francisco, USA) Check out our upcoming events schedule at www.revolution-computing.com/events __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a list with a list
Dear list, I have two lists: one with data and one with TRUE/FALSE values for data I want to further analyze (see example below). I have been able to use a for loop to extract the data that I want to keep, but think that there probably exists a way to do it without a loop. Any ideas? #sample data set.seed(100) example = list(letters[1:10], letters[1:10], letters[1:10]) ind = list(as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE))) #loop method for (i in 1:length(example)) print(example[[i]][ind[[i]]]) #result [1] c g i [1] a b e f j [1] a b c d g h i Thank you for the help. Greg -- Greg Hirson ghir...@ucdavis.edu Graduate Student Agricultural and Environmental Chemistry 1106 Robert Mondavi Institute North One Shields Avenue Davis, CA 95616 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a list with a list
Your example has one more element in 'ind' than 'example' and that is what causes the error, but this should be close. #sample data set.seed(100) example = list(letters[1:10], letters[1:10], letters[1:10]) ind = list(as.logical(sample(0:1, 10, rep=TRUE)), +as.logical(sample(0:1, 10, rep=TRUE)), +as.logical(sample(0:1, 10, rep=TRUE)), +as.logical(sample(0:1, 10, rep=TRUE))) mapply(function(a,b)a[b], example, ind) [[1]] [1] c g i [[2]] [1] a b e f j [[3]] [1] a b c d g h i [[4]] [1] b d e f h i Warning message: In mapply(function(a, b) a[b], example, ind) : longer argument not a multiple of length of shorter On Fri, Jun 26, 2009 at 2:27 PM, Greg Hirson ghir...@ucdavis.edu wrote: Dear list, I have two lists: one with data and one with TRUE/FALSE values for data I want to further analyze (see example below). I have been able to use a for loop to extract the data that I want to keep, but think that there probably exists a way to do it without a loop. Any ideas? #sample data set.seed(100) example = list(letters[1:10], letters[1:10], letters[1:10]) ind = list(as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE))) #loop method for (i in 1:length(example)) print(example[[i]][ind[[i]]]) #result [1] c g i [1] a b e f j [1] a b c d g h i Thank you for the help. Greg -- Greg Hirson ghir...@ucdavis.edu Graduate Student Agricultural and Environmental Chemistry 1106 Robert Mondavi Institute North One Shields Avenue Davis, CA 95616 __ R-help@r-project.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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] Indexing a list with a list
Jim, Thank you for your help. I had a feeling it would be an apply function that could help. Kind regards, Greg jim holtman wrote: Your example has one more element in 'ind' than 'example' and that is what causes the error, but this should be close. #sample data set.seed(100) example = list(letters[1:10], letters[1:10], letters[1:10]) ind = list(as.logical(sample(0:1, 10, rep=TRUE)), +as.logical(sample(0:1, 10, rep=TRUE)), +as.logical(sample(0:1, 10, rep=TRUE)), +as.logical(sample(0:1, 10, rep=TRUE))) mapply(function(a,b)a[b], example, ind) [[1]] [1] c g i [[2]] [1] a b e f j [[3]] [1] a b c d g h i [[4]] [1] b d e f h i Warning message: In mapply(function(a, b) a[b], example, ind) : longer argument not a multiple of length of shorter On Fri, Jun 26, 2009 at 2:27 PM, Greg Hirson ghir...@ucdavis.edu mailto:ghir...@ucdavis.edu wrote: Dear list, I have two lists: one with data and one with TRUE/FALSE values for data I want to further analyze (see example below). I have been able to use a for loop to extract the data that I want to keep, but think that there probably exists a way to do it without a loop. Any ideas? #sample data set.seed(100) example = list(letters[1:10], letters[1:10], letters[1:10]) ind = list(as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE))) #loop method for (i in 1:length(example)) print(example[[i]][ind[[i]]]) #result [1] c g i [1] a b e f j [1] a b c d g h i Thank you for the help. Greg -- Greg Hirson ghir...@ucdavis.edu mailto:ghir...@ucdavis.edu Graduate Student Agricultural and Environmental Chemistry 1106 Robert Mondavi Institute North One Shields Avenue Davis, CA 95616 __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html http://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? -- Greg Hirson ghir...@ucdavis.edu Graduate Student Agricultural and Environmental Chemistry 1106 Robert Mondavi Institute North One Shields Avenue Davis, CA 95616 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a list with a list
You have four elements in ind and three elements in example, then, try this
mapply('[', example, ind[1:3])
On Fri, Jun 26, 2009 at 3:27 PM, Greg Hirson ghir...@ucdavis.edu wrote:
Dear list,
I have two lists: one with data and one with TRUE/FALSE values for data I
want to further analyze (see example below). I have been able to use a for
loop to extract the data that I want to keep, but think that there probably
exists a way to do it without a loop. Any ideas?
#sample data
set.seed(100)
example = list(letters[1:10], letters[1:10], letters[1:10])
ind = list(as.logical(sample(0:1, 10, rep=TRUE)), as.logical(sample(0:1,
10, rep=TRUE)), as.logical(sample(0:1, 10, rep=TRUE)),
as.logical(sample(0:1, 10, rep=TRUE)))
#loop method
for (i in 1:length(example)) print(example[[i]][ind[[i]]])
#result
[1] c g i
[1] a b e f j
[1] a b c d g h i
Thank you for the help.
Greg
--
Greg Hirson
ghir...@ucdavis.edu
Graduate Student
Agricultural and Environmental Chemistry
1106 Robert Mondavi Institute North
One Shields Avenue
Davis, CA 95616
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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[R] (performance) time in Windows vs Linux
Hi, all.
