Re: [R] Authoring a book
//people.ee.ethz.ch/~oetiker/lshort/lshort.pdf It really was a not too short intro. I'll have a look at it. Yes definitly not too short. But it states in LaTeX in 133 min ... Seems to be for Linux only. My server is Windows, even if I have the rest of the components. Hm at the projects' homepage its stated that it is OS independent: http://sourceforge.net/projects/uniwakka/ I also dont see a hint that it is Linux dependent in the installation notes. Maybe you think its Linux stuff because of the *.tar.gz ending? You can unzip those files also on Windows, its just a packing file format. For unpacking on a windows machine use either 7zip: http://www.7-zip.org/ or the total commander: http://www.ghisler.com . The first is a free packing program and the second is the best ever windows file manager (if you want to open it there just double click at the archive). Stefan Grosse The unpacked instl. notes: Wakka Installation Not much to it (as long as it works, ahem). Unpack/upload the distribution files into a directory that can be accessed via the web. Then go to the corresponding URL. A web-based installer will walk you through the rest. Example: If your website, say, http://www.mysite.com, is mapped to the directory /home/jdoe/www/, and you place the Wakka distribution files into /home/jdoe/www/wakka/, you should go to http://www.mysite.com/wakka/. Note that Wakka distributions normally unpack into directories that include the version in their name; you'll probably want to rename those to just wakka -- or, if you're on a unixoid system, set up a symbolic link. During first installs, the installer will try to create a file called wakka.config.php in your Wakka directory. In order to do this, you will need to either make the Wakka directory writable by the web server, or create a new (empty) file called wakka.config.php which is writable by the web server. If the installer still fails to create the file, it will dump the file's contents which you can then upload manually. IMPORTANT: for installing or upgrading Wakka, do NOT access any of the files contained in the setup/ subdirectory. They're used by the web-based installer/updater, but you should really just access the Wakka directory itself, and it will (or at least should) work perfectly. Detailed instructions are available at http://www.wakkawiki.com/WakkaInstallation. - Hendrik Mans [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] tktoplevel tkgetSaveFile options
Dear list, Previously I posted these question to R-SIG-GUI. Perhaps here is a better place. 1. Is there some option for mantaining the tktoplevel window always on top? 2. Is there some option to eliminate the border icons maximize, minimize and close of a tktoplevel window? 3. Is possible to avoid the warning message (or changing its contents) in tkgetSaveFile when the file to save already exists? Thanks, Manel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot tick marks and line thickness
Deepayan Sarkar deepayan.sarkar at gmail.com writes: Right. the grid call shouldn't be necessary, axis.line controls the panel borders. And tck can be a vector, to get rid of the ugly bumps on top: xyplot(x ~ x | g, type = l, lwd = lwd, scales = list(tck = c(-1, 0)), par.settings = list(axis.line = list(lwd = lwd), strip.border = list(lwd = lwd))) The tickmarks look strange for me (Windows, R 2.3.1, all packages updated today) See http://www.menne-biomed.de/uni/gridit.png. This was created on screen, saved as a bitmap, and 3x enlarged lower portion. Dieter __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 in Nature
Hi all, We've just had a paper accepted for publication in Nature. We used R for 95% of our analyses (one of my co-authors sneaked in some GenStat when I wasn't looking.). The preprint is available from the Nature web site, in the open peer-review trial section. I searched Nature for previous references to R Development Core Team, and I received no hits. So I tentatively conclude that our paper is the first Nature paper to cite R. A great many thanks to the R Development Core Team for R, and Prof. Bates for lmer. Cheers, Simon. (I'm off to the pub to celebrate.) -- Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat. Centre for Resource and Environmental Studies The Australian National University Canberra ACT 0200 Australia T: +61 2 6125 7800 email: Simon.Blomberg_at_anu.edu.au F: +61 2 6125 0757 CRICOS Provider # 00120C The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. - John Tukey. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] exact Wilcoxon signed rank test with ties and the no longer under development exactRanksumTests package
Dear List, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately the procedure cannot calculate the exact values with ties. Is there any other package that is providing a similiar test? Or is there an easy work out how to take the ties into account? (Or a chance that the correction is taken into account for the stats package?) Stefan Grosse Take the following example from Bortz/Lienert/Boehnke: x1-c(9,14,8,11,14,10,8,14,12,14,13,9,15,12,9) x2-c(13,15,9,12,16,10,8,13,12,16,9,10,16,12,9) # exactRankTests package: wilcox.exact(x1,x2,paired=TRUE) Exact Wilcoxon signed rank test data: x1 and x2 V = 13, p-value = 0.1367 alternative hypothesis: true mu is not equal to 0 # wilcox.test by stats package: wilcox.test(x1,x2,paired=TRUE,exact=TRUE) Wilcoxon signed rank test with continuity correction data: x1 and x2 V = 13, p-value = 0.1436 alternative hypothesis: true mu is not equal to 0 Warning messages: 1: cannot compute exact p-value with ties in: wilcox.test.default(x1, x2, paired = TRUE, exact = TRUE) 2: cannot compute exact p-value with zeroes in: wilcox.test.default(x1, x2, paired = TRUE, exact = TRUE) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot tick marks and line thickness
Piet Bell skreiv: The publisher of our paper has requested: 1. all tick marks should point inwards instead of outwards. Point him to William S. Cleveland’s excellent book /The Elements of Graphing Data/, where Cleveland strongly recommends that tick marks should point *outwards* ‘because ticks that point inward can obscure data’. See the discussion on pages 31–35, and especially figure 2.12 and 2.13. 2. All lines should be thicker (lines, axes, boxes, etc. Everything). Lines is easy...I used: lwd=1.5 but what about the lines of the axes, and the lines that build up the plot itself?? I find that library(Hmisc) setps(filename) # Or setpdf. You might also want to add 'color=TRUE'. ... plotting commands ... dev.off() usually gives much better-looking plots, and with thicker lines. -- Karl Ove Hufthammer E-mail and Jabber: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Check values in colums matrix
Dear all, I would like to thank everybody who replied for their useful suggestions. Maybe, I am going through the book statistics to teach (fresh) myself. Wish you have a nice weekend. Regards, Muhammad Subianto On this day 24/08/2006 18:59, Muhammad Subianto wrote: Dear all, I apologize if my question is quite simple. I have a dataset (20 columns 1000 rows) which some of columns have the same value and the others have different values. Here are some piece of my dataset: obj - cbind(c(1,1,1,4,0,0,1,4,-1), c(0,1,1,4,1,0,1,4,-1), c(1,1,1,4,2,0,1,4,-1), c(1,1,1,4,3,0,1,4,-1), c(1,1,1,4,6,0,1,5,-1), c(1,1,1,4,6,0,1,6,-1), c(1,1,1,4,6,0,1,7,-1), c(1,1,1,4,6,0,1,8,-1)) obj.tr - t(obj) obj.tr obj.tr [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [1,]11140014 -1 [2,]01141014 -1 [3,]11142014 -1 [4,]11143014 -1 [5,]11146015 -1 [6,]11146016 -1 [7,]11146017 -1 [8,]11146018 -1 How can I do to check columns 2,3,4,6,7 and 9 have the same value, and columns 1,5 and 8 have different values. Best, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Authoring a book
Dear Tom,
To add a few things to explore:
- I'd definitely go with LaTeX. Depending on how much formatting control you
want, though, and if your coworkers are reluctant to jump into LaTeX, you
might start with reStructuredText (http://docutils.sourceforge.net/rst.html)
or text2tags (http://txt2tags.sourceforge.net/). With both, you can produce
LaTeX, but innitially at least it allows you to write text with structure
using markup that is a lot simpler than latex.
- I'd definitely use a version control system. Instead of CVS or SVN, though,
I'd suggest you take a look at some of the distributed ones, in particular
Bazaar-NG (http://bazaar-vcs.org), Mercurial
(http://www.selenic.com/mercurial/wiki/index.cgi) or Darcs
(http://abridgegame.org/darcs/).
These three are probably among the most mature ones (though oppinions will
vary, of course; I have some notes and links at:
http://www.ligarto.org/rdiaz/VersionControl.html).
What I like about any of these is that I think they provide you essentially
all SVN can provide (except for the user-base and years of existence of SVN)
plus a lot more. For instance, if you often work without access to the remote
repository, with any of these three systems you can enjoy all the benifits of
version control. Cherry-picking is easier with any of these than with
CVS/SVN, and Darcs in particular excels at it.
- For bibliography, I find CiteULike (http://www.citeulike.org/) fabulous.
Needs internet access, and might not work with the journals/data bases that
you use, though. It can export as bibtex.
- If you find outliners useful (or absolutely essential) then you might want
to look at Leo (http://webpages.charter.net/edreamleo/front.html). Leo is
agnostic regarding whether you write LaTeX, plain text, or R code (though it
has great support for some languages such as Python or rst), and you can use
Leo and still edit files in your editor of choice (I use Leo for working with
fairly large latex files that I edit under Emacs). However, for this to work,
all of you should agree to use Leo (or at least not disturb the sentinel
lines that leo uses).
Hope this helps (or at least provides entertaining links :-).
R.
On Thursday 24 August 2006 21:10, Tom Backer Johnsen wrote:
Stefan Grosse wrote:
I think Peter Dalgaard is right.
Since you are able to use R I believe you will be very fast in learning
LaTeX.
I think it needs less then a week to learn the most common LaTeX
commands. And setting up a wiki and trying then to convert this into a
printable document format plus learning the wiki syntax is probably more
time consuming. Beside this R is able to work perfectly together with
LaTeX, it creates LaTeX output and is doing excellent graphics in the
EPS/PS format.
The best introduction for LaTeX is the not so short introduction:
http://people.ee.ethz.ch/~oetiker/lshort/lshort.pdf
It really was a not too short intro. I'll have a look at it.
If you still are not convinced have a look at UniWakkaWiki:
http://uniwakka.sourceforge.net/HomePage
It is a Wiki for Science and University purposes and claims to be able
to export to Openoffice as well as to LaTeX.
Looks interesting and I really like the concept, but how stable is it?
It looks rather fresh from the web page, but I may be wrong. A
bibliography function is really a big advantage, so ... perhaps.
Tom
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Re: [R] exact Wilcoxon signed rank test with ties and the no longerunder development exactRanksumTests package
On Fri, 25 Aug 2006, Stefan Grosse wrote: Dear List, Stefan, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately the procedure cannot calculate the exact values with ties. indeed, this is the only gap to be filled in `coin', I just haven't had the time to implement this. And of course, `exactRankTests' is still available and will be available in the future. So you _can_ use it! The message just means that I'm not going to add _new_ features to `exactRankTests' and that we as the authors of both packages believe that the `coin'-way of doing things is more appropriate. Hope that helps sorry for the confusion! Torsten ps: please cc emails about packages to the maintainer. Is there any other package that is providing a similiar test? Or is there an easy work out how to take the ties into account? (Or a chance that the correction is taken into account for the stats package?) Stefan Grosse Take the following example from Bortz/Lienert/Boehnke: x1-c(9,14,8,11,14,10,8,14,12,14,13,9,15,12,9) x2-c(13,15,9,12,16,10,8,13,12,16,9,10,16,12,9) # exactRankTests package: wilcox.exact(x1,x2,paired=TRUE) Exact Wilcoxon signed rank test data: x1 and x2 V = 13, p-value = 0.1367 alternative hypothesis: true mu is not equal to 0 # wilcox.test by stats package: wilcox.test(x1,x2,paired=TRUE,exact=TRUE) Wilcoxon signed rank test with continuity correction data: x1 and x2 V = 13, p-value = 0.1436 alternative hypothesis: true mu is not equal to 0 Warning messages: 1: cannot compute exact p-value with ties in: wilcox.test.default(x1, x2, paired = TRUE, exact = TRUE) 2: cannot compute exact p-value with zeroes in: wilcox.test.default(x1, x2, paired = TRUE, exact = TRUE) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R in Nature
Simon Blomberg wrote: Hi all, We've just had a paper accepted for publication in Nature. We used R for 95% of our analyses (one of my co-authors sneaked in some GenStat when I wasn't looking.). The preprint is available from the Nature web site, in the open peer-review trial section. I searched Nature for previous references to R Development Core Team, and I received no hits. So I tentatively conclude that our paper is the first Nature paper to cite R. A great many thanks to the R Development Core Team for R, and Prof. Bates for lmer. Cheers, Simon. (I'm off to the pub to celebrate.) Congratulations but I cannot find your article on http://blogs.nature.com/nature/peerreview/trial Can you post valid link to it? Thanks in advance. Andrej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R in Nature
Andrej Kastrin [EMAIL PROTECTED] writes: Simon Blomberg wrote: Hi all, We've just had a paper accepted for publication in Nature. We used R for 95% of our analyses (one of my co-authors sneaked in some GenStat when I wasn't looking.). The preprint is available from the Nature web site, in the open peer-review trial section. I searched Nature for previous references to R Development Core Team, and I received no hits. So I tentatively conclude that our paper is the first Nature paper to cite R. A great many thanks to the R Development Core Team for R, and Prof. Bates for lmer. Cheers, Simon. (I'm off to the pub to celebrate.) Congratulations but I cannot find your article on http://blogs.nature.com/nature/peerreview/trial Can you post valid link to it? Thanks in advance. Andrej Found it in the Ecology section. http://blogs.nature.com/nature/peerreview/trial/Fisher%20manuscript.pdf (BTW: The year in the reference is wrong, R 2.3.1 is from 2006.) -- 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 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] extremely slow recursion in R?
Thomas Lumley wrote:
On Thu, 24 Aug 2006, Jason Liao wrote:
I recently coded a recursion algorithm in R and ir ran a few days
without returning any result. So I decided to try a simple case of
computing binomial coefficient using recusrive relationship
choose(n,k) = choose(n-1, k)+choose(n-1,k-1)
I implemented in R and Fortran 90 the same algorithm (code follows).
The R code finishes 31 minutes and the Fortran 90 program finishes in 6
seconds. So the Fortran program is 310 times faster. I thus wondered if
there is room for speeding up recursion in R. Thanks.
Recursive code that computes the same case many times can often be sped up
by memoization, eg
memo-new.env(hash=TRUE)
chewse-function(n,k) {
if (n==k) return(1)
if(k==1) return(n)
if(exists(paste(n,k),memo,inherits=FALSE))
return(get(paste(n,k),memo))
rval-chewse(n-1,k)+chewse(n-1,k-1)
assign(paste(n,k),rval,envir=memo)
return(rval)
}
This idea was discussed in an early Programmers' Niche article by Bill
Venables in R News.
However, I'm surprised that you're surprised that compiled Fortran 90 is
310 times faster than interpreted R. That would be about what I would
expect for code that isn't making use of vectorized functions in R.
-thomas
maybe someone's interested:
I made the same observation of seemingly very slow recursion recently:
just for fun I used the (in)famously inefficient
fib - function(n = 1) {
if (n 2)
fn - 1
else
fn - fib(n - 1) + fib(n - 2)
fn
}
for calculating the fibonacci numbers and compared `fib(30)' (about
1.3e6 recursive function calls ...) to some other languages (times in sec):
language time
==
C 0.034 (compiled, using integers)
Ocaml 0.041 (compiled, using integers)
Ocaml 0.048 (interpreted, using integers)
C 0.059 (compiled, using floats)
Lua1.1
ruby 3.4
R 21
octave120
apart from octave (which seems to have a _real_ issue with recursive
function calls), R is by far the slowest in this list and still a factor
7-20 slower than the interpreter based Lua and ruby. the speed loss
compared to C or Ocaml is about a factor of 350-600 here (Ocaml keeps
its speed more or less in this simple example even in 'script mode',
which is remarkable, I think (and it usually loses only a factor of
about 7 or so in script mode compared to the compiled variant)
for the specialists the bad performance of R in this situation might not
be surprising, but I was not aware that recursive function calls are
seemingly as expensive as explicit loops (where the execution time ratio
of R to C again is of the same order, i.e. =~ 400).
of course, such comparsions don't make too much sense: the reason to use
R will definitely not be its speed (which, luckily, often does not
matter), but the combination of flexible language, the number of
available libraries and the good 2D graphics.
joerg
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Re: [R] Using a 'for' loop : there should be a better way in R
--- Gabor Grothendieck [EMAIL PROTECTED]
wrote:
Use cbind to create a two column matrix, mat,
and multiply that by the appropriate inflation
factors.
Then use rowsum to sum the rows according to the
id grouping factor.
inf.fac - list(year1 = 1, year2 = 5, year3 = 10)
mat - cbind(s1 = df1$cat1 + df1$cat2, s2 = df1$cat3
+ df1$cat4)
rowsum(mat * unlist(inf.fac[df1$year]), df1$id)
Thanks very much. It took me a few minutes to see
what was happening but it is lovely. I would never
have thought of using a list like that.
On 8/24/06, John Kane [EMAIL PROTECTED] wrote:
I need to apply a yearly inflation factor to some
wages and supply some simple sums by work
category. I
have gone at it with a brute force for loop
approach
which seems okay as it is a small dataset. It
looks
a bit inelegant and given all the warnings in the
Intro to R, etc, about using loops I wondered if
anyone could suggest something a bit simpler or
more
efficent?
Example:
cat1 - c( 1,1,6,1,1,5)
cat2 - c( 1,2,3,4,5,6)
cat3 - c( 5,4,6,7,8,8)
cat4 - c( 1,2,1,2,1,2)
years - c( 'year1', 'year2', 'year3', 'year3',
'year1', 'year1')
id - c('a','a','b','c','c','a')
df1 - data.frame(id,years,cat1,cat2, cat3, cat4)
nn - levels(df1$id)# levels for outer loop
hh - levels(df1$years) # levels for inter loop
mp - c(1, 5, 10) # inflation factor
tt - data.frame(matrix(NA, length(nn), 2))
names(tt) - c(s1,s2)
rownames(tt) - nn
for (i in 1:length(nn)){
scat - data.frame(matrix(NA, length(hh),2))
dd1 - subset(df1, id==nn[i])
for (j in 1:length(hh)){
dd2 - subset(dd1, dd1$years==hh[j])
s1 - sum(dd2$cat1,dd2$cat2, na.rm=T)
s2 - sum(dd2$cat3,dd2$cat4,na.rm=T)
scat[j,] - c(s1,s2) *mp[j]# multiply by the
inflation factor
}
crush - apply(scat, 2, sum)
tt[i,] - crush
}
tt
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[R] plot question
Hi everyone, I have what may appear to be a newbie question, but I have looked everywhere I can think to look and I cannot find an answer. On page 35 of An Introduction to R the following command appears: plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE). What is the do.points argument? I know what it does (suppresses printing of the points) but where can I find help on it? I want to be able to explain it fully to my students. Thanks for your help, Cathy -- Dr. Cathy Carter Department of Geography University of Maryland A LeFrak Hall 301.405.4620 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot question
Take a look at ?plot.stepfun. 'ecdf()' returns an object of class ecdf inheriting from class stepfun and 'plot.ecdf()' calls 'plot.stepfun'. -roger Catherine Carter wrote: Hi everyone, I have what may appear to be a newbie question, but I have looked everywhere I can think to look and I cannot find an answer. On page 35 of An Introduction to R the following command appears: plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE). What is the do.points argument? I know what it does (suppresses printing of the points) but where can I find help on it? I want to be able to explain it fully to my students. Thanks for your help, Cathy -- Roger D. Peng | http://www.biostat.jhsph.edu/~rpeng/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] extremely slow recursion in R?
Thank you, Thomas and Joerg! Joerg's example is extremely useful. In
fact I had wanted to compare R with other interpreting language myself.
I heard that Ruby is not fast among scripting languages. I thus believe
that there is room to improve R's recursive function.
Jason
--- Joerg van den Hoff [EMAIL PROTECTED] wrote:
maybe someone's interested:
I made the same observation of seemingly very slow recursion
recently:
just for fun I used the (in)famously inefficient
fib - function(n = 1) {
if (n 2)
fn - 1
else
fn - fib(n - 1) + fib(n - 2)
fn
}
for calculating the fibonacci numbers and compared `fib(30)' (about
1.3e6 recursive function calls ...) to some other languages (times in
sec):
language time
==
C 0.034 (compiled, using integers)
Ocaml 0.041 (compiled, using integers)
Ocaml 0.048 (interpreted, using integers)
C 0.059 (compiled, using floats)
Lua1.1
ruby 3.4
R 21
octave120
apart from octave (which seems to have a _real_ issue with recursive
function calls), R is by far the slowest in this list and still a
factor
7-20 slower than the interpreter based Lua and ruby. the speed loss
compared to C or Ocaml is about a factor of 350-600 here (Ocaml keeps
its speed more or less in this simple example even in 'script mode',
which is remarkable, I think (and it usually loses only a factor of
about 7 or so in script mode compared to the compiled variant)
for the specialists the bad performance of R in this situation might
not
be surprising, but I was not aware that recursive function calls are
seemingly as expensive as explicit loops (where the execution time
ratio
of R to C again is of the same order, i.e. =~ 400).
of course, such comparsions don't make too much sense: the reason to
use
R will definitely not be its speed (which, luckily, often does not
matter), but the combination of flexible language, the number of
available libraries and the good 2D graphics.
joerg
Jason Liao, http://www.geocities.com/jg_liao
Department of Epidemiology and Biostatistics
Drexel University School of Public Health
245 N. 15th Street, Mail Stop 660
Philadelphia, PA 19102-1192
phone 215-762-3934
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Re: [R] plot question
On Fri, 2006-08-25 at 09:08 -0400, Catherine Carter wrote: Hi everyone, I have what may appear to be a newbie question, but I have looked everywhere I can think to look and I cannot find an answer. On page 35 of An Introduction to R the following command appears: plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE). What is the do.points argument? I know what it does (suppresses printing of the points) but where can I find help on it? I want to be able to explain it fully to my students. Thanks for your help, Cathy A couple of options: 1. If you are aware of how R uses method dispatch, then you might know that the plot() function is a generic method and that plot.ecdf() is the specific method that is called when the initial argument is of class ecdf, as is the case above. Thus, using ?plot.ecdf will get you to the help page, where there is a notation in the Arguments section that the '...' arguments are then passed to plot.stepfun(). 'do.points' is passed in this fashion, so using ?plot.stepfun will then get you to the help page where 'do.points' is defined as: logical; if true, also draw points at the (xlim restricted) knot locations. 2. Using: RSiteSearch(do.points, restrict = functions) will search the online function documentation, bringing up a browser window, where the first link gets you to ?plot.stepfun. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot question
Thank you! That is exactly what I needed. Roger D. Peng wrote: Take a look at ?plot.stepfun. 'ecdf()' returns an object of class ecdf inheriting from class stepfun and 'plot.ecdf()' calls 'plot.stepfun'. -roger Catherine Carter wrote: Hi everyone, I have what may appear to be a newbie question, but I have looked everywhere I can think to look and I cannot find an answer. On page 35 of An Introduction to R the following command appears: plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE). What is the do.points argument? I know what it does (suppresses printing of the points) but where can I find help on it? I want to be able to explain it fully to my students. Thanks for your help, Cathy -- Dr. Cathy Carter Department of Geography University of Maryland A LeFrak Hall 301.405.4620 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] correlation between 3 vectors
Hi R-users, I am trying to calculate the correlation between 3 vectors of numbers. Is there any other solution then passing by kendall.w function which need ranks in input? Thanks, Bianca Dimulescu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] extremely slow recursion in R?
i tried fib(30):
R: 8.99s
Python (interpreted): 0.96s
Python (interpreted but using psyco library): 0.03s
C++: 0.015s
cheers
Damien
--- On Fri 08/25, Joerg van den Hoff [EMAIL PROTECTED] wrote:
From: Joerg van den Hoff [mailto: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED], r-help@stat.math.ethz.ch
Date: Fri, 25 Aug 2006 13:52:04 +0200
Subject: Re: [R] extremely slow recursion in R?
Thomas Lumley wrote:br On Thu, 24 Aug 2006, Jason Liao wrote:br br I
recently coded a recursion algorithm in R and ir ran a few daysbr without
returning any result. So I decided to try a simple case ofbr computing
binomial coefficient using recusrive relationshipbrbr choose(n,k) =
choose(n-1, k)+choose(n-1,k-1)brbr I implemented in R and Fortran 90
the same algorithm (code follows).br The R code finishes 31 minutes and the
Fortran 90 program finishes in 6br seconds. So the Fortran program is 310
times faster. I thus wondered ifbr there is room for speeding up recursion
in R. Thanks.brbr br Recursive code that computes the same case many
times can often be sped up br by memoization, egbr br
memo-new.env(hash=TRUE)br chewse-function(n,k) {br if (n==k)
return(1)br if(k==1) return(n)br br
if(exists(paste(n,k),memo,inherits=FALSE))br
return(get(paste(n,k),memo))br
rval-chewse(n-1,k)+chewse(n-1,k-1)br
assign(paste(n,k),rval,envir=memo)br return(rval)br }br br
This idea was discussed in an early Programmers' Niche article by Bill br
Venables in R News.br br However, I'm surprised that you're surprised
that compiled Fortran 90 is br 310 times faster than interpreted R. That
would be about what I would br expect for code that isn't making use of
vectorized functions in R.br br br -thomasbrbrbrmaybe
someone's interested:brI made the same observation of seemingly very slow
recursion recently: brjust for fun I used the (in)famously
inefficientbrbrfib - function(n = 1) {brif (n 2)br fn -
1brelsebr fn - fib(n - 1) + fib(n - 2)brfnbr}brbrfor
calculating the fibonacci numbers and compared `fib(30)' (about br1.3e6
recursive function calls ...) to some other languages (times in
sec):brbrlanguage timebr==brC !
0.034 (compiled, using integers)brOcaml 0.041 (compiled, using
integers)brOcaml 0.048 (interpreted, using integers)brC
0.059 (compiled, using floats)brLua1.1brruby 3.4brR
21broctave120brbrapart from octave (which seems to have a _real_
issue with recursive brfunction calls), R is by far the slowest in this list
and still a factor br7-20 slower than the interpreter based Lua and ruby. the
speed loss brcompared to C or Ocaml is about a factor of 350-600 here (Ocaml
keeps brits speed more or less in this simple example even in 'script mode',
brwhich is remarkable, I think (and it usually loses only a factor of
brabout 7 or so in script mode compared to the compiled variant)brbrfor
the specialists the bad performance of R in this situation might not brbe
surprising, but I was not aware that recursive function calls are brseemingly
as expensive as explicit loops (where the execution time rati!
o
brof R to C again is of the same order, i.e. =~ 400).brbrof course, such
comparsions don't make too much sense: the reason to use brR will definitely
not be its speed (which, luckily, often does not brmatter), but the
combination of flexible language, the number of bravailable libraries and the
good 2D
graphics.brbrbrbrjoergbrbr__brR-help@stat.math.ethz.ch
mailing listbrhttps://stat.ethz.ch/mailman/listinfo/r-helpbrPLEASE do read
the posting guide http://www.R-project.org/posting-guide.htmlbrand provide
commented, minimal, self-contained, reproducible code.br
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Total (un)standardized effects in SEM?
Dear Rense,
(This question was posted a few days ago when I wasn't reading my email.)