I began to migrate my R codes from Windows to Linux and surprised me
with an old question. I simplified the problem and made a little test to
compare times at same
computer and the Linux time is worse (not so little) than Windows time:
28 vs 53 seconds.
I make an example (below) to facilitate all to see the difference.
I also build from source (it's my first time) a version of R to compare with
the distributed (compiled) R version. The times are similar to the other Linux
version.
I supposed R on Linux should be faster (32 and 64 bit) than windows version. Is
this difference because 64 bit R version is slower than 32 bits one? I started
the machine in both sittuations and checked free memory.
Tecnichal details:
Machine: Intel Core 2 Duo DDR2 4 Gb RAM
Windows version: XP Professional - 32 bits
R version: 2.9* binaries
Linux version: Ubuntu 8* (Hardy) - 64 bits
R version: 2.9* binaries and 2.9* compiled from source
Thanks to all,
Cezar Freitas
#code
N = 5
n = 15000
#makes data
dad = as.data.frame(cbind(sample(N,N,replace=FALSE), rpois(N,30)))
names(dad) = c(id,age)
aux = as.data.frame(cbind(sample(N,n,replace=FALSE), round(runif(n),4)))
names(aux) = c(id,score)
#calculates time
set.seed(790) #to be equal to everyone
system.time({
dad$score = 0
subdad = subset(dad, id%in%aux$id)
for(k in 1:(dim(subdad)[1])){
temp = aux$score[aux$id==subdad$id[k]]
if(length(temp)) subdad$score[k] = temp
}
})
#windows time
# user system elapsed
# 27.81 0.00 27.82
#linux usual compilation time
# user system elapsed
# 52.635 0.016 52.748
#linux (my compilation) time
# user system elapsed
# 52.567 0.016 52.588
#==END OF CODE
Veja quais são os assuntos do momento no Yahoo! +Buscados
http://br.maisbuscados.yahoo.com
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[R] 50993 point distance matrix, too big to as.matrix, looking for another way to calculate point-level summary
Hello, Im working on a 50933 point count bird abundance dataset. I've succeeded in calculating a distance matrix for this entire set, but I don't have sufficient memory to convert this to a matrix, as below... abun.dist - dist(abun.mat[1:50993,1:235) test - rowMeans(as.matrix(abun.dist)) Error in matrix(0, size, size) : too many elements specified ive been able to run a hclust() clustering procedure, due to the fact that hclust() makes a call to fortran code, but id like to be able to generate a calinski index for each of the clusters to assess the validity. Unfortunately, all the validation routines I have found are all native R code, and usually call as.matrix, resulting in the same error i receive above. What I'd like to figure out is how to just go through, one point at a time, and calculate the values i need. But I've been unable to come up with code to call the correct positions in the dist vector, can anyone suggest some code that might do this? Thanks... ...leif -- -- First they ignore you, then they laugh at you, then they fight you, then you win - Mohandas Gandhi [[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] Automatically placing a legend in an area with the most white space...
At one point I believe I heard of an R package that would automatically find the most empty space in a plot, and then that answer could then be used to intelligently place a legend. I would like to try to apply that R package to the contrived example shown below, so thank you for any hints or tips that can be provided. x = seq(0, 1000, by = 1) y1_vals-rnorm(1000, mean = 0, sd = 50) y2_vals-rnorm(1000, mean = 25, sd = 40) y3_vals-rnorm(1000, mean = 115, sd = 40) plot(x_vals, y1_vals, pch=22, col=dark red, bg = dark red, cex =0.3) points(x_vals, y2_vals, pch=22, col=dark blue, bg = dark blue, cex =0.3) points(x_vals, y3_vals, pch=22, col=dark green, bg = dark green, cex =0.3) abline(h=0, col=dark red, lwd=1.5, lty=6) abline(h=25, col=dark blue, lwd=1.5, lty=6) abline(h=115, col=dark green, lwd=1.5, lty=6) legend(topright, c(Best Guess 1 , Best Guess 2, Best Guess 3), bg=white, lwd = 2, title = Randomness:) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] (performance) time in Windows vs Linux
On Fri, Jun 26, 2009 at 12:23:35PM -0700, Cézar Freitas wrote: I supposed R on Linux should be faster (32 and 64 bit) than windows version. Is this difference because 64 bit R version is slower than 32 bits one? I started the machine in both sittuations and checked free memory. I suspect that the compiler is to blame. Download Intel's C and C++ compiler for linux (it is free for personal use), try to compile R with it, and see what results you get (and report them here!). Of course, if you have the time and are willing to tinker :) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] gradient fill of a grid.polygon
Dear list,
Following a recent enquiry, I've been playing with the idea of creating a
colour gradient for a polygon, using the Grid package. The idea is to draw a
number of stripes of different colours, using the grid.clip function. Below
is my current attempt at this,
library(grid)
rotate.polygon - function(g, angle=0){ # utility function, works fine
matR - matrix(c(cos(angle), -sin(angle), sin(angle), cos(angle)),
nrow = 2)
gravity.x = unit(mean(g$x),npc)
gravity.y = unit(mean(g$y),npc)
x.center = convertX(g$x - gravity.x ,npc,TRUE)
y.center = convertY(g$y - gravity.y ,npc,TRUE)
new.xy - matrix(c(x.center, y.center),
ncol=2) %*% matR
editGrob(g, x=unit(new.xy[,1],npc) + gravity.x,
y=unit(new.xy[,2],npc) + gravity.y)
}
gradient.polygon - function(g, n=100,
cols=colorRampPalette(c(#E41A1C, #377EB8, #4DAF4A, #984EA3))(n),
alpha=0.5,
stripe=FALSE, angle=0){
vp = viewport(angle = angle)
g = rotate.polygon(g, - angle)
gx - grobWidth(g)
gy - grobHeight(g)
dx - unit(convertX(gx, npc, valueOnly = TRUE)/(n-1), npc) # width of
the stripes
startx - min(g$x)
starty - min(g$y)
for(ii in seq(1, n)){
grid.clip(x= startx + (ii-1) * dx , y=starty,
width= 1.0*dx, # fudge factor of 1.2 seems needed to overlap well
height=gy,
just=bottom)
if(stripe){
if(ii%%2)# plotting only every other slice
grid.draw(editGrob(g, gp=gpar(fill=cols[ii], col=cols[ii], alpha=alpha)))
}else{
grid.draw(editGrob(g, gp=gpar(fill=cols[ii], col=cols[ii], alpha=alpha)))
}
}
}
g -
polygonGrob(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(fill=NA,
col=grey90))
g2 -
polygonGrob(x=c(0, 0.5, 1), y=c(0.5, 0, 0.5), gp=gpar(fill=NA,
col=grey90))
grid.rect(gp=gpar(fill=black))
grid.rect(y=1, gp=gpar(fill=white))
gradient.polygon(g)
gradient.polygon(g2)
Now, I have a (large) number of issues here,
1) the stripes don't exactly blend well, at least on the quartz device. I
can add a fudge factor for their width but then the overlap might be
visible if alpha is not unity.
2) a serious flaw is the rotation that's not working at the moment. My
initial thought was to rotate the grob (triangle here), paint it with the
gradient, and plot the result in a rotated viewport with opposite rotation
angle. I'm still hopeful it might work, but see 3).
3) each stripe is directly plotted, as opposed to saved as a grob. I tried
to use either a gList or a gTree to store the output of the for loop, but I
didn't understand why the result wouldn't appear on screen with grid.draw.
What's the best structure to hold together these different slices? It is
presumably this object that I should be plotting eventually in the right
orientation.
Sorry for the length of this email, I hope it's clear with the code above.
Best regards,
baptiste
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[R] Heteroskedasticity and Autocorrelation in SemiPar package
Hi all, Does anyone know how to report heteroskedasticity and autocorrelation-consistent standard errors when using the spm command in SemiPar package? Suppose the original command is sp1-spm(y~x1+x2+f(x3), random=~1,group=id) Any suggestion would be greatly appreciated. Thanks, Susan [[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] Alternate error structures in lme4?
Hi R users, The nlme library enabled several alternate error structures useful for longitudinal or repeated-measures data. For example, a continuous AR(1) process: model_2 = update(model_1, correlation = corCAR1(form = ~ time | subject)) Does anybody know if this is available in lme4? Thank you Ben -- View this message in context: http://www.nabble.com/Alternate-error-structures-in-lme4--tp24226643p24226643.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] (performance) time in Windows vs Linux
Hi there, I have both systems on a DELL 64bit machine. I compiled R 2.9.0 on both systems, to get 64bits capability. Surpriselly, on Linux (Ubuntu with I installed 3 month ago) I spent 41s to run the same test you did, and less time (35s) under Vista. In fact I had noticed that I not have gained time when running under linux (I had done jobs that run for several days). But somethings I gain with memory managment, because for some programs or steps, windows say that memory is full, while Linux run up to the end of the job. good luck milton On Fri, Jun 26, 2009 at 4:01 PM, Zeljko Vrba zv...@ifi.uio.no wrote: On Fri, Jun 26, 2009 at 12:23:35PM -0700, Cézar Freitas wrote: I supposed R on Linux should be faster (32 and 64 bit) than windows version. Is this difference because 64 bit R version is slower than 32 bits one? I started the machine in both sittuations and checked free memory. I suspect that the compiler is to blame. Download Intel's C and C++ compiler for linux (it is free for personal use), try to compile R with it, and see what results you get (and report them here!). Of course, if you have the time and are willing to tinker :) __ R-help@r-project.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] (performance) time in Windows vs Linux
Yes, under 64-bit it is sometimes slower and it highly depends on the
problem and the compiler you have. Note also that nobody managed to get
a 64-bit Windows R binary compiled with gcc so far.
Remember, 10 years ago there was the SUN Ultra Sparc III and above
architecture, and gcc was known to produce extremely inefficient 64-bit
binaries for that platform. Things got somewhat better in the meantime.
With the tests I used 32-bit R compiled with gcc was roughly 10% slower
on Windows than under Linux - but as I said, it depends on the problem.
Trying loops versus matrix operations is a way to specify a two very
different problems, for example.
Uwe Ligges
Cézar Freitas wrote:
Hi, all.
I began to migrate my R codes from Windows to Linux and surprised me
with an old question. I simplified the problem and made a little test to
compare times at same
computer and the Linux time is worse (not so little) than Windows time:
28 vs 53 seconds.
I make an example (below) to facilitate all to see the difference.
I also build from source (it's my first time) a version of R to compare with
the distributed (compiled) R version. The times are similar to the other Linux
version.
I supposed R on Linux should be faster (32 and 64 bit) than windows version. Is
this difference because 64 bit R version is slower than 32 bits one? I started
the machine in both sittuations and checked free memory.