So-called effect decompositions are simple functions of the structural
coefficients of the model, which in a model fit by sem() are contained in
the $A component of the returned object. (See ?sem.) One approach,
therefore, would be to put the coefficients in the appropriate locations of
the estimated Beta, Gamma, Lamda-x, and Lambda-y matrices of the LISREL
model, and then to compute the effects in the usual manner.
It should be possible to do this for the RAM formulation of the model as
well, simply by distinguishing exogenous from endogenous variables. Here's
an illustration using model C in the LISREL 7 Manual, pp. 169-177, for the
Wheaton et al. stability of alienation data (a common example--I happen to
have an old LISREL manual handy):
S.wh - matrix(c(
+11.834, 0,0,0, 0,0,
+ 6.947,9.364, 0,0, 0,0,
+ 6.819,5.091, 12.532, 0, 0,0,
+ 4.783,5.028,7.495,9.986,0,0,
+-3.839, -3.889, -3.841, -3.625, 9.610, 0,
+-2.190, -1.883, -2.175, -1.878, 3.552,4.503),
+ 6, 6)
rownames(S.wh) - colnames(S.wh) -
+ c('Anomia67','Powerless67','Anomia71','Powerless71','Education','SEI')
model.wh - specify.model()
1: Alienation67 - Anomia67, NA, 1
2: Alienation67 - Powerless67, lam1, NA
3: Alienation71 - Anomia71, NA, 1
4: Alienation71 - Powerless71, lam2, NA
5: SES- Education, NA, 1
6: SES- SEI, lam3, NA
7: SES- Alienation67, gam1, NA
8: Alienation67 - Alienation71, beta, NA
9: SES- Alienation71, gam2, NA
10: Anomia67 - Anomia67, the1, NA
11: Anomia71 - Anomia71, the3, NA
12: Powerless67- Powerless67, the2, NA
13: Powerless71- Powerless71, the4, NA
14: Education - Education, thd1, NA
15: SEI- SEI, thd2, NA
16: Anomia67 - Anomia71, the13, NA
17: Alienation67 - Alienation67, psi1, NA
18: Alienation71 - Alienation71, psi2, NA
19: SES- SES, phi,NA
20:
Read 19 records
sem.wh - sem(model.wh, S.wh, 932)
summary(sem.wh)
Model Chisquare = 6.3349 Df = 5 Pr(Chisq) = 0.27498
Chisquare (null model) = 17973 Df = 15
Goodness-of-fit index = 0.99773
Adjusted goodness-of-fit index = 0.99046
RMSEA index = 0.016934 90 % CI: (NA, 0.05092)
Bentler-Bonnett NFI = 0.99965
Tucker-Lewis NNFI = 0.99978
Bentler CFI = 0.3
BIC = -27.852
Normalized Residuals
Min. 1st Qu.Median Mean 3rd Qu. Max.
-9.57e-01 -1.34e-01 -4.24e-02 -9.17e-02 6.43e-05 5.47e-01
Parameter Estimates
Estimate Std Error z value Pr(|z|)
lam1 1.02656 0.053424 19.2152 0.e+00 Powerless67 --- Alienation67
lam2 0.97089 0.049608 19.5712 0.e+00 Powerless71 --- Alienation71
lam3 0.51632 0.042247 12.2214 0.e+00 SEI --- SES
gam1 -0.54981 0.054298 -10.1258 0.e+00 Alienation67 --- SES
beta 0.61732 0.049486 12.4746 0.e+00 Alienation71 --- Alienation67
gam2 -0.21151 0.049862 -4.2419 2.2164e-05 Alienation71 --- SES
the1 5.06546 0.373464 13.5635 0.e+00 Anomia67 -- Anomia67
the3 4.81176 0.397345 12.1098 0.e+00 Anomia71 -- Anomia71
the2 2.21438 0.3197406.9256 4.3423e-12 Powerless67 -- Powerless67
the4 2.68322 0.3312748.0997 4.4409e-16 Powerless71 -- Powerless71
thd1 2.73051 0.5177375.2739 1.3353e-07 Education -- Education
thd2 2.66905 0.182260 14.6442 0.e+00 SEI -- SEI
the13 1.88739 0.2416277.8112 5.7732e-15 Anomia71 -- Anomia67
psi1 4.70477 0.427511 11.0050 0.e+00 Alienation67 -- Alienation67
psi2 3.86642 0.343971 11.2406 0.e+00 Alienation71 -- Alienation71
phi6.87948 0.659208 10.4360 0.e+00 SES -- SES
Iterations = 58
A - sem.wh$A # structural coefficients
exog - apply(A, 1, function(x) all(x == 0))
endog - !exog
(B - A[endog, endog, drop=FALSE]) # direct effects, endogenous -
endogenous
Anomia67 Powerless67 Anomia71 Powerless71 Education SEI
Anomia670 00 0 0 0
Powerless67 0 00 0 0 0
Anomia710 00 0 0 0
Powerless71 0 00 0 0 0
Education 0 00 0 0 0
SEI 0 00 0 0 0
Alienation670 00 0 0 0
Alienation710 00
Re: [R] extremely slow recursion in R?
On Fri, 25 Aug 2006, Joerg van den Hoff wrote:
maybe someone's interested:
I made the same observation of seemingly very slow recursion recently: just
for fun I used the (in)famously inefficient
fib - function(n = 1) {
if (n 2)
fn - 1
else
fn - fib(n - 1) + fib(n - 2)
fn
}
for calculating the fibonacci numbers and compared `fib(30)' (about 1.3e6
recursive function calls ...) to some other languages (times in sec):
language time
==
C 0.034 (compiled, using integers)
Ocaml 0.041 (compiled, using integers)
Ocaml 0.048 (interpreted, using integers)
C 0.059 (compiled, using floats)
Lua1.1
ruby 3.4
R 21
octave120
apart from octave (which seems to have a _real_ issue with recursive function
calls), R is by far the slowest in this list and still a factor 7-20 slower
than the interpreter based Lua and ruby. the speed loss compared to C or
Ocaml is about a factor of 350-600 here (Ocaml keeps its speed more or less
in this simple example even in 'script mode', which is remarkable, I think
(and it usually loses only a factor of about 7 or so in script mode compared
to the compiled variant)
for the specialists the bad performance of R in this situation might not be
surprising, but I was not aware that recursive function calls are seemingly
as expensive as explicit loops (where the execution time ratio of R to C
again is of the same order, i.e. =~ 400).
Recursive function calls are likely to be *more* expensive than explicit
loops, because of the memory and function call overhead. This is true in
most languages, and is why tail recursion optimization is a basic feature
of compilers for functional languages.
of course, such comparsions don't make too much sense: the reason to use R
will definitely not be its speed (which, luckily, often does not matter), but
the combination of flexible language, the number of available libraries and
the good 2D graphics.
The benchmarks are interesting (and reinforce my feeling that I need to
look at ocaml sometime). They might be more interesting on a problem for
which recursion was a sensible solution -- something simple like printing
the nodes of a tree, or a more complicated problem like depth-first search
in a graph.
It's certainly true that people don't pick R over other minority languages
like Ocaml for its blazing speed, but for other reasons. As programming
blogger Steve Yegge has observed when comparing advantages and
disadvantages of various languages: ML and Haskell are both from Planet
Weird. And the people of that planet evidently do not believe in
libraries.
More to the point, the ratio of speeds is much lower for many uses. For
example, I recently translated to C an R program that computed the
conditional score estimator for a Cox model with measurement error, and
the C version ran about five times faster. This was about the same speed
increase than I had already obtained in the interpreted code by replacing
a bisection search with uniroot().
In some cases the ratio may even be less than 1 -- since R can use
optimized BLAS routines you might expect that some matrix operations would
be faster in interpreted R than in C code written by the average user. I
have a second-hand report from a student here that this was actually the
case -- he wrote compiled matrix routines and his code got slower.
It would certainly be nice if there were a faster statistical language
with nice features, whether by speeding up R or by adding statistical
features to something that compiles to native code, like Common Lisp or
ocaml, or by taking better advantage of multicore computers as they
proliferate. These problems do need more people to work on them
(especially as Bell Labs isn't in the statistical computing business any
more). There's a conference in New Zealand in February and abstract
submissions are still open.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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[R] increasing the # of socket connections
Dear R-helpers, using snow on socket connections, I ran into the following error cl - makeSOCKcluster(hosts) Error in socketConnection(port = port, server = TRUE, blocking = TRUE : all connections are in use with showConnections(all=T) showing 50 open connections. As - for administrative reasons - I would prefer to use snow's SOCK capabilities (instead of MPI and the like), I wonder if there is a way to increase the number of simultaneous open connections allowed in an R session (~100 would be sufficient). Any help/hints are greatly appreciated, best regards, Marc -- Dipl. Inform. Med. Marc Kirchner Interdisciplinary Centre for Scientific Computing (IWR) Multidimensional Image Processing INF 368 University of Heidelberg D-69120 Heidelberg Tel: ++49-6221-54 87 97 Fax: ++49-6221-54 88 50 [EMAIL PROTECTED] signature.asc Description: Digital signature __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] biglm 0.4
biglm fits linear and generalized linear models to large data sets, using bounded memory. What's New: generalized linear models. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Plot y ~ x under condition of variable a and b
Hi All, I want to plot y~ x under the condition of variable a and b. Followed is the dataset: plotid lndenlnvol source 369 9037.0 10.419002 -4.101039226 S 370 9037.0 9.840548 -2.432385723 S 371 9037.0 8.973351 -1.374842169 S 372 9037.0 8.242756 -0.813800113 S 373 9037.0 8.006368 -0.366743413 S 374 9037.0 7.396335 -0.041375532 S 375 9037.0 6.194405 0.744573249 S 376 9038.0 10.417209 -2.938129138 S 377 9038.0 9.709296 -1.906228589 S 378 9038.0 8.581107 -1.187441385 S 379 9038.0 7.539027 -0.748873856 S 380 9038.0 6.866933 -0.228547521 S 381 9038.0 6.672033 0.222818889 S 382 9038.0 6.380123 0.863026089 S 11003.1 7.281089 5.563470357 P 21003.1 7.165854 5.587837467 P 31003.1 7.126938 5.604757978 P 41003.1 6.833951 5.709078555 P 560 3.1 6.634462 5.678818058 P 610 3.2 7.052830 5.534234273 P 710 3.2 6.905777 5.559511276 P 810 3.2 6.885776 5.590614404 P 910 3.2 6.685106 5.716040812 P 10103.2 6.495349 5.631784504 P 11103.3 6.697376 5.414815010 P 12103.3 6.553336 5.441823472 P 13103.3 6.581116 5.455788329 P 14103.3 6.279641 5.543868038 P 15103.3 6.119298 5.528003301 P 16103.4 7.035589 5.783924732 P 17103.4 6.875624 5.798852319 P 18103.4 6.812445 5.807787244 P I used par.plot(lnvol~lnden|source,data=dat,sub=as.factor(plotid),col=T); It gave good plots, but it put the different data sources to separated graphs, i.e. S and P. What I want is to plot them on the same graph. If anyone has the experience in doing plotting like this, please kindly give me some hints. Thanks! Jen. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot y ~ x under condition of variable a and b [Broadcast ]
It's the |source in your formula that tells lattice to separate them.
If you drop that, you'll get all points without S and P distinguished at
all. If you add a groups argument, you should get them presented with
different colors/symbols/etc. depending on your trellis settings (warning:
untested code):
par.plot(lnvol~lnden, groups = source,data=dat,sub=as.factor(plotid),col=T)
Hope this helps,
Matt Wiener
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of jennystadt
Sent: Friday, August 25, 2006 11:28 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Plot y ~ x under condition of variable a and b [Broadcast]
Hi All,
I want to plot y~ x under the condition of variable a and b. Followed is the
dataset:
plotid lndenlnvol source
369 9037.0 10.419002 -4.101039226 S
370 9037.0 9.840548 -2.432385723 S
371 9037.0 8.973351 -1.374842169 S
372 9037.0 8.242756 -0.813800113 S
373 9037.0 8.006368 -0.366743413 S
374 9037.0 7.396335 -0.041375532 S
375 9037.0 6.194405 0.744573249 S
376 9038.0 10.417209 -2.938129138 S
377 9038.0 9.709296 -1.906228589 S
378 9038.0 8.581107 -1.187441385 S
379 9038.0 7.539027 -0.748873856 S
380 9038.0 6.866933 -0.228547521 S
381 9038.0 6.672033 0.222818889 S
382 9038.0 6.380123 0.863026089 S
11003.1 7.281089 5.563470357 P
21003.1 7.165854 5.587837467 P
31003.1 7.126938 5.604757978 P
41003.1 6.833951 5.709078555 P
560 3.1 6.634462 5.678818058 P
610 3.2 7.052830 5.534234273 P
710 3.2 6.905777 5.559511276 P
810 3.2 6.885776 5.590614404 P
910 3.2 6.685106 5.716040812 P
10103.2 6.495349 5.631784504 P
11103.3 6.697376 5.414815010 P
12103.3 6.553336 5.441823472 P
13103.3 6.581116 5.455788329 P
14103.3 6.279641 5.543868038 P
15103.3 6.119298 5.528003301 P
16103.4 7.035589 5.783924732 P
17103.4 6.875624 5.798852319 P
18103.4 6.812445 5.807787244 P
I used par.plot(lnvol~lnden|source,data=dat,sub=as.factor(plotid),col=T);
It gave good plots, but it put the different data sources to separated
graphs, i.e. S and P. What I want is to plot them on the same graph. If
anyone has the experience in doing plotting like this, please kindly give me
some hints. Thanks!