Tecnichal details:
Machine: Intel Core 2 Duo DDR2 4 Gb RAM
Windows version: XP Professional - 32 bits
R version: 2.9* binaries
Linux version: Ubuntu 8* (Hardy) - 64 bits
R version: 2.9* binaries and 2.9* compiled from source
Thanks to all,
Cezar Freitas
#code
N = 5
n = 15000
#makes data
dad = as.data.frame(cbind(sample(N,N,replace=FALSE), rpois(N,30)))
names(dad) = c(id,age)
aux = as.data.frame(cbind(sample(N,n,replace=FALSE), round(runif(n),4)))
names(aux) = c(id,score)
#calculates time
set.seed(790) #to be equal to everyone
system.time({
dad$score = 0
subdad = subset(dad, id%in%aux$id)
for(k in 1:(dim(subdad)[1])){
temp = aux$score[aux$id==subdad$id[k]]
if(length(temp)) subdad$score[k] = temp
}
})
#windows time
# user system elapsed
# 27.810.00 27.82
#linux usual compilation time
# user system elapsed
# 52.635 0.016 52.748
#linux (my compilation) time
# user system elapsed
# 52.567 0.016 52.588
#==END OF CODE
Veja quais são os assuntos do momento no Yahoo! +Buscados
http://br.maisbuscados.yahoo.com
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[R] How to read a specific dataset, not the entire data, from HDF5?
Hi. Recently I am working for a project to generate massive numeric data. After storing them in HDF5 using PyTables, we are trying to use R for data analysis and visualisation. Surprising to me, however, I could not find a R package to allow the reading of a specific dataset (or its slide) in a HDF5 file. I found I can read and write only the entire data using hdf5 package and I could not locate rhdf5 in bioconductor. Each dataset is huge, so selecting a specific dataset is a mandatory function for my project. Could you give me a hint for the solution? Thanks. -- DH __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] changing default arguments of a function and return the modified function as a result
Dear R-users,
I am trying to develop a function that takes another function as an argument,
changes its default values and returns a list of things, among which the
initial function with its default arguments changed. An example of what i
will like to obtain below:
## initial function
myfun - function(x, a=19, b=21){ return(a * x + b) }
## this is the function i will like to create
## (does not work as it is written here)
mysecond.fun - function(a, b, c = myfun(a=2, b=15)){
return(list(a=a, b=b c=c))
}
So I would be able to call:
mysecond.fun$c(x=12)
And this will be equivalent of calling:
myfun(x=12, a=2, b=15 ) ## i.e. i have changed the default values of myfun and
## stored it in a new function mysecond.fun$c
Any help will be greatly appreciated!
Miguel Bernal.
Current address:
Ocean Modeling group,
Institute of Marine and Coastal Sciences
University of Rutgers
71 Dudley Road, New Brusnkwick,
New Jersey 08901, USA
email: mber...@marine.rutgers.edu
phone: +1 732 932 3692
Fax: +1 732 932 8578
-
Permanent address:
Instituto Español de Oceanografía
Centro Oceanográfico de Cádiz
Puerto Pesquero, Muelle de Levante, s/n
Apdo. 2609, 11006 Cádiz, Spain
email: miguel.ber...@cd.ieo.es
phone: +34 956 294189
Fax: +34 956 294232
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Re: [R] changing the loss function in the logistic regression?
Hi Frank, Thanks for your help! I want to incorporate lift score as the optimization objective. How to do that in logistic regression? Thanks! On Fri, Jun 26, 2009 at 7:47 AM, Frank E Harrell Jrf.harr...@vanderbilt.edu wrote: Michael wrote: Hi all, Is there a way to change the loss function in the logistic regression? Or we could provide a customized loss function in the logistic regression so we could use that loss function in the Cross Validation in logistic regression? Thanks a lot! The goal is to use a loss function that yields optimality, with a sensible definition of optimality. For many purposes, maximum likelihood or penalized maximum likelihood is optimum. So don't change the optimality criteria just because you are cross-validating a different measure. By the way, it's often not a good idea to cross-validate a different measure. At least the accuracy index should be information-preserving. Deviance, log-likelihood, and AIC are your friends. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Where can I find information on how to subsample a time series?
I suspect I'm looking in the wrong places, so guidance to the relevant documentation would be as welcome as a little code snippet. I have time series data stored in a MySQL database. There is the usual DATE field, along with a double precision number: there are daily values (including only normal working days: Monday through Friday). I actually have to do a couple things here. Because of how the result is to be used, I need to first create two time series. The first is the delta between 22 working days, and the second is the delta between 66 working days. I have hundreds of these datasets, and some go back 30 years. I need to estimate the correlation between 22 day deltas (i.e. is the delta for one month correlated with that of the previous month) and between the 22 day delta and the 66 day delta that ends the day before the the first day of the 22 day delta. However, I KNOW the statistical properties of the time series are not constant (so the usual assumptions do not apply to the entire series). Therefore, I want to subsample finely enough to get a reasonably sensible correlation and examine how that changes through time. (There are no tests of significance here: I just want to explore just how much the properties of these series change through time). I have C++ code, admittedly not written particularly efficiently, that does this. The question is, is it possible to do this reasonably efficiently using R? Thanks Ted [[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] Heteroskedasticity and Autocorrelation in SemiPar package
On Fri, Jun 26, 2009 at 10:11 PM, Susan Chensuen...@yahoo.com wrote: Does anyone know how to report heteroskedasticity and autocorrelation-consistent standard errors when using the spm command in SemiPar package? Suppose the original command is sp1-spm(y~x1+x2+f(x3), random=~1,group=id) There's HAC() in library(gmm), which is a modification from the sandwich::vcovHAC(), designed to accept as input any vectors (and not only specific objects, such as lm). The sandwich::vcovHAC() can also be relatively easily extended to accept other fits as input (not only lm or glm, for example), but this would require some understanding of the internals of both SemiPar and sandwich. You might enquire if the author of SemiPar is interested in adding such functionality. For technical details, check the first vignette of sandwich. Once you get HAC covariance matrix for your fit, you can use either lmtest::coeftest or library(car) to obtain the HAC std errs. Liviu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] changing the loss function in the logistic regression?