Jen.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] fitting a gaussian to some x,y data
I apologize if this is redundant. I've been Googling, searching the archive and reading the help all morning and I am not getting closer to my goal. I have a series of data( xi, yi). It is not evenly sampled and it is messy (meaning that there is a lot of scatter in the data). I want to fit a normal distribution (i.e. a gaussian) to the data in order to find the center. (The data has a loose normal look to it.) I have done this with polynomial fitting in R with nls but want to try it with a Gaussian (I am a student astronomer and have not a lot of experience in statistics). In looking at the fitdistr function, it seems to take as input a bunch of x values and it will fit a gaussian to the histogram. That is not what I need to do, I want to fit a normal distribution to the x,y values and get out the parameters of the fit. I'm fooling with nls but it seems to want the data in some other format or something because it is complaining about singular gradient. I'm sure this isn't hard and the answer must be out there somewhere but I can't find it. I appreciate any assistance. Cheers, Michael filepath - system.file(data, infile , package=datasets) mm -read.table(filepath) mm dmk - data.frame( x=mm$V1, y=mm$V2) attach(dmk) plot(x,y) #ymk -nls(y~c*x^2+b*x+a,start=list(a=1,b=1,c=1),trace=TRUE) ymk -nls(y~a*exp(-x^2/sig),start=list(a=1,sig=1),trace=TRUE) co = coef(ymk) cmk - list(a=co[[1]], b=co[[2]], c=co[[3]] ) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] increasing the # of socket connections
On Fri, 25 Aug 2006, Marc Kirchner wrote: As - for administrative reasons - I would prefer to use snow's SOCK capabilities (instead of MPI and the like), I wonder if there is a way to increase the number of simultaneous open connections allowed in an R session (~100 would be sufficient). You need to change #define NCONNECTIONS 50 at the top of src/main/connections.c and recompile. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] increasing the # of socket connections
On Fri, 25 Aug 2006, Marc Kirchner wrote: Dear R-helpers, using snow on socket connections, I ran into the following error cl - makeSOCKcluster(hosts) Error in socketConnection(port = port, server = TRUE, blocking = TRUE : all connections are in use with showConnections(all=T) showing 50 open connections. As - for administrative reasons - I would prefer to use snow's SOCK capabilities (instead of MPI and the like), I wonder if there is a way to increase the number of simultaneous open connections allowed in an R session (~100 would be sufficient). If you really need them open at once (and are not just forgetting to close them), then change the constant in the file and recompile. Any help/hints are greatly appreciated, Is this the appropriate list: not on my reading of the posting guide! best regards, Marc -- Brian D. Ripley, [EMAIL PROTECTED] 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot with different symbols and colors?
Dear List, I try to make a xyplot with different colors and symbols, I came this far: library(DAAG) xyplot(csoa~it|sex*agegp,data=tinting,groups=target,pch=list(1,2),auto.key=list(space = right)) this produces a plot with different colors and symbols but unfortunately the legend does only follow the color scheme and not the different symbols. Any suggestions what to change? And how do I change or switch of that background colors in each plots title? Stefan Grosse __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 get back POSIXct format after calculating with hist() results
On Fri, 25 Aug 2006, [EMAIL PROTECTED] wrote: Hi, I have a casting/formatting question on hist.POSIXt: The histogram plot from POSIXct works perfect (with help of Prof. Ripley -thanks!). When processing the hist(plot=FALSE) output and then plotting the results over the x-axis (bins) coming from hist(), I lose the date/time labels, getting instead integers displayed. Trying to cast the $breaks with as.POSIXct gives silly results with StarTrek-like years. But you actually called as.POSIXct(as.Date()) on them, which was `silly'. class(login_bins$mids) - class(logins$DateTime) should do the trick. What is the proper way to get back the old YY-MM-DD hh:mm:ss labels ? Peter str(logins) `data.frame': 25792 obs. of 1 variable: $ DateTime:'POSIXct', format: chr 2006-08-01 00:00:02 2006-08-01 00:00:25 2006-08-01 00:00:41 ... login_bins-hist(samples$DateTime, hours, freq=TRUE,plot=FALSE) str(login_bins) List of 7 $ breaks : num [1:337] 1.15e+09 1.15e+09 1.15e+09 1.15e+09 1.15e+09 ... $ counts : int [1:336] 6 10 45 25 40 87 257 356 309 214 ... $ intensities: num [1:336] 0.000233 0.000388 0.001744 0.000969 0.001550 ... $ density: num [1:336] 6.46e-08 1.08e-07 4.85e-07 2.69e-07 4.31e-07 ... $ mids : num [1:336] 1.15e+09 1.15e+09 1.15e+09 1.15e+09 1.15e+09 ... $ xname : chr logins$DateTime $ equidist : logi TRUE - attr(*, class)= chr histogram ... Then I do some calculations with the login_bins$counts, resulting in a new vector: str(userabs) int [1:336] 1 -3 34 39 68 127 347 641 844 943 ... When I now plot this y over the x-axis coming from the previous histogram x-axis with plot(login_bins$mids, userabs) I get the x-axis of labelled with the nums 115440, 115480... (number of seconds since 1970 ?) Trying to convert this gives funny results. as.POSIXct(as.Date(login_bins$mids[1])) [1] 2568-10-12 02:00:00 W. Europe Daylight Time as.POSIXct(as.Date(login_bins$mids[2])) [1] 2578-08-21 02:00:00 W. Europe Daylight Time __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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, [EMAIL PROTECTED] 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 get back POSIXct format after calculating with hist() results
yep, it works! impressingly elegant. R seems worth to dig into the details. I'll try harder. Thank You very much Peter -Ursprüngliche Nachricht- Von: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 25. August 2006 17:53 An: Marx Peter Cc: r-help@stat.math.ethz.ch Betreff: Re: [R] How to get back POSIXct format after calculating with hist() results On Fri, 25 Aug 2006, [EMAIL PROTECTED] wrote: Hi, I have a casting/formatting question on hist.POSIXt: The histogram plot from POSIXct works perfect (with help of Prof. Ripley -thanks!). When processing the hist(plot=FALSE) output and then plotting the results over the x-axis (bins) coming from hist(), I lose the date/time labels, getting instead integers displayed. Trying to cast the $breaks with as.POSIXct gives silly results with StarTrek-like years. But you actually called as.POSIXct(as.Date()) on them, which was `silly'. class(login_bins$mids) - class(logins$DateTime) should do the trick. What is the proper way to get back the old YY-MM-DD hh:mm:ss labels ? Peter str(logins) `data.frame': 25792 obs. of 1 variable: $ DateTime:'POSIXct', format: chr 2006-08-01 00:00:02 2006-08-01 00:00:25 2006-08-01 00:00:41 ... login_bins-hist(samples$DateTime, hours, freq=TRUE,plot=FALSE) str(login_bins) List of 7 $ breaks : num [1:337] 1.15e+09 1.15e+09 1.15e+09 1.15e+09 1.15e+09 ... $ counts : int [1:336] 6 10 45 25 40 87 257 356 309 214 ... $ intensities: num [1:336] 0.000233 0.000388 0.001744 0.000969 0.001550 ... $ density: num [1:336] 6.46e-08 1.08e-07 4.85e-07 2.69e-07 4.31e-07 ... $ mids : num [1:336] 1.15e+09 1.15e+09 1.15e+09 1.15e+09 1.15e+09 ... $ xname : chr logins$DateTime $ equidist : logi TRUE - attr(*, class)= chr histogram ... Then I do some calculations with the login_bins$counts, resulting in a new vector: str(userabs) int [1:336] 1 -3 34 39 68 127 347 641 844 943 ... When I now plot this y over the x-axis coming from the previous histogram x-axis with plot(login_bins$mids, userabs) I get the x-axis of labelled with the nums 115440, 115480... (number of seconds since 1970 ?) Trying to convert this gives funny results. as.POSIXct(as.Date(login_bins$mids[1])) [1] 2568-10-12 02:00:00 W. Europe Daylight Time as.POSIXct(as.Date(login_bins$mids[2])) [1] 2578-08-21 02:00:00 W. Europe Daylight Time __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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, [EMAIL PROTECTED] 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot with different symbols and colors?
On Fri, 25 Aug 2006 17:51:24 +0200, Stefan Grosse [EMAIL PROTECTED] wrote: Dear List, I try to make a xyplot with different colors and symbols, I came this far: library(DAAG) xyplot(csoa~it|sex*agegp,data=tinting,groups=target,pch=list(1,2),auto.key=list(space = right)) this produces a plot with different colors and symbols but unfortunately the legend does only follow the color scheme and not the different symbols. Any suggestions what to change? I think that 'pch' is passed only to the panel function, and the key uses whatever the current trellis.par.set() is, if you don't define it explicitly (all explained in ?xyplot). I find it easier to call trellis.par.set before the xyplot: R trellis.par.set(superpose.symbol=list(pch=c(1, 2))) Warning message: Note: The default device has been opened to honour attempt to modify trellis settings in: trellis.par.set(superpose.symbol = list(pch = c(1, 2))) R xyplot(csoa~it|sex*agegp,data=tinting,groups=target,auto.key=list(space=right)) Cheers, -- Seb __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 critical values
Adam D. I. Kramer wrote: Hello, I have an HLM-like situation: a data frame with rows representing individuals, and columns representing some statistics for the individual (e.g., a mean and variance for that individual, plus the number of observations for that person). I'm interested in computing a confidence interval around each individual's mean, but to do that, I would need to know the critical t-value for each individual. Ergo, is there any sort of function that returns a critical t-value for a given p/alpha and df? ?qt qt(p=.05, df=100) [1] -1.66023 Thanks, Adam Kramer __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot y ~ x under condition of variable a and b[Broadcast ]
Thanks Matthew. This does help. But from the discription of the function and
examples, we are not able to use groups argument and the trellis settings you
mentioned.
Jen.
- Original Message -
From: Wiener, Matthew, [EMAIL PROTECTED]
Sent: 2006-08-25, 09:46:44
To: jennystadt; r-help@stat.math.ethz.ch, [EMAIL PROTECTED];
r-help@stat.math.ethz.ch
Subject: RE: [R] Plot y ~ x under condition of variable a and b[Broadcast ]
It's the |source in your formula that tells lattice to separate them.
If you drop that, you'll get all points without S and P distinguished at
all. If you add a groups argument, you should get them presented with
different colors/symbols/etc. depending on your trellis settings (warning:
untested code):
par.plot(lnvol~lnden, groups = source,data=dat,sub=as.factor(plotid),col=T)
Hope this helps,
Matt Wiener
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of jennystadt
Sent: Friday, August 25, 2006 11:28 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Plot y ~ x under condition of variable a and b [Broadcast]
Hi All,
I want to plot y~ x under the condition of variable a and b. Followed is the
dataset:
plotid lndenlnvol source
369 9037.0 10.419002 -4.101039226 S
370 9037.0 9.840548 -2.432385723 S
371 9037.0 8.973351 -1.374842169 S
372 9037.0 8.242756 -0.813800113 S
373 9037.0 8.006368 -0.366743413 S
374 9037.0 7.396335 -0.041375532 S
375 9037.0 6.194405 0.744573249 S
376 9038.0 10.417209 -2.938129138 S
377 9038.0 9.709296 -1.906228589 S
378 9038.0 8.581107 -1.187441385 S
379 9038.0 7.539027 -0.748873856 S
380 9038.0 6.866933 -0.228547521 S
381 9038.0 6.672033 0.222818889 S
382 9038.0 6.380123 0.863026089 S
11003.1 7.281089 5.563470357 P
21003.1 7.165854 5.587837467 P
31003.1 7.126938 5.604757978 P
41003.1 6.833951 5.709078555 P
560 3.1 6.634462 5.678818058 P
610 3.2 7.052830 5.534234273 P
710 3.2 6.905777 5.559511276 P
810 3.2 6.885776 5.590614404 P
910 3.2 6.685106 5.716040812 P
10103.2 6.495349 5.631784504 P
11103.3 6.697376 5.414815010 P
12103.3 6.553336 5.441823472 P
13103.3 6.581116 5.455788329 P
14103.3 6.279641 5.543868038 P
15103.3 6.119298 5.528003301 P
16103.4 7.035589 5.783924732 P
17103.4 6.875624 5.798852319 P
18103.4 6.812445 5.807787244 P
I used par.plot(lnvol~lnden|source,data=dat,sub=as.factor(plotid),col=T);
It gave good plots, but it put the different data sources to separated
graphs, i.e. S and P. What I want is to plot them on the same graph. If
anyone has the experience in doing plotting like this, please kindly give me
some hints. Thanks!
Jen.
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Re: [R] Check values in colums matrix
Hi and if speed is an issue and object is numeric matrix something like function(x) which(colSums(abs(diff(x)))==0) is a little bit quicker Cheers Petr On 25 Aug 2006 at 13:39, [EMAIL PROTECTED] wrote: Date sent: Fri, 25 Aug 2006 13:39:48 +1000 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED] Copies to: r-help@stat.math.ethz.ch Subject:Re: [R] Check values in colums matrix As a minor footnote to both of these, I would add that both assume that all the columns of the dataset are numeric. It doesn't cost much to generalize it to cover any matrix structure, of any mode: constantColmuns - function(Xmat) which(apply(Xmat, 2, function(z) length(unique(z)) == 1)) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Berton Gunter Sent: Friday, 25 August 2006 9:37 AM To: 'Gabor Grothendieck'; 'Muhammad Subianto' Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Check values in colums matrix Absolutely. But do note that if the values in obj are the product of numerical computations then columns of equal values may turn out to be only **nearly** equal and so the sd may turn out to be **nearly** 0 and not exactly 0. This is a standard issue in numerical computation, of course, and has been commented on in this list at least dozens of times, but it's still a gotcha for the unwary (so now dozens +1). -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gabor Grothendieck Sent: Thursday, August 24, 2006 4:28 PM To: Muhammad Subianto Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Check values in colums matrix Try sd(obj.tr) which will give a vector of standard deviations, one per column. A column's entry will be zero if and only if all values in the column are the same. On 8/24/06, Muhammad Subianto [EMAIL PROTECTED] wrote:Dear all,I apologize if my question is quite simple.I have a dataset (20 columns 1000 rows) whichsome of columns have the same value and the others have different values.Here are some piece of my dataset: obj - cbind(c(1,1,1,4,0,0,1,4,-1), c(0,1,1,4,1,0,1,4,-1),c(1,1,1,4,2,0,1,4,-1), c(1,1,1,4,3,0,1,4,-1), c(1,1,1,4,6,0,1,5,-1),c(1,1,1,4,6,0,1,6,-1), c(1,1,1,4,6,0,1,7,-1), c(1,1,1,4,6,0,1,8,-1))obj.tr - t(obj)obj.tr obj.tr[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [1,]11140014 -1[2,]01 141014 -1[3,]11142 014 -1[4,]11143014 -1[5,]11146015 -1 [6,]11146016 -1[7,]11 146017 -1[8,]11146 018 -1 How can I do to check columns 2,3,4,6,7 and 9 havethe same value, and columns 1,5 and 8 have different values. Best, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Need help with difficulty loading page www.bioconductor.org
I didn't see a response to this, and am not an expert, but wanted to point you in the right direction. You should ask this question on the bioconductor mailing list. The site is very much alive at this end (though I'm very close to it, physically), so you'll need to provide some more information about the problems you're experiencing. At a minimum it would help to include the output of the R command sessionInfo() If you're using Windows you might also look at mailing list threads dealing with server proxies. Searchable archives are at http://dir.gmane.org/gmane.science.biology.informatics.conductor this link http://article.gmane.org/gmane.science.biology.informatics.conductor/2152/match=internet2 and searching for internet2 might be a good start. Martin Debashis Bhattacharya [EMAIL PROTECTED] writes: The page is either too busy, or there is something seriously wrong with access to this page. Most of the time, trying to reach www.bioconductor.org results in failure. Only once in a blue moon, do I get through. In fact, thus far, I have not been able to install bioconductor, since the first source(...) command from the R command window -- following instruction on www.bioconductor.org page, that I did manage to reach, one time -- has failed, every time. Please help. Debashis Bhattacharya. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot with different symbols and colors?
R trellis.par.set(superpose.symbol=list(pch=c(1, 2))) Warning message: Note: The default device has been opened to honour attempt to modify trellis settings in: trellis.par.set(superpose.symbol = list(pch = c(1, 2))) R xyplot(csoa~it|sex*agegp,data=tinting,groups=target,auto.key=list(space=right)) Thanks that did it! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot with different symbols and colors?
On 8/25/06, Sebastian P. Luque [EMAIL PROTECTED] wrote: On Fri, 25 Aug 2006 17:51:24 +0200, Stefan Grosse [EMAIL PROTECTED] wrote: Dear List, I try to make a xyplot with different colors and symbols, I came this far: library(DAAG) xyplot(csoa~it|sex*agegp,data=tinting,groups=target,pch=list(1,2),auto.key=list(space = right)) this produces a plot with different colors and symbols but unfortunately the legend does only follow the color scheme and not the different symbols. Any suggestions what to change? I think that 'pch' is passed only to the panel function, and the key uses whatever the current trellis.par.set() is, if you don't define it explicitly (all explained in ?xyplot). I find it easier to call trellis.par.set before the xyplot: R trellis.par.set(superpose.symbol=list(pch=c(1, 2))) Warning message: Note: The default device has been opened to honour attempt to modify trellis settings in: trellis.par.set(superpose.symbol = list(pch = c(1, 2))) R xyplot(csoa~it|sex*agegp,data=tinting,groups=target,auto.key=list(space=right)) Also, see the entry for 'par.settings' in help(xyplot). -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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.squared in Weighted Least Square using the Lm Function
Hello all, I am using the function lm to do my weighted least square regression. model-lm(Y~X1+X2, weight=w) What I am confused is the r.squared. It does not seem that the r.squared for the weighted case is an ordinary 1-RSS/TSS. What is that precisely? Is the r.squared measure comparable to that obtained by the ordinary least square? I also notice that model$res is the unweighted residual while summary(model)$res is the weighted residual Thank you in advance for helping! Charles __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] fitting a gaussian to some x,y data
Michael Koppelman wrote: I apologize if this is redundant. I've been Googling, searching the archive and reading the help all morning and I am not getting closer to my goal. I have a series of data( xi, yi). It is not evenly sampled and it is messy (meaning that there is a lot of scatter in the data). I want to fit a normal distribution (i.e. a gaussian) to the data in order to find the center. (The data has a loose normal look to it.) I have done this with polynomial fitting in R with nls but want to try it with a Gaussian (I am a student astronomer and have not a lot of experience in statistics). In looking at the fitdistr function, it seems to take as input a bunch of x values and it will fit a gaussian to the histogram. That is not what I need to do, I want to fit a normal distribution to the x,y values and get out the parameters of the fit. I'm fooling with nls but it seems to want the data in some other format or something because it is complaining about singular gradient. I'm sure this isn't hard and the answer must be out there somewhere but I can't find it. I appreciate any assistance. Fitting a distribution simply means estimating the parameters of that distribution. For a Gaussian distribution this problem is trivial. The parameters are the mean vector mu and the covariance matrix Sigma. The (optimal; maximum-likelihood-adjusted-to-be-unbiased) estimates of these are the obvious ones --- the sample mean and the sample covariance. In R you most easily get these by o cbinding your ``x'' and ``y'' vectors into matrix: xy - cbind(x,y) o applying mean() to this matrix: mu.hat - apply(xy,2,mean) o calling upon var(): Sigma.hat - var(xy) That is all there is to it. If all you really want is an estimate of the ``centre'' then this estimate is just mu.hat; you don't need to ``fit a distribution'' at all. From your description of the problem there may well be other issues --- lack of independence of your data, biased sample, outliers, skewness, god knows what. Such issues should be dealt with as carefully as possible. It may be the case that you should be doing some sort of robust estimation. Or it may be the case that this data set is hopeless and should be thrown away and a new data set collected in a manner that will actually yield some information. But ``fitting a distribution'' is *NOT* the issue. I don't mean to be rude, but this question illustrates the point that you really shouldn't be *doing* statistics unless you *understand* some statistics. At the very best you'll waste a great deal of time. cheers, Rolf Turner [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R.squared in Weighted Least Square using the Lm Function
On Fri, 25 Aug 2006, Charles wrote:
Hello all,
I am using the function lm to do my weighted least
square regression.
model-lm(Y~X1+X2, weight=w)
What I am confused is the r.squared.
What r.squared? There is no r.squared in that object, but it is
calculated by the summary method.
It does not seem that the r.squared for the weighted
case is an ordinary 1-RSS/TSS.
What is that precisely?
Precisely that, with weights in the SS. The code is
r - z$residuals
f - z$fitted
w - z$weights
if (is.null(w)) {
mss - if (attr(z$terms, intercept))
sum((f - mean(f))^2)
else sum(f^2)
rss - sum(r^2)
} else {
mss - if (attr(z$terms, intercept)) {
m - sum(w * f/sum(w))
sum(w * (f - m)^2)
}
else sum(w * f^2)
rss - sum(w * r^2)
r - sqrt(w) * r
}
ans$r.squared - mss/(mss + rss)
That's the great thing about R: you can answer your own question by
reading the code for yourself.
Is the r.squared measure comparable to that obtained
by the ordinary least square?
I also notice that
model$res is the unweighted residual while
summary(model)$res is the weighted residual
Yes, as documented with added emphasis in ?summary.lm .
--
Brian D. Ripley, [EMAIL PROTECTED]
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
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[R] Modifying the embed-results
Hi, Here is a vector and the result from the embed-command: VECTOR=c(0,3,6,3,11,2,4,3,7,6,4,5,10,2,3,5,8) embed(VECTOR, dimension=5) [,1] [,2] [,3] [,4] [,5] [1,] 113630 [2,]2 11363 [3,]42 1136 [4,]342 113 [5,]7342 11 [6,]67342 [7,]46734 [8,]54673 [9,] 105467 [10,]2 10546 [11,]32 1054 [12,]532 105 [13,]8532 10 Is there a way to little modify the algorithm so that the result looks like this: [1] 0 3 6 11 2 - beginning from the first number of the VECTOR [1] 3 6 11 2 4 - beginning from the second number of the VECTOR etc [1] 6 3 11 2 4 [1] 3 11 2 4 7 [1] 11 2 4 3 7 [1] 2 4 3 7 6 [1] 4 3 7 6 5 [1] 3 7 6 4 5 [1] 7 6 4 5 10 [1] 6 4 5 10 2 [1] 4 5 10 2 3 [1] 5 10 2 3 8 [1] 10 2 3 5 8 Every row consists of next five unique(!) member of the VECTOR. I made this example result with a time consuming algorithm which uses for-loops and whiles. How to do this better?? Thanks in advance! Atte Tenkanen University of Turku __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] issues with Sweave and inclusion of graphics in a document
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
savePlot is just an internal version of dev.copy, part of the support for
the menus on the windows() graphics device.
It is described in `An Introduction to R' (the most basic R manual).
the most basic R manual doesn't quite answer my question.
by itself, dev.copy doesn't copy the width and height of the device whereas
savePlot
copies whatever is displayed on the screen giving me
what-you-see-is-what-you-save
capabilities (but only under the Windows OS).
i can get pretty close to this in linux by writing a function to save the
plot to a pdf device:
label=first.ar.1, results=hide=
# no savePlot in Linux ... so write my own function
savePlotAsPdf- function(pdfname, from=dev.cur()) {
from- from
pdf(pdfname, width=width, height=height)
to- dev.cur()
dev.set(from)
dev.copy(which=to)
dev.off()
}
# a long AR process is best viewed in a wide window ...
# width height are now variables
width- 20; height- 5
x11(width=width, height=height)
sp- make.ar.1(alpha=.5, n=800)
plot(sp, type=l, col=blue)
# width height via dynamic scoping in savePlotAsPdf
savePlotAsPdf(ar.pdf)
@
the only (minor) inconvenience is that i have to specify width and height as
variables to
take advantage of dynamic scoping in order to minimize mess.
while this is a workaround, via dev.copy, as you pointed out, it still doesn't
answer why
Sweave doesn't like x11() devices (or windows() devices under the Windows OS
for that
matter) within figure environments.
perhaps this is a question for the package maintainer? but i was hoping that
all the avid
Sweave users would pitch in with how they work with graphics in practice.
here is a revised .Rnw example illustrating the above:
% example-linux2.Rnw
\documentclass[a4paper]{article}
\begin{document}
\noindent This is an example of how I can use \texttt{Sweave} under Linux.
echo=false,results=hide=
# create a simple AR process:
make.ar.1- function(alpha=1,n=300) {
Z- rnorm(n);
Y- numeric(n);
Y[1]- Z[1];
for (i in 2:n) Y[i]- alpha*Y[i-1]+Z[i];
return(Y)
}
@
\texttt{Sweave} doesn't like the [EMAIL PROTECTED](width=width, height=height)@
command
in the following code chunk if it is placed within a \texttt{figure}
environment.
Instead, I have to save the plot to a file and then I use [EMAIL PROTECTED]@ in
the \texttt{figure} environment. This isn't a bad thing, as it allows me to
fine-tune
the \LaTeX\ graphics placement.
label=first.ar.1, results=hide=
# no savePlot in Linux ... so write my own function
savePlotAsPdf- function(pdfname, from=dev.cur()) {
from- from
pdf(pdfname, width=width, height=height)
to- dev.cur()
dev.set(from)
dev.copy(which=to)
dev.off()
}
# a long AR process is best viewed in a wide window ...
# width height are now variables
width- 20; height- 5
x11(width=width, height=height)
sp- make.ar.1(alpha=.5, n=800)
plot(sp, type=l, col=blue)
# width height via dynamic scoping in savePlotAsPdf
savePlotAsPdf(ar.pdf)
@
\begin{figure}
\begin{center}
% i retain direct control over graphics in LaTeX; i can fine-tune the
% the graphics placement as much as i want:
\includegraphics[width=14.5cm]{./ar.pdf}
\caption{An AR(1) process of length~\protect\Sexpr{length(sp)}
is best viewed in a wide window.}
\end{center}
\end{figure}
Under X, then,
\begin{itemize}
\item Why doesn't \texttt{Sweave} like [EMAIL PROTECTED](width=width,
height=height)@?
\end{itemize}
There are, however, advantages to doing things this way:
\begin{itemize}
\item I can save the plot to a file without writing any other code;
\item I can include the saved plot in my \LaTeX\ figure, allowing me to
fine-tune with the [EMAIL PROTECTED]@ command.
\end{itemize}
\end{document}
% example-linux2.Rnw
On Sat, 19 Aug 2006, Thomas Harte wrote:
the problem is a little hard to explain; the .Rnw files (below)
probably do a better job, but here goes ...
Sweave doesn't like it when i size a graphical device in a code
chunk using either, e.g.:
windows(width=20, height=5)
in Windows, or, e.g.
x11(width=20, height=5)
under X, when i then plot something in said device and try to
include this graphical output in the resulting document.