Michael wrote: Hi Frank, Thanks for your help! I want to incorporate lift score as the optimization objective. How to do that in logistic regression? Thanks! Please re-read my note. Models should be fitted using proper scoring rules. Otherwise the resulting fit is bogus. Thanks Frank On Fri, Jun 26, 2009 at 7:47 AM, Frank E Harrell Jrf.harr...@vanderbilt.edu wrote: Michael wrote: Hi all, Is there a way to change the loss function in the logistic regression? Or we could provide a customized loss function in the logistic regression so we could use that loss function in the Cross Validation in logistic regression? Thanks a lot! The goal is to use a loss function that yields optimality, with a sensible definition of optimality. For many purposes, maximum likelihood or penalized maximum likelihood is optimum. So don't change the optimality criteria just because you are cross-validating a different measure. By the way, it's often not a good idea to cross-validate a different measure. At least the accuracy index should be information-preserving. Deviance, log-likelihood, and AIC are your friends. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] correspondence analysis question
Hi, I have a matrix of 31 rows and 8 columns. The rows represent attributes of a product and the columns represent segments in a market. The cell values are utilities scaled so that the sum of the utilities across attributes for a segment equals 100. I want to find which attributes are closely related to which segments. I ran a CA on the matrix and got an interesting plot. For example, attributes 4 and 6 are close to segment 1 and rows 1 and 2 are close to segment 7. Is there any way to calculate a measure of closeness from the ca output so that I can develop groups of attributes for each segment? I'd like to be able to have a list of attributes for each segment and a weight on how important (close?) an attribute is for that segment. Any suggestions? Thanks, Walt -- Walter R. Paczkowski, Ph.D. Data Analytics Corp. 44 Hamilton Lane Plainsboro, NJ 08536 (V) 609-936-8999 (F) 609-936-3733 dataanalyt...@earthlink.net www.dataanalyticscorp.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 avoid ifelse statement converting factor to character
Craig P. Pyrame wrote: Stavros Macrakis wrote: It gives me a headache, too! I think you'll have to wait for a more expert user than me to supply explanations of these behaviors and their rationales. Thanks, Stavros. I hope someone with expertise will shed more light on this, and in the meantime I'll try to learn more from the manuals. Well first, you (and others) should shake any bees relating to raw objects out of your bonnet. The raw type should be viewed as an add-on enabling access to low-level data structures, not as a fundamental data type in R. The original data modes are logical, numeric, character, with upwards coercion rules (and some further twiddles for storage modes of numerics). One particular aspect is that in the original system, logical works as a wild card type that can be coerced to all the others (hence NA has mode logical, which causes some confusion at first notice). Why does this fail: r = as.raw(TRUE) ifelse(TRUE, r, r) = error This gives an error which I take for saying that raw cannot be coerced to logical, but yes it can: as.logical(r) = TRUE Wrong. Or rather, irrelevant. The error is ifelse(TRUE,r,r) Error in ans[test !nas] - rep(yes, length.out = length(ans))[test : incompatible types (from raw to logical) in subassignment type fix Notice subassignment type fix. This corresponds to this sort of operation r - as.raw(TRUE) x - TRUE x[1] - r Error in x[1] - r : incompatible types (from raw to logical) in subassignment type fix Now, notice that when the coercion is bigger to smaller the fix is to coerce on the left side y - rep(TRUE,5) y[1] - 5 y [1] 5 1 1 1 1 So the thing that might be expected here would be that x be coerced to raw before the assignment, but this is not implemented. I.e. you could have had the effect of. x - as.raw(x) x[1] - r x [1] 01 This is designed not to happen automatically. I didn't write the code but I suppose one possible reason is that the raw mode does not allow missing values (or negative numbers for that matter, everything is 0-255). It also cuts the other way - no subassignment fixes are defined for raw: x - as.raw(1:10) x[3] - NA Error in x[3] - NA : incompatible types (from logical to raw) in subassignment type fix x[3] - TRUE Error in x[3] - TRUE : incompatible types (from logical to raw) in subassignment type fix x[3] - 4 Error in x[3] - 4 : incompatible types (from double to raw) in subassignment type fix x[3] - 4L Error in x[3] - 4L : incompatible types (from integer to raw) in subassignment type fix and raw can even be used as the condition vector in ifelse: ifelse(r, 1, 2) = 1 Yes, but that's a different issue. Ifelse has the C language convention that anything nonzero is TRUE. ifelse(pi,2,1) [1] 2 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] changing default arguments of a function and return the modified function as a result
On Jun 26, 2009, at 12:52 PM, Miguel Bernal wrote:
Dear R-users,
I am trying to develop a function that takes another function as an
argument,
changes its default values and returns a list of things, among which
the
initial function with its default arguments changed. An example of
what i
will like to obtain below:
## initial function
myfun - function(x, a=19, b=21){ return(a * x + b) }
## this is the function i will like to create
## (does not work as it is written here)
mysecond.fun - function(a, b, c = myfun(a=2, b=15)){
^^
Two things: a) I doubt that is possible in that manner and b) you
couldn't possibly get myfun to work, because you are not passing it
the correct number of arguments.
return(list(a=a, b=b c=c))
^
Third concern, missing comma.