Sweave does not object to my writing code chunks in the above
manner, so long as i do not wish to include the code in a LaTeX
figure environment.
oftentimes i want to do precisely what Sweave doesn't appear
to allow. for example, with time-series data, i want to see a
[R] tcltk command to figure out which widget is on focus (or clicked)
Hi, I'm making an interface, where a Tcl/Tk window have few listbox widgets. I need to select separate parameters from separate listboxes. It is clear how to get cursor selection value, once you know which listbox widget you clicked. The problem is I can't figure out which one tcltk command to use to get an information which listbox widget is in focus (or clicked). Thank you, Vlad __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 contrast with factorial experiment
Hello, Ted, Thank you for your help! So I can not contrast the mean yields between sections 1-8 and 9-11 under Trt but I can contrast mean yields for sections 1-3 and 6-11 because there exists significant interaction between two factors (Trt:section4, Trt:section5). Could I use the commands below to test the difference between sections 1-3 and 6-11 ? contrasts(section)-c(-2,-2,-2,0,0,1,1,1,1,1,1) newobj-lm(log2(yield)~treat*section) summary(newobj) Call: lm(formula = log2(yield) ~ treat * section) Residuals: Min 1Q Median 3Q Max -0.49647 -0.14913 -0.01521 0.17471 0.51105 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 6.288400.05003 125.682 2e-16 *** treatTrt1.221220.07076 17.259 2e-16 *** section10.178310.03911 4.559 4.08e-05 *** section2 -0.231020.16595 -1.392 0.17087 section32.381700.16595 14.352 2e-16 *** section43.368340.16595 20.298 2e-16 *** section5 -1.568730.16595 -9.453 3.67e-12 *** section6 -0.415220.16595 -2.502 0.01613 * section7 -0.899430.16595 -5.420 2.38e-06 *** section80.095220.16595 0.574 0.56901 section9 -0.787840.16595 -4.748 2.21e-05 *** section10 0.748210.16595 4.509 4.79e-05 *** treatTrt:section1 0.101010.05532 1.826 0.07461 . treatTrt:section2 0.272700.23468 1.162 0.25151 treatTrt:section3 -1.222100.23468 -5.207 4.85e-06 *** treatTrt:section4 -1.391870.23468 -5.931 4.26e-07 *** treatTrt:section5 -0.761370.23468 -3.244 0.00225 ** treatTrt:section6 0.073200.23468 0.312 0.75658 treatTrt:section7 0.331080.23468 1.411 0.16535 treatTrt:section8 -0.136860.23468 -0.583 0.56276 treatTrt:section9 0.220860.23468 0.941 0.35180 treatTrt:section10 -0.144760.23468 -0.617 0.54054 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 0.2874 on 44 degrees of freedom Multiple R-Squared: 0.973, Adjusted R-squared: 0.9601 F-statistic: 75.55 on 21 and 44 DF, p-value: 2.2e-16 How can I infer that there is significant difference between sections 1-3 and sections 6-11 for the Trt from the output above? Joshua Quoting (Ted Harding) [EMAIL PROTECTED]: On 24-Aug-06 [EMAIL PROTECTED] wrote: Hello, R users, I have two factors (treat, section) anova design experiment where there are 3 replicates. The objective of the experiment is to test if there is significant difference of yield between top (section 9 to 11) and bottom (section 9 to 11) [I think you mean sections 1 to 8] of the fruit tree under treatment. I found that there are interaction between two factors. I wonder if I can contrast means from levels of one factor (section) under another factor (treat)? if so, how to do it in R and how to interpret the output? I think you would be well advised to look at a plot of the data. For example, let Y stand for yield, R for replicate, T for treat and S for section. ix-(T==Trt);plot(S[ix],Y[ix],col=red,ylim=c(0,1000)) ix-(T==Ctl);points(S[ix],Y[ix],col=blue) From this it is clear that sections 4 and 5 are in a class of their own. Also, in sections 1-3 and 6-11 the Ctl yields are not only lower, but have smaller (in some cases hardly any) variance, compared with the Trt yields. The variances for sections 7,8,9,10,11 are greater than for 1,2,3,6 without great change in mean value. While there is an evident difference between Trt yields and Ctrl yields for sections 1-3 and 6-11, this is not so for sections 4 and 5. This sort of behaviour no doubt provides some reasons for the interaction you observed. You seem to have a quite complex phenomenon here! To some extent the problems with variance can be diminished by working with logarithms. Compare the previous plot with ix-(T==Trt);plot(S[ix],log10(Y[ix]),col=red,ylim=c(0,3)) ix-(T==Ctl);points(S[ix],log10(Y[ix]),col=blue) (you have used log2() in your commands). The above observations can be seen reflected in R if you look at the output of summary(obj) where in particular: treatTrt:section2 -1.116910.33189 -3.365 0.001595 ** treatTrt:section3 -0.456340.33189 -1.375 0.176099 treatTrt:section4 -1.566270.33189 -4.719 2.42e-05 *** treatTrt:section5 -1.736040.33189 -5.231 4.48e-06 *** treatTrt:section6 -0.913110.33189 -2.751 0.008588 ** treatTrt:section7 -0.078530.33189 -0.237 0.814055 treatTrt:section8 0.179350.33189 0.540 0.591654 treatTrt:section9 -0.288590.33189 -0.870 0.389277 treatTrt:section10 0.069130.33189 0.208 0.835972 treatTrt:section11 -0.296490.33189 -0.893 0.376543 which, precisely, contrasts means from levels of one factor (section) under another factor (treat), and shows that most of the interaction arises in sections 4 and 5.
[R] How to iteratively extract elements out of a list
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
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[R] tick.number for date in xyplot
I would like a tick mark for each month; for example, xyplot(runif(365)~I(as.Date(1999-01-01) + 1:365), scales=list(x=list(format=%b %Y,tick.number=12))) I know I could make x numeric and use 'at' and 'labels', but I was wondering if there is a more direct route I'm missing. (In particular, one that doesn't have to be modified for new data). Thanks, Ben __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword remove AND library but got timed out.(why?)) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Thanks in adv. Horace __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] fitting a gaussian to some x,y data
Thank you. Yes, I do feel that I am under-qualified to even ask questions of y'all. Plus I'm an astronomer, which doesn't help! ;) I'll try again. I have two columns of data, the first column (x) is a distance (or length or separation) and the second column (y) is a flux (or number of counts or brightness) at that distance. Thus, when you plot y vs. x you get a spatial profile which shows how bright this thing is as a function of position. (See the small, attached PNG file. You can see there is a vague gaussian shape to the data.) This is measured data from a bizarre technique which yields data that is not evenly-spaced in x and it does not represent a pure mathematical function (i.e. it is not a point spread function or something like that), it represents the actual, non-uniform shape of an astronomical object. We are making the assumption that the shape of this object can be roughly represented with a gaussian. I want to fit a gaussian to this with the purpose of determining systematically the center of the normal-like shape of the spatial feature. I have successfully done so in R with a polynomial but I can't figure out how to do it with a gaussian. Does that make sense? Thanks! Michael On Aug 25, 2006, at 2:04 PM, MARK LEEDS wrote: hi : i'm not clear on what you mean but someone else might be ? if you say ( x,y), then it sounds like you are talking about a bivariate normal distribution. to fit a regular one dimensional gaussian distribution, you can't have 2 dimensional data. i honestly don't mean to sound rude but i think you should explain what you want to do more clearly because I don't think I am the only one that will be confused by what you said. send out a clearer email and you will get quite a response because there are a lot of really smart people ( compared to me ) on this list that love to help. it's an amazing list in that sense. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] rgl: exporting to pdf or png does not work
On 8/24/2006 4:34 PM, Gaspard Lequeux wrote:
Hej,
On Thu, 24 Aug 2006, Duncan Murdoch wrote:
On 8/24/2006 11:19 AM, Gaspard Lequeux wrote:
On Wed, 23 Aug 2006, Duncan Murdoch wrote:
On 8/23/2006 5:15 PM, Gaspard Lequeux wrote:
When exporting a image from rgl, the following error is encountered:
rgl.postscript('testing.pdf', fmt=pdf)
RGL: ERROR: can't bind glx context to window
RGL: ERROR: can't bind glx context to window
Warning messages:
1: X11 protocol error: GLXBadContextState
2: X11 protocol error: GLXBadContextState
The pdf file is created and is readable, but all the labels are gone.
Taking a snapshot (to png) gives 'failed' and no file is created.
Version of rgl used: 0.67-2 (2006-07-11)
Version of R used: R 2.3.1; i486-pc-linux-gnu; 2006-07-13 01:31:16;
Running Debian GNU/Linux testing (Etch).
That looks like an X11 error to me, not something that I'm very likely
to be able to fix. If you can debug the error, it would be helpful.
Actually after upgrading everything (debian testing (etch)) and restarting
X, I didn't get that error anymore. It was worse: R crashed:
library(rgl);triangles3d(c(1,,2,3),c(1,2,4),c(1,3,5));rgl.postscript('testing.pdf','pdf')
X Error of failed request: GLXBadContextState
Major opcode of failed request: 142 (GLX)
Minor opcode of failed request: 5 (X_GLXMakeCurrent)
Serial number of failed request: 85
Current serial number in output stream: 85
[EMAIL PROTECTED]:~/seqanal$
I downloaded the source package (debian testing (etch), rgl-0.67-2).
In rgl-0.67-2/src/ I changed the following files:
rglview.cpp, around line 587. Commenting the function call gl2psBeginPage
removed the crash (but also no pdf output...)
I enabled this function again and went to gl2ps.c, to the function
gl2psBeginPage. At the end of that function, around line 4426, commenting
out the line
glRenderMode(GL_FEEDBACK);
removes the R crash, but of course still no pdf output (well, only the
background).
GL_FEEDBACK is defined in /usr/include/GL/gl.h as:
/* Render Mode */
#define GL_FEEDBACK 0x1C01
#define GL_RENDER 0x1C00
#define GL_SELECT 0x1C02
Trying glRenderMode(GL_RENDER) removed the crash, but still only the
background in the pdf.
If someone has some suggestions about what to do next...
gl2ps is a separate project, whose source has been included into rgl.
You can see the gl2ps project page at http://www.geuz.org/gl2ps/.
We're using version 1.2.2, which is a couple of years old. The current
stable release of gl2ps is 1.3.1. It might fix your problem.
I don't know if we modified gl2ps.c or gl2ps.h when they were included,
but they haven't been modified since. (Daniel put them in, based on a
patch from Albrecht Gebhardt, according to the log.)
It would be helpful to know:
1. Is the rgl source identical to 1.2.2?
Yes. The version of gl2ps in rgl is identical to gl2ps version 1.2.2.
2. Does rgl work if 1.3.1 is dropped in instead?
No:
In version 1.3.1:
#define GL2PS_PS 0
#define GL2PS_EPS 1
#define GL2PS_TEX 2
#define GL2PS_PDF 3
#define GL2PS_SVG 4
#define GL2PS_PGF 5
while in version 1.2.2:
#define GL2PS_PS 1
#define GL2PS_EPS 2
#define GL2PS_TEX 3
#define GL2PS_PDF 4
Thus rgl.postscript('probeer.pdf','tex') should be used to generate a pdf.
The pdf has still no characters (axes annotations).
In R/enum.R
The last line (line 54)
rgl.enum (postscripttype, ps=1, eps=2, tex=3, pdf=4)
should be
rgl.enum (postscripttype, ps=0, eps=1, tex=2, pdf=3)
and mayebe add svg and pgf...
3. Does 1.3.1 fix the bug you're seeing?
No. Same error.
The error occurs also on ubuntu dapper. On that ubuntu machine, when
installing the libgl1-mesa-swrast, the packages libgl1-mesa
libgl1-mesa-dri and x-window-system-core are removed. rgl.postscript
doesn't produce any errors anymore, the pdf is created but no text (axes
decorations) is written to the pdf.
On debian testing, libgl1-mesa-swx11 can be installed. This removes the
follwing packages:
freeglut3-dev libgl1-mesa-dev libgl1-mesa-dri libgl1-mesa-glx
libglitz-glx1-dev libglitz1-dev libglu1-mesa-dev libglui-dev libglut3-dev
x-window-system-core xlibmesa-gl-dev xorg
but R doesn't crash anymore and the figure is written to file (still
without axes annotations).
Reinstal libgl1-mesa-glx removes libgl1-mesa-swx11 and the R crash
returns.
So it seems the bug is really triggered by libgl1-mesa. I filled in a bug
report for the debian package libgl1-mesa-glx.
I'll look into these at some point, but probably not this week.
Thanks. No hurry however, as I can still use the classical screenshots.
The figures will probable not have to be published, as the expected
results are not attained.
Thanks for your work on this. I'll put the gl2ps 1.3.1 code into rgl
with the changes you found.
Duncan Murdoch
[R] horizontal direct product
II am translating some gauss code into R, and gauss has a matrix product function called the horizontal direct product (*~), which is some sort of variant on the Kronecker product. For example if x is 2x2 and y is 2x2 the horizontal direct product, z, of x and y is defined (in the Gauss manual) as: row 1 = x11*y11 x11*y12 x12*y11 x12*y12 row 2 = x21*y21 x21*y22 x22*y21 x22*y22 Or in R code if: x - matrix(seq(1,4,by=1),2,2, byrow=TRUE) y - matrix(seq(5,8,by=1),2,2, byrow=TRUE) The resulting matrix, if I had an operator, would be the following matrix z, here formed in a contrived manner: z.1 - c(5, 6, 10, 12) z.2 - c(21,24,28,32) z - rbind(z.1,z.2) I realize that this is just the first and last row of x%*%y when x and y are two by two but this won't generalize with larger matrices. Any ideas about whether this can be done with existing R functions in a general way short of writing my own function? Thanks Luke Luke Keele Department of Political Science Ohio State University [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem in library.dynam problems on Linux
I don't know if mount -l shows something interesting below.
We're wondering if there is an issue with the gcc compiler, which has been
updated since we last updated R. Do you see anything that could be wrong with
the info below?
We plan on updating to R 2.3.1, hoping that will clear things up.