}
So I would be able to call:
mysecond.fun$c(x=12)
Not in this language. Maybe in some other. Maybe you are in the wrong
help list? You should instead provide a specification of the inputs
and outputs you expect rather than a hashed up view of how you think R
should accept functional arguments.
Perhaps something like this:
myfun - function(x, a=19, b=21){ return(a * x + b) }
mysecond.fun - function(x, a, b){ c - myfun(x, a=2, b=15)
list(a=a, b=b, c=c) }
mysecond.fun(x=12, 2 , 3)
$a
[1] 2
$b
[1] 3
$c
[1] 39
# c= 2 * 12 + 15
And this will be equivalent of calling:
myfun(x=12, a=2, b=15 ) ## i.e. i have changed the default values of
myfun and
## stored it in a new function mysecond.fun$c
Any help will be greatly appreciated!
Miguel Bernal.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Alternate error structures in lme4?
bamsel wrote: Hi R users, The nlme library enabled several alternate error structures useful for longitudinal or repeated-measures data. For example, a continuous AR(1) process: model_2 = update(model_1, correlation = corCAR1(form = ~ time | subject)) Does anybody know if this is available in lme4? It's not, and probably won't be any time soon; the author (Doug Bates) has stated that it's not at the top of his list, and he's a very busy guy. (1) Is there a reason that you can't use nlme? (2) Further messages should probably be directed to the r-sig-mixed-mod...@r-project.org mailing list. Ben Bolker -- View this message in context: http://www.nabble.com/Alternate-error-structures-in-lme4--tp24226643p24227667.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] ANOVA with means and SDs as input
Le jeudi 25 juin 2009 à 12:15 +0200, Sebastian Stegmann a écrit : Dear R-community, I'm struggling with a paper that reports only fragmented results of a 2by2by3 experimental design. However, Means and SDs for all cells are given. Does anyone know a package/function that helps computing an ANOVA with only Means and SDs as input (resulting in some sort of effect size and significance test)? The reason why I'm interested is simple: I'm conducting a meta-analysis. If I included only what the authors would like to show the world, the results would be somewhat biased... What you aim to do is not, as far as I can tell, possible using already-programmed solutions in R packages. Some packages offer very close solutions (e. g. lme()) but lack some necessary options (e. g. the ability to fix intra-level variance in lme() : varFixed() works as documented, not as what its name suggests...). Why not use some established meta-analysis package such as meta or rmeta ? You may have to recompute some variances or effect sizes, but it can usually be done. If you aim to go a bit farther (i. e. more than 2-groups comparisons), the newly-released metafor package seems quite interesting. Wolfgang Viechtbauer aimed to offer a tool for meta-regression, and the current stage of the package (0.5.something IIRC) seems to be quite useable, if somewhat counter-intuitive in places. I'd recommend to have a look at it. WV didn't think useful to include the vignettes and documentation for mima(), predecessor of metafor, which explains quite nicely the motivation and algorithms he used. Too bad... However, this documentation is still available on his Web site. Be aware that the current main interface to the rma() function (workhorse of the package) does not (yet ?) follow the usual format for regression-like functions common to many R packages (the (formula, data, subset, etc...) format). For example, you may make meta-regression, but you'll have to code the model matrix yourself. Similarly, the necessary bricks for (e. g.) indirect comparisons are here, but you'll have to build them yourself (bring your wheelbarrow an some mortar...). An alternative to be seriously considered : Bayesian modeling. Meta-analysis is one of the rare medical domains where journal reviewers won't immediately reject a paper at the sight of the B***sian word... As far as I know, psychological journals are not as asinine. Here again, some manual work will be in order. Some tutorial material is readily available (google Bayesian meta analysis tutorial, which should lead you to the various Statistics in Medicine tutorials, among other things). I've combed through the web and my various books on R, but it's not that easy to find. Or is it? Variance analysis through manual decomposition of sum of squares ? Hugh ... I did that when I first learned anova(so long a time ago that I am uncomfortable to think of it. In these times, even a pocket calculator was too expensive for a random student (therefore forbidden in exams), and I learned to cope with slide rule...). Tiring and messy. Furthermore, manual algorithms (i. e. simple, non-iterative algorithms) exists only for the simplest (e. g. one-way anova) or ideal (e. g. balanced two-way anova with equal group sizes) cases ; furthermore, modern critiques of the bases of such works cast some doubts on their foundations (e. g. random- and mixed-effect anova)... Therefore, the art is fast forgotten, if not totally lost. If you insist, I might try to unearth my copy of Winer's handbook (ca. 1959) and look up specific questions. HTH, Emmanuel Charpentier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] changing default arguments of a function and return the modified function as a result
Is this what you want?
myfun - function(x, a=19, b=21){ return(a * x + b) }
mysecond.fun - function(a, b, cc=myfun, cc.args=list(a=2,b=15) ){
list(a=a, b=b, cc = function(x) cc(x, cc.args$a, cc.args$b))
}
mysecond.fun(a=1,b=2)$cc(x=12)
It may be that you're after a Curry (*) function, as in,
Curry - # from roxygen
function (f, ..., .left=TRUE)
{
.orig = list(...)
function(...){
if(.left) {args - c(.orig, list(...))} else {args - c(list(...),
.orig)}
do.call(f, args)
}
}
I believe there are some recent discussions on currying in the archives.
(*): http://en.wikipedia.org/wiki/Currying
HTH,
baptiste
2009/6/26 Miguel Bernal mber...@marine.rutgers.edu
Dear R-users,
I am trying to develop a function that takes another function as an
argument,
changes its default values and returns a list of things, among which the
initial function with its default arguments changed. An example of what i
will like to obtain below:
## initial function
myfun - function(x, a=19, b=21){ return(a * x + b) }
## this is the function i will like to create
## (does not work as it is written here)
mysecond.fun - function(a, b, c = myfun(a=2, b=15)){
return(list(a=a, b=b c=c))
}
So I would be able to call:
mysecond.fun$c(x=12)
And this will be equivalent of calling:
myfun(x=12, a=2, b=15 ) ## i.e. i have changed the default values of myfun
and
## stored it in a new function mysecond.fun$c
Any help will be greatly appreciated!
Miguel Bernal.
Current address:
Ocean Modeling group,
Institute of Marine and Coastal Sciences
University of Rutgers
71 Dudley Road, New Brusnkwick,
New Jersey 08901, USA
email: mber...@marine.rutgers.edu
phone: +1 732 932 3692
Fax: +1 732 932 8578
-
Permanent address:
Instituto Español de Oceanografía
Centro Oceanográfico de Cádiz
Puerto Pesquero, Muelle de Levante, s/n
Apdo. 2609, 11006 Cádiz, Spain
email: miguel.ber...@cd.ieo.es
phone: +34 956 294189
Fax: +34 956 294232
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
__
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Where can I find information on how to subsample a time series?
assuming you pull the data you want into x and y: w...@ubuntu:~$ R library(fts) x - fts() y - fts() xy.cor.200 - moving.cor(x,y,200) tail(xy.cor.200) [,1] 2012-03-12 -0.3009635 2012-03-13 -0.2923489 2012-03-14 -0.2824015 2012-03-15 -0.2662689 2012-03-16 -0.2566354 2012-03-17 -0.2537089 2012-03-18 -0.2490421 2012-03-19 -0.2391911 2012-03-20 -0.2263381 2012-03-21 -0.2113029 which is just using c++ to do the calculation. here is the template function for correlation that fts uses: http://github.com/armstrtw/tslib/blob/5b0fe2fc5ecb393d1dca097c2c19008227eb6c7e/tslib/vector.summary/cor.hpp -Whit On Fri, Jun 26, 2009 at 4:57 PM, Ted Byersr.ted.by...@gmail.com wrote: I suspect I'm looking in the wrong places, so guidance to the relevant documentation would be as welcome as a little code snippet. I have time series data stored in a MySQL database. There is the usual DATE field, along with a double precision number: there are daily values (including only normal working days: Monday through Friday). I actually have to do a couple things here. Because of how the result is to be used, I need to first create two time series. The first is the delta between 22 working days, and the second is the delta between 66 working days. I have hundreds of these datasets, and some go back 30 years. I need to estimate the correlation between 22 day deltas (i.e. is the delta for one month correlated with that of the previous month) and between the 22 day delta and the 66 day delta that ends the day before the the first day of the 22 day delta. However, I KNOW the statistical properties of the time series are not constant (so the usual assumptions do not apply to the entire series). Therefore, I want to subsample finely enough to get a reasonably sensible correlation and examine how that changes through time. (There are no tests of significance here: I just want to explore just how much the properties of these series change through time). I have C++ code, admittedly not written particularly efficiently, that does this. The question is, is it possible to do this reasonably efficiently using R? Thanks Ted [[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 avoid ifelse statement converting factor to character
Dear Peter, Thank you for the explanations. Some further questions are inlined below. Best regards, Craig Peter Dalgaard wrote: Craig P. Pyrame wrote: Stavros Macrakis wrote: It gives me a headache, too! I think you'll have to wait for a more expert user than me to supply explanations of these behaviors and their rationales. Thanks, Stavros. I hope someone with expertise will shed more light on this, and in the meantime I'll try to learn more from the manuals. Well first, you (and others) should shake any bees relating to raw objects out of your bonnet. The raw type should be viewed as an add-on enabling access to low-level data structures, not as a fundamental data type in R. The original data modes are logical, numeric, character, with upwards coercion rules (and some further twiddles for storage modes of numerics). I think I've seen examples in the list's archives where downwards coercion rules were used, but I am not sure. One particular aspect is that in the original system, logical works as a wild card type that can be coerced to all the others (hence NA has mode logical, which causes some confusion at first notice). That's right, I find it confusing at first. Why does this fail: r = as.raw(TRUE) ifelse(TRUE, r, r) = error This gives an error which I take for saying that raw cannot be coerced to logical, but yes it can: as.logical(r) = TRUE Wrong. Or rather, irrelevant. The error is ifelse(TRUE,r,r) Error in ans[test !nas] - rep(yes, length.out = length(ans))[test : incompatible types (from raw to logical) in subassignment type fix Notice subassignment type fix. This corresponds to this sort of operation r - as.raw(TRUE) x - TRUE x[1] - r Error in x[1] - r : incompatible types (from raw to logical) in subassignment type fix Now, notice that when the coercion is bigger to smaller the fix is to coerce on the left side What do you mean by coercion from bigger to smaller? What I thought should happen in the example above is to have x coerced