Thanks for your help, Jason
[EMAIL PROTECTED] haplo.stats]$ which gcc
/usr/bin/gcc
[EMAIL PROTECTED] haplo.stats]$ ls -al /usr/bin/gcc
-rwxr-xr-x 2 root root 103336 Mar 8 15:18 /usr/bin/gcc
[EMAIL PROTECTED] haplo.stats]$ ls -al /usr/local/biotools/r
total 36
drwxr-xr-x 4 magee rcf 4096 May 1 13:17 .
drwxr-xr-x 23 magee rcf 4096 Jun 21 15:25 ..
lrwxrwxrwx 1 root root7 May 1 13:17 current - R-2.2.1
drwxr-xr-x 5 magee rcf 4096 Jan 3 2005 R-2.0.1
drwxr-xr-x 5 magee rcf 4096 Feb 22 2006 R-2.2.1
[EMAIL PROTECTED] rpack]$ gcc -dumpversion
3.4.5
[EMAIL PROTECTED] rpack]$ mount -l
/dev/sda5 on / type ext3 (rw) [/]
none on /proc type proc (rw)
none on /sys type sysfs (rw)
none on /dev/pts type devpts (rw,gid=5,mode=620)
usbfs on /proc/bus/usb type usbfs (rw)
/dev/sda1 on /boot type ext3 (rw) [/boot]
none on /dev/shm type tmpfs (rw)
/dev/sda10 on /local1 type ext3 (rw) [/local1]
/dev/sda9 on /tmp type ext3 (rw) [/tmp]
/dev/sda7 on /usr type ext3 (rw) [/usr]
/dev/sda8 on /var type ext3 (rw) [/var]
none on /proc/sys/fs/binfmt_misc type binfmt_misc (rw)
sunrpc on /var/lib/nfs/rpc_pipefs type rpc_pipefs (rw)
rcfcluster-nfs2:/home on /home type nfs
(rw,rsize=8192,wsize=8192,intr,timeo=15,addr=172.23.70.7)
rcfcluster-nfs2:/data on /data type nfs
(rw,rsize=8192,wsize=8192,intr,timeo=15,addr=172.23.70.7)
rcfcluster-nfs2:/scratch on /scratch type nfs
(rw,rsize=8192,wsize=8192,intr,timeo=15,addr=172.23.70.7)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Peter Dalgaard
Sent: Thursday, August 24, 2006 6:02 PM
To: Sinnwell, Jason P.
Cc: 'r-help@stat.math.ethz.ch'
Subject: Re: [R] Problem in library.dynam problems on Linux
Sinnwell, Jason P. [EMAIL PROTECTED] writes:
We have R 2.2.1 installed on a Linux cluster that seems to
have problems loading either of our shared object libraries
for packages. This seems to be happening on both local and
global versions of packages that we install. However, we
have only noticed this problem in the past 3 months on this R
installation, whereas some users had success before then. It
could be that something on our system changed, but I am not
an admin so I wouldn't know where to look. Can anyone help
with this problem?
## A LOCAL INSTALLATION OF HAPLO.STATS APPEARS SUCCESSFUL
[EMAIL PROTECTED] rpack]$ R CMD INSTALL -l
/home/sinnwell/rdir/tmplib
haplo.stats
* Installing *source* package 'haplo.stats' ...
** libs
make: `haplo.stats.so' is up to date.
** R
** data
** demo
** inst
** preparing package for lazy loading
** help
Building/Updating help pages for package 'haplo.stats'
Formats: text html latex example
** building package indices ...
* DONE (haplo.stats)
## TRY AND LOAD THE LOCALLY INSTALLED LIBRARY, YET THE
SYSTEM COMMAND
SHOWS THE .so FILE IS THERE
R library(haplo.stats, lib.loc=/home/sinnwell/rdir/tmplib/)
Error in dyn.load(x, as.logical(local), as.logical(now)) :
unable to load shared library
'/home/sinnwell/rdir/tmplib/haplo.stats/libs/haplo.stats.so':
/home/sinnwell/rdir/tmplib/haplo.stats/libs/haplo.stats.so: cannot
open shared object file: No such file or directory Error in
library(haplo.stats, lib.loc = /home/sinnwell/rdir/tmplib/) :
.First.lib failed for 'haplo.stats'
R system('ls -al /home/sinnwell/rdir/tmplib/haplo.stats/libs')
total 88
drwxr-xr-x 2 sinnwell sinnwell 4096 Aug 24 15:14 .
drwxr-xr-x 13 sinnwell sinnwell 4096 Aug 24 15:14 ..
-rwxr-xr-x 1 sinnwell sinnwell 61566 Aug 24 15:14 haplo.stats.so
H.. Does mount -l show something interesting on the
filesystem containing the .so file? (e.g., option noexec).
--
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
~~ - ([EMAIL PROTECTED]) FAX:
(+45) 35327907
__
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https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
see ?detach Horace Tso [EMAIL PROTECTED] wrote: Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword remove AND library but got timed out.(why?)) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Thanks in adv. Horace __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] polychor error
Dear Janet, Performing a traceback after the error gives a hint: tmp.pcc-polychor(tmp.mat, ML=T, std.err=T) traceback() 8: stop(at least one element of , sQuote(lower), is larger than , sQuote(upper)) 7: checkmvArgs(lower = lower, upper = upper, mean = mean, corr = corr, sigma = sigma) 6: pmvnorm(lower = c(row.cuts[i], col.cuts[j]), upper = c(row.cuts[i + 1], col.cuts[j + 1]), corr = R) 5: binBvn(rho, row.cuts, col.cuts) 4: fn(par, ...) 3: function (par) fn(par, ...)(c(0.422816748044139, -2.20343446037221, -2.2163627792244, -1.79075055316993, -1.11077161663679, -0.323037731981826, 0.664036943094355, -2.71305188847272, -1.58338678422633, -0.534011853182102, 0.65365619155084 )) 2: optim(c(optimise(f, interval = c(-1, 1))$minimum, rc, cc), f, control = control, hessian = std.err) 1: polychor(tmp.mat, ML = T, std.err = T) The parameters are in the order correlation, row thresholds (of which there are 6), column thresholds (of which there are 4), and at this point in the maximization of the likelihood the first row threshold (-2.20343446037221) is *above* the second threshold (-2.2163627792244), causing pmvnorm() to complain. This can happen when the problem is ill-conditioned, since optim() doesn't constrain the order of the thresholds. So, why is the problem ill-conditioned? Here is your contingency table: tmp.mat 3 4 5 6 7 1 0 2 0 0 0 2 0 0 1 1 0 3 0 0 5 1 1 4 0 5 10 12 2 5 0 5 27 29 11 6 1 3 20 57 31 7 0 1 9 34 32 There are only two observations in the first row, two in the second row, and one in the first column. You're expecting a lot out of ML to get estimates of the first couple of thresholds for rows and the first for columns. One approach would be to eliminate one or more sparse rows or columns; e.g., polychor(tmp.mat[-1,], ML = TRUE, std.err = TRUE) Polychoric Correlation, ML est. = 0.3932 (0.05719) Test of bivariate normality: Chisquare = 14.21, df = 19, p = 0.7712 Row Thresholds Threshold Std.Err. 1 -2.4410 0.24270 2 -1.8730 0.14280 3 -1.1440 0.09236 4 -0.3353 0.07408 50.6636 0.07857 Column Thresholds Threshold Std.Err. 1 -2.6500 0.30910 2 -1.6420 0.12100 3 -0.5494 0.07672 40.6508 0.07833 polychor(tmp.mat[,-1], ML = TRUE, std.err = TRUE) Polychoric Correlation, ML est. = 0.4364 (0.05504) Test of bivariate normality: Chisquare = 14.85, df = 17, p = 0.6062 Row Thresholds Threshold Std.Err. 1 -2.4940 0.26020 2 -2.2080 0.19160 3 -1.7850 0.13400 4 -1.1090 0.09113 5 -0.3154 0.07371 60.6625 0.07821 Column Thresholds Threshold Std.Err. 1 -1.6160 0.11970 2 -0.5341 0.07639 30.6507 0.07800 A more defensible alternative would be to collapse sparse rows or columns. BTW, you *can* get estimated standard-errors from polychor() for your original table for the two-step estimator: polychor(tmp.mat, ML = FALSE, std.err = TRUE) Polychoric Correlation, 2-step est. = 0.4228 (0.05298) Test of bivariate normality: Chisquare = 19.22, df = 23, p = 0.6883 That's because the two-step estimator estimates the thresholds from the marginal distributions of the variables rather than from their joint distribution. So, is this a bug or a feature? I suppose that it's a bug to allow the thresholds to get out of order, though to constrain the optimization to prevent that from happening is probably not worth the effort and could cause some strange results. On the other hand, the error tells you something about the data, so maybe it's a feature. I noticed that you posted another version of this question two days before this one. I apologize for the slow response -- I wasn't able to read my email for a few days and it's taken me most of today to catch up. Regards, John original message --- Janet Rosenbaum jrosenba at rand.org Thu Aug 24 00:41:16 CEST 2006 Hi. Does anyone know whether the following error is a result of a bug or a feature? I can eliminate the error by making ML=F, but I would like to see the values of the cut-points and their variance. Is there anything that I can do? tmp.vec-c(0, 0, 0 , 0 ,0 , 1, 0, 2, 0 , 0, 5 ,5 ,3 ,1, 0 , 1, 5, 10, 27, 20, 9, 0, 1, 1, 12, 29, 57, 34, 0, 0, 1, 2, 11, 31, 32) tmp.mat-matrix(tmp.vec, nrow=7) rownames(tmp.mat)-1:7 colnames(tmp.mat)-3:7 tmp.pcc-polychor(tmp.mat, ML=T, std.err=T) Error in checkmvArgs(lower = lower, upper = upper, mean = mean, corr = corr, : at least one element of 'lower' is larger than 'upper' Thanks, Janet John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
try detach(package:zoo) Sachin Horace Tso [EMAIL PROTECTED] wrote: Sachin, I did try that, ex detach(zoo) Error in detach(zoo) : invalid name detach(zoo) Error in detach(zoo) : invalid name But zoo has been loaded, sessionInfo() Version 2.3.1 (2006-06-01) i386-pc-mingw32 attached base packages: [1] methods datasets stats tcltk utils graphics [7] grDevices base other attached packages: tseries quadprog zoo MASS Rpad 0.10-1 1.4-8 1.2-0 7.2-27.1 1.1.1 Thks, H. Sachin J 8/25/2006 12:56 PM see ?detach Horace Tso wrote: Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword remove AND library but got timed out.(why?)) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Thanks in adv. Horace __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
Sachin, I did try that, ex detach(zoo) Error in detach(zoo) : invalid name detach(zoo) Error in detach(zoo) : invalid name But zoo has been loaded, sessionInfo() Version 2.3.1 (2006-06-01) i386-pc-mingw32 attached base packages: [1] methods datasets stats tcltk utils graphics [7] grDevices base other attached packages: tseries quadprogzoo MASS Rpad 0.10-11.4-81.2-0 7.2-27.11.1.1 Thks, H. Sachin J [EMAIL PROTECTED] 8/25/2006 12:56 PM see ?detach Horace Tso [EMAIL PROTECTED] wrote: Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword remove AND library but got timed out.(why?)) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Thanks in adv. Horace __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
On Fri, 25 Aug 2006, Horace Tso wrote: Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword remove AND library but got timed out.(why?)) Probably because the site is offline. How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. Do you mean 'package' (see detach and unLoadNamespace) or 'library' (see dyn.unload and library.dynam.unload)? A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Are you restoring a workspace containing .First? Always try with --vanilla to check. -- Brian D. Ripley, [EMAIL PROTECTED] 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
Aah, that works. The missing package:... H. Sachin J [EMAIL PROTECTED] 8/25/2006 1:16 PM try detach(package:zoo) Sachin Horace Tso [EMAIL PROTECTED] wrote: Sachin, I did try that, ex detach(zoo) Error in detach(zoo) : invalid name detach(zoo) Error in detach(zoo) : invalid name But zoo has been loaded, sessionInfo() Version 2.3.1 (2006-06-01) i386-pc-mingw32 attached base packages: [1] methods datasets stats tcltk utils graphics [7] grDevices base other attached packages: tseries quadprog zoo MASS Rpad 0.10-1 1.4-8 1.2-0 7.2-27.1 1.1.1 Thks, H. Sachin J 8/25/2006 12:56 PM see ?detach Horace Tso wrote: Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword remove AND library but got timed out.(why?)) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Thanks in adv. Horace __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] horizontal direct product
maybe something like:
%*~% - function(x, y){
n - nrow(x)
out - matrix(0, n, ncol(x) * ncol(y))
for(i in 1:n)
out[i, ] - c(y[i, ] %o% x[i, ])
out
}
x - matrix(1:4, 2, 2, TRUE)
y - matrix(5:8, 2, 2, TRUE)
x %*~% y
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
Quoting Luke Keele [EMAIL PROTECTED]:
II am translating some gauss code into R, and gauss has a matrix
product function called the horizontal direct product (*~), which is
some sort of variant on the Kronecker product.
For example if x is 2x2 and y is 2x2
the horizontal direct product, z, of x and y is defined (in the Gauss
manual) as:
row 1 = x11*y11 x11*y12 x12*y11 x12*y12
row 2 = x21*y21 x21*y22 x22*y21 x22*y22
Or in R code if:
x - matrix(seq(1,4,by=1),2,2, byrow=TRUE)
y - matrix(seq(5,8,by=1),2,2, byrow=TRUE)
The resulting matrix, if I had an operator, would be the following
matrix z, here formed in a contrived manner:
z.1 - c(5, 6, 10, 12)
z.2 - c(21,24,28,32)
z - rbind(z.1,z.2)
I realize that this is just the first and last row of x%*%y when x
and y are two by two but this won't generalize with larger
matrices. Any ideas about whether this can be done with existing R
functions in a general way short of writing my own function?