upwards to logical. y - rep(TRUE,5) y[1] - 5 y [1] 5 1 1 1 1 So the thing that might be expected here would be that x be coerced to raw before the assignment, No, I would expect upwards coercion, just as in the example immediately above. I would expect r to be coerced to logical. but this is not implemented. I.e. you could have had the effect of. x - as.raw(x) x[1] - r x [1] 01 No, I would rather expect to have the effect of: x[1] - as.logical(r) This is designed not to happen automatically. I didn't write the code but I suppose one possible reason is that the raw mode does not allow missing values (or negative numbers for that matter, everything is 0-255). It also cuts the other way - no subassignment fixes are defined for raw: x - as.raw(1:10) x[3] - NA Error in x[3] - NA : incompatible types (from logical to raw) in subassignment type fix x[3] - TRUE Error in x[3] - TRUE : incompatible types (from logical to raw) in subassignment type fix x[3] - 4 Error in x[3] - 4 : incompatible types (from double to raw) in subassignment type fix x[3] - 4L Error in x[3] - 4L : incompatible types (from integer to raw) in subassignment type fix I don't understand any of these. If you can do the following: as.raw(TRUE) = 1 as.raw(4L) = 04 as.raw(-1) = 00, with a warning as.raw(NA) = 00, with a warning Why can't this be done automatically. What would be wrong in having this done automatically, with a warning if needed? and raw can even be used as the condition vector in ifelse: ifelse(r, 1, 2) = 1 Yes, but that's a different issue. Ifelse has the C language convention that anything nonzero is TRUE. ifelse(pi,2,1) [1] 2 That's a news for me. The man page does not explain that! It says test: an object which can be coerced to logical mode and also Missing values in test give missing values in the result. If ifelse were working the C way, I would expect 1 as the result here: ifelse(as.raw(1), 1, 2) = error as.raw(1) == 0 = FALSE The raw 1 is nonzero. Best regards, Craig __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Constrained corr matrix closest to a given corr matrix
Dear All! Is there any code to find a constrained correlation matrix closest in some sense (preferably in the sense of the geometric approximation) to a given correlation matrix? By constrained I mean with some elements constrained, or, simply, set to zero. I have read in a paper that says this can be accomplished with Lagrange multipliers and an optimization routine. Thanks a lot! Serguei Kaniovski __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] changing default arguments of a function and return the modified function as a result
That was indeed what i was looking for (thanks also for the currying cite). I
also wanted flexibility on the number of arguments you can pass to the
function, which can be achieved by:
myfun - function(x, a=19, b=21){ return(a * x + b) }
mysecond.fun -
function(dumb1, dumb2, cc=myfun, cc.args=list(a=2, b=5)){
fun.args - formals(cc)
fun.args[match(names(cc.args), names(fun.args))] - cc.args
formals(cc) - fun.args
list(a=dumb1, b=dumb2, cc = cc)
}
test - mysecond.fun(1, 2)
myfun(2) ## 59
test$cc(2) ## 9
Thanks both of you for your replies,
Miguel.
On Friday 26 June 2009 18:20:54 baptiste auguie wrote:
Is this what you want?
myfun - function(x, a=19, b=21){ return(a * x + b) }
mysecond.fun - function(a, b, cc=myfun, cc.args=list(a=2,b=15) ){
list(a=a, b=b, cc = function(x) cc(x, cc.args$a, cc.args$b))
}
mysecond.fun(a=1,b=2)$cc(x=12)
It may be that you're after a Curry (*) function, as in,
Curry - # from roxygen
function (f, ..., .left=TRUE)
{
.orig = list(...)
function(...){
if(.left) {args - c(.orig, list(...))} else {args - c(list(...),
.orig)}
do.call(f, args)
}
}
I believe there are some recent discussions on currying in the archives.
(*): http://en.wikipedia.org/wiki/Currying
HTH,
baptiste
2009/6/26 Miguel Bernal mber...@marine.rutgers.edu
Dear R-users,
I am trying to develop a function that takes another function as an
argument,
changes its default values and returns a list of things, among which the
initial function with its default arguments changed. An example of what i
will like to obtain below:
## initial function
myfun - function(x, a=19, b=21){ return(a * x + b) }
## this is the function i will like to create
## (does not work as it is written here)
mysecond.fun - function(a, b, c = myfun(a=2, b=15)){
return(list(a=a, b=b c=c))
}
So I would be able to call:
mysecond.fun$c(x=12)
And this will be equivalent of calling:
myfun(x=12, a=2, b=15 ) ## i.e. i have changed the default values of
myfun and
## stored it in a new function mysecond.fun$c
Any help will be greatly appreciated!
Miguel Bernal.
Current address:
Ocean Modeling group,
Institute of Marine and Coastal Sciences
University of Rutgers
71 Dudley Road, New Brusnkwick,
New Jersey 08901, USA
email: mber...@marine.rutgers.edu
phone: +1 732 932 3692
Fax: +1 732 932 8578
-
Permanent address:
Instituto Español de Oceanografía
Centro Oceanográfico de Cádiz
Puerto Pesquero, Muelle de Levante, s/n
Apdo. 2609, 11006 Cádiz, Spain
email: miguel.ber...@cd.ieo.es
phone: +34 956 294189
Fax: +34 956 294232
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Current address:
Ocean Modeling group,
Institute of Marine and Coastal Sciences
University of Rutgers
71 Dudley Road, New Brusnkwick,
New Jersey 08901, USA
email: mber...@marine.rutgers.edu
phone: +1 732 932 3692
Fax: +1 732 932 8578
-
Permanent address:
Instituto Español de Oceanografía
Centro Oceanográfico de Cádiz
Puerto Pesquero, Muelle de Levante, s/n
Apdo. 2609, 11006 Cádiz, Spain
email: miguel.ber...@cd.ieo.es
phone: +34 956 294189
Fax: +34 956 294232
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