Thanks
Luke
Luke Keele
Department of Political Science
Ohio State University
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] increasing the # of socket connections
On Fri, 25 Aug 2006, Prof Brian Ripley wrote: On Fri, 25 Aug 2006, Marc Kirchner wrote: Dear R-helpers, using snow on socket connections, I ran into the following error cl - makeSOCKcluster(hosts) Error in socketConnection(port = port, server = TRUE, blocking = TRUE : all connections are in use with showConnections(all=T) showing 50 open connections. As - for administrative reasons - I would prefer to use snow's SOCK capabilities (instead of MPI and the like), I wonder if there is a way to increase the number of simultaneous open connections allowed in an R session (~100 would be sufficient). If you really need them open at once (and are not just forgetting to close them), then change the constant in the file and recompile. Any help/hints are greatly appreciated, Is this the appropriate list: not on my reading of the posting guide! Would you have asked that question if the answer had been options(max.connections=100)? I for one would not. Best, luke -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: [EMAIL PROTECTED] Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Quickie : unload library
Hi, There are two ways that I know of: detach(package:zoo) or use detach() with the number of that package's place in the search list. library(zoo) search() [1] .GlobalEnvpackage:zoo package:methods [4] package:graphics package:grDevices package:utils [7] package:datasets package:ecoutils package:ecodist [10] package:stats package:gtkDevice Autoloads [13] package:base detach(package:zoo) search() [1] .GlobalEnvpackage:methods package:graphics [4] package:grDevices package:utils package:datasets [7] package:ecoutils package:ecodist package:stats [10] package:gtkDevice Autoloads package:base OR library(zoo) search() [1] .GlobalEnvpackage:zoo package:methods [4] package:graphics package:grDevices package:utils [7] package:datasets package:ecoutils package:ecodist [10] package:stats package:gtkDevice Autoloads [13] package:base detach(2) search() [1] .GlobalEnvpackage:methods package:graphics [4] package:grDevices package:utils package:datasets [7] package:ecoutils package:ecodist package:stats [10] package:gtkDevice Autoloads package:base Sarah -- Sarah Goslee http://www.stringpage.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] fitting a gaussian to some x,y data
hi michael : ( i stupidly didn't send the initial email to the whole list so they will have to read from the bottom ), it's clearer now but fitting a gaussian still may not be the right thing to do in this case. someone else can answer better but it sounds to me like you want to do some kind of regression ( i dont know whether it would be linear or glm or what ) using distance versus brightness. i have no clue about astronomy but if you think there is a relationship between them then this is probably the more appropriate approach to take. again, don't take my word in stone. someone else will likely reply. - Original Message - From: Michael Koppelman [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, August 25, 2006 3:35 PM Subject: Re: [R] fitting a gaussian to some x,y data Thank you. Yes, I do feel that I am under-qualified to even ask questions of y'all. Plus I'm an astronomer, which doesn't help! ;) I'll try again. I have two columns of data, the first column (x) is a distance (or length or separation) and the second column (y) is a flux (or number of counts or brightness) at that distance. Thus, when you plot y vs. x you get a spatial profile which shows how bright this thing is as a function of position. (See the small, attached PNG file. You can see there is a vague gaussian shape to the data.) This is measured data from a bizarre technique which yields data that is not evenly-spaced in x and it does not represent a pure mathematical function (i.e. it is not a point spread function or something like that), it represents the actual, non-uniform shape of an astronomical object. We are making the assumption that the shape of this object can be roughly represented with a gaussian. I want to fit a gaussian to this with the purpose of determining systematically the center of the normal-like shape of the spatial feature. I have successfully done so in R with a polynomial but I can't figure out how to do it with a gaussian. Does that make sense? Thanks! Michael On Aug 25, 2006, at 2:04 PM, MARK LEEDS wrote: hi : i'm not clear on what you mean but someone else might be ? if you say ( x,y), then it sounds like you are talking about a bivariate normal distribution. to fit a regular one dimensional gaussian distribution, you can't have 2 dimensional data. i honestly don't mean to sound rude but i think you should explain what you want to do more clearly because I don't think I am the only one that will be confused by what you said. send out a clearer email and you will get quite a response because there are a lot of really smart people ( compared to me ) on this list that love to help. it's an amazing list in that sense. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] tick.number for date in xyplot
Try this:
xyplot(runif(365)~I(as.Date(1999-01-01) + 1:365),
scales=list(x=list(at=seq(as.Date('1999-1-1'), length=12, by='1
month'), labels=NULL)))
On 8/25/06, Benjamin Tyner [EMAIL PROTECTED] wrote:
I would like a tick mark for each month; for example,
xyplot(runif(365)~I(as.Date(1999-01-01) + 1:365),
scales=list(x=list(format=%b %Y,tick.number=12)))
I know I could make x numeric and use 'at' and 'labels', but I was
wondering if there is a more direct route I'm missing. (In particular,
one that doesn't have to be modified for new data).
Thanks,
Ben
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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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] How to iteratively extract elements out of a list
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/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 you are trying to solve?
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Problem with geeglm
event.nab.2 is 0/1 and I dichotomized va to get va.2 to see if I could get geeglm to work. glm has no problem with the data but geeglm chokes. Each subject (patient.id) has at most 2 observations and more than 3/4 of the subjects have 2 observations. I have even worse problems trying to use glmmPQL from MASS and worse still trying to use lmer from lme4. But I figured a marginal model would work. (geeglm seems to work OK with most of the explanatory factors in the data set I have but a couple of them show similar problems.) summary(glm(event.nab.2~va.2,family=binomial(link=logit),data=test)) Call: glm(formula = event.nab.2 ~ va.2, family = binomial(link = logit), data = test) Deviance Residuals: Min 1Q Median 3Q Max -0.3787 -0.3787 -0.2246 -0.2246 2.7177 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)-2.5993 0.1804 -14.41 2e-16 *** va.2(84, Inf] -1.0685 0.3435 -3.11 0.00187 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 363.19 on 958 degrees of freedom Residual deviance: 352.28 on 957 degrees of freedom AIC: 356.28 Number of Fisher Scoring iterations: 6 summary(geeglm(event.nab.2~va.2,family=binomial(link=logit),id=patient.id,cor=exch,data=test)) Error in geese.fit(xx, yy, id, offset, soffset, w, waves = waves, zsca, : nrow(zsca) and length(y) not match head(test) patient.id event.nab.2 va.2 1 1 0 (-Inf,84] 2 1 0 (-Inf,84] 3 2 0 (84, Inf] 4 2 0 (84, Inf] 5 3 0 (84, Inf] 6 3 0 (84, Inf] I'm using R 2.3.1 and the latest version of geepack. I get a similar error message if I use va which is continuous. I don't know what the error message from geeglm/geese means. Rick B. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] horizontal direct product
On Fri, 25 Aug 2006, Luke Keele wrote:
II am translating some gauss code into R, and gauss has a matrix
product function called the horizontal direct product (*~), which is
some sort of variant on the Kronecker product.
For example if x is 2x2 and y is 2x2
the horizontal direct product, z, of x and y is defined (in the Gauss
manual) as:
row 1 = x11*y11 x11*y12 x12*y11 x12*y12
row 2 = x21*y21 x21*y22 x22*y21 x22*y22
It looks as though
%~% - function (A, B)
{
m - ncol(A)
n - ncol(B)
A[, rep(1:m, each = n)] * B[, rep(1:n, m)]
}
would do it.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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Re: [R] R in Nature
AAh. Then my hypothesis has been rejected. Oh well! Cheers, Simon. Simon, Congratulations! It used to be that R Ihaka and R Gentleman R: A Language for Data Analysis and Graphics Journal of Computational and Graphical Statistics, 1996 was used to cite R. I see a handful of these in Nature from around 2003. Chuck On Fri, 25 Aug 2006, Simon Blomberg wrote: Hi all, We've just had a paper accepted for publication in Nature. We used R for 95% of our analyses (one of my co-authors sneaked in some GenStat when I wasn't looking.). The preprint is available from the Nature web site, in the open peer-review trial section. I searched Nature for previous references to R Development Core Team, and I received no hits. So I tentatively conclude that our paper is the first Nature paper to cite R. A great many thanks to the R Development Core Team for R, and Prof. Bates for lmer. Cheers, Simon. (I'm off to the pub to celebrate.) -- Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat. Centre for Resource and Environmental Studies The Australian National University Canberra ACT 0200 Australia T: +61 2 6125 7800 email: Simon.Blomberg_at_anu.edu.au F: +61 2 6125 0757 CRICOS Provider # 00120C The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. - John Tukey. [ Part 3.91: Included Message ] Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717 -- Simon Blomberg, B.Sc.(Hons.), Ph.D, M.App.Stat. Centre for Resource and Environmental Studies The Australian National University Canberra ACT 0200 Australia T: +61 2 6125 7800 F: +61 2 6125 0757 CRICOS Provider # 00120C __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] generating an expression for a formula automatically
Thank you for your answers yesterday. I now have another question!
Suppose that instead of creating a formula for a regression model I
simply wanted to add the variables. I believe I cannot use the
as.formula anymore. Again I tried to use expression to no avail. I get
an expression but I can't use it.
fit.sum - function(x) {
fo - expression(paste(x, collapse = +))
eval( fo)
}
fit.sum(c(x1,x2))
Basically what I need is to learn how to use variables when what is
given to me are their names (character list).
Thanks again, Maria
Sundar Dorai-Raj wrote:
Maria Montez wrote:
Hi!
I would like to be able to create formulas automatically. For
example, I want to be able to create a function that takes on two
values: resp and x, and then creates the proper formula to regress
resp on x.
My code:
fit.main - function(resp,x) {
form - expression(paste(resp, ~ ,paste(x,sep=,collapse= +
),sep=))
z - lm(eval(form))
z
}
main - fit.main(y,c(x1,x2,x3,x4))
and I get this error:
Error in terms.default(formula, data = data) :
no terms component
Any suggestions?
Thanks, Maria
__
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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.
Hi, Maria,
Try
regr - paste(x, collapse = +)
form - as.formula(sprintf(%s ~ %s, resp, regr))
HTH,
--sundar
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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] generating an expression for a formula automatically
Try this and note we must pay attention to which environment
the expression is evaluated in.
fit.sum - function(x, env = parent.frame())
eval(parse(text = paste(x, collapse = +)), env = env)
# test
x1 - x2 - 1
fit.sum(c(x1,x2))
On 8/25/06, Maria Montez [EMAIL PROTECTED] wrote:
Thank you for your answers yesterday. I now have another question!
Suppose that instead of creating a formula for a regression model I
simply wanted to add the variables. I believe I cannot use the
as.formula anymore. Again I tried to use expression to no avail. I get
an expression but I can't use it.
fit.sum - function(x) {
fo - expression(paste(x, collapse = +))
eval( fo)
}
fit.sum(c(x1,x2))
Basically what I need is to learn how to use variables when what is
given to me are their names (character list).
Thanks again, Maria
Sundar Dorai-Raj wrote:
Maria Montez wrote:
Hi!
I would like to be able to create formulas automatically. For
example, I want to be able to create a function that takes on two
values: resp and x, and then creates the proper formula to regress
resp on x.
My code:
fit.main - function(resp,x) {
form - expression(paste(resp, ~ ,paste(x,sep=,collapse= +
),sep=))
z - lm(eval(form))
z
}
main - fit.main(y,c(x1,x2,x3,x4))
and I get this error:
Error in terms.default(formula, data = data) :
no terms component
Any suggestions?
Thanks, Maria
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hi, Maria,
Try
regr - paste(x, collapse = +)
form - as.formula(sprintf(%s ~ %s, resp, regr))
HTH,
--sundar
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quickie : unload library
You likely want the answer that Sarah has already given but in addition you might also want to look at the thread below, the point being that detaching a package still leaves portions: https://www.stat.math.ethz.ch/pipermail/r-help/2006-July/109056.html On 8/25/06, Sarah Goslee [EMAIL PROTECTED] wrote: Hi, There are two ways that I know of: detach(package:zoo) or use detach() with the number of that package's place in the search list. library(zoo) search() [1] .GlobalEnvpackage:zoo package:methods [4] package:graphics package:grDevices package:utils [7] package:datasets package:ecoutils package:ecodist [10] package:stats package:gtkDevice Autoloads [13] package:base detach(package:zoo) search() [1] .GlobalEnvpackage:methods package:graphics [4] package:grDevices package:utils package:datasets [7] package:ecoutils package:ecodist package:stats [10] package:gtkDevice Autoloads package:base OR library(zoo) search() [1] .GlobalEnvpackage:zoo package:methods [4] package:graphics package:grDevices package:utils [7] package:datasets package:ecoutils package:ecodist [10] package:stats package:gtkDevice Autoloads [13] package:base detach(2) search() [1] .GlobalEnvpackage:methods package:graphics [4] package:grDevices package:utils package:datasets [7] package:ecoutils package:ecodist package:stats [10] package:gtkDevice Autoloads package:base Sarah __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Memory usage decreases drastically after save workspace, quit, restart, load workspace
Dear all, I have the following problem: - I have written a program in R which runs out of physical memory on my MacBook Pro with 1.5 GB RAM - The memory usage is much higher then I would expect from the actual data in my global environment - When I save the workspace, quit R, restart R and load the workspace again, memory usage is much less then before (~50 MB instead of ~1.5 GB), although nothing seems to be missing. - It doesn't seem to be possible to reduce the amount of memory used by calling gc() - the behavior is independent of if I use R in command line or from GUI, so I can exclude a problem of the Mac GUI Any suggestions what the problem might be or how I could debug this? Thanks in advance for any help. Best regards, Klaus platform i386-apple-darwin8.6.1 arch i386 os darwin8.6.1 system i386, darwin8.6.1 status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 iteratively extract elements out of a list
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/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 you are trying to solve?
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/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 iteratively extract elements out of a list
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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and provide commented, minimal, self-contained, reproducible code.
