[R] install raster package on ubuntu
Dear all, I am running R 2.8.1. under ubuntu, and I need to install the package raster but I get the following error: install.packages(raster) Warning in install.packages(raster) : argument 'lib' is missing: using '/usr/local/lib/R/site-library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done Warning message: In install.packages(raster) : package raster is not available I had installed other packages without problem (like sp, maptools, spatstat and adehabitat). I tryed the install.packages as root. Any hint are welcome. Bests, milton brazil-toronto sessionInfo() R version 2.8.1 (2008-12-22) x86_64-pc-linux-gnu locale: LC_CTYPE=en_CA.UTF-8;LC_NUMERIC=C;LC_TIME=en_CA.UTF-8;LC_COLLATE=en_CA.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_CA.UTF-8;LC_PAPER=en_CA.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_CA.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tcltk_2.8.1 tools_2.8.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] install raster package on ubuntu
Oops. I found and are working fine #install.packages(raster, repos=http://R-Forge.R-project.orghttp://r-forge.r-project.org/ ) Thanks a lot, milton On Sat, Apr 18, 2009 at 2:00 AM, milton ruser milton.ru...@gmail.comwrote: Dear all, I am running R 2.8.1. under ubuntu, and I need to install the package raster but I get the following error: install.packages(raster) Warning in install.packages(raster) : argument 'lib' is missing: using '/usr/local/lib/R/site-library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done Warning message: In install.packages(raster) : package raster is not available I had installed other packages without problem (like sp, maptools, spatstat and adehabitat). I tryed the install.packages as root. Any hint are welcome. Bests, milton brazil-toronto sessionInfo() R version 2.8.1 (2008-12-22) x86_64-pc-linux-gnu locale: LC_CTYPE=en_CA.UTF-8;LC_NUMERIC=C;LC_TIME=en_CA.UTF-8;LC_COLLATE=en_CA.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_CA.UTF-8;LC_PAPER=en_CA.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_CA.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tcltk_2.8.1 tools_2.8.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tinn-R (Version 2.2.0.2) dose not support R 2.9.0 very well
Hi, I used Tinn-R for a long time, but had real headaches on Vista (commands not being sent, etc...) that were not resolved since R 2.8 or so. Since then, I have switched to Eclipse/StatET. The setup requires more effort, but it is much easier to manage your own packages. With the Eclipse IDE and StatET functionality, it also becomes much easier to refactor your code, if you are playing around with more than one or two .R files. I have also found that with some tinkering, Eclipse/StatET also works on XP Pro 32, Vista 32, and Ubuntu amd64. Read this thread for more details and google for the StatET website to download. Having gotten used to StatET, it would be extremely painful to go without it... Good luck. http://www.nabble.com/Eclipse-and-StatET-Howto-(also-added-Subversion,-Rtools)-td22764049.html#a22764049 LI Qi-2 wrote: Hi£¬ I found that Tinn-R (Version 2.2.0.2) dose not support R 2.9.0 very well. Maybe, somebody who have solved this problem can help me. Thanks for your attention! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Tinn-R-%28Version-2.2.0.2%29-dose-not-support-R-2.9.0-very-well-tp23109020p23109862.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Need Help with R.oo and multiple inheritance
This is what I'm trying to do: HasPersistence - an Interface class which implements things like WriteToDisk(), ReadFromDisk() B isA A C isA A, but also implements HasPersistence My solution in R.oo is to have a root class called A which implements IsPersistent() by returning FALSE, but is overrideable by C to return TRUE. C will also implement WriteToDisk() and ReadFromDisk(). This solution will work, but is kludgy, because it is not directly obvious that IsPersistent(), WriteToDisk(), and ReadFromDisk() go together in a conceptual HasPersistence. Is there a more R.oo-clean-way of multiply inheriting like this, so if someone else reads my code, the concept of HasPersistence is more obvious? In other words, could I actually have a HasPersistence class and have C inherit from A and implement HasPersistence also? - Ken -- View this message in context: http://www.nabble.com/Need-Help-with-R.oo-and-%22multiple-inheritance%22-tp23109913p23109913.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Binomial simulation
Not being entirely sure what you mean, I think rbinom(1000, 10, .25) may be what you want. Hi, Thanks for your reply. It is close to that but I need to know the probabilty of how many judges pick a certain brand. Just say x= 6 judges pick brand A which has P=0.25. Using R it would be: dbinom(6,10,.25) [1] 0.016222 Probability of six judges choosing brand A. Hence not very likely. I have been asked to do this for all values of x = 1 to 10. But the question says to simulate 1000 trials for each x value. I'm not sure how to construct the simulation. regards Brendan -- View this message in context: http://www.nabble.com/Binomial-simulation-tp23106347p23109522.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Dummy (factor) based on a pair of variables
Dear All! my data is on pairs of countries, i and j, e.g.: y,i,j 1,AUT,BEL 2,AUT,GER 3,BEL,GER I would like to create a dummy (indicator) variable for use in regression (using factor?), such that it takes the value of 1 if the country is in the pair (i.e. EITHER an i-country OR an j-country). Thank you for your help, Serguei Austrian Institute of Economic Research (WIFO) P.O.Box 91 Tel.: +43-1-7982601-231 1103 Vienna, AustriaFax: +43-1-7989386 Mail: serguei.kaniov...@wifo.ac.at http://www.wifo.ac.at/Serguei.Kaniovski [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tinn-R (Version 2.2.0.2) dose not support R 2.9.0 very well
I also had some problems with tinn-R and switched to Notepad++ with NppToR. It works with R 2.9.0, though I have to admit that I have not extensively tested yet. Since I am not using Vista, I cannot tell anything about its Vista performance. The relevant links are http://www.stat.tamu.edu/~aredd/site/ http://notepad-plus.sourceforge.net Ken-JP wrote: Hi, I used Tinn-R for a long time, but had real headaches on Vista (commands not being sent, etc...) that were not resolved since R 2.8 or so. Since then, I have switched to Eclipse/StatET. The setup requires more effort, but it is much easier to manage your own packages. With the Eclipse IDE and StatET functionality, it also becomes much easier to refactor your code, if you are playing around with more than one or two .R files. I have also found that with some tinkering, Eclipse/StatET also works on XP Pro 32, Vista 32, and Ubuntu amd64. Read this thread for more details and google for the StatET website to download. Having gotten used to StatET, it would be extremely painful to go without it... Good luck. http://www.nabble.com/Eclipse-and-StatET-Howto-(also-added-Subversion,-Rtools)-td22764049.html#a22764049 LI Qi-2 wrote: Hi£¬ I found that Tinn-R (Version 2.2.0.2) dose not support R 2.9.0 very well. Maybe, somebody who have solved this problem can help me. Thanks for your attention! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Erich Neuwirth, University of Vienna Faculty of Computer Science Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-39464 Fax: +43-1-4277-39459 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Dummy (factor) based on a pair of variables
On Sat, 2009-04-18 at 08:55 +0200, Serguei Kaniovski wrote: Dear All! my data is on pairs of countries, i and j, e.g.: y,i,j 1,AUT,BEL 2,AUT,GER 3,BEL,GER I would like to create a dummy (indicator) variable for use in regression (using factor?), such that it takes the value of 1 if the country is in the pair (i.e. EITHER an i-country OR an j-country). Thank you for your help, Serguei Hi Serguei, If I understand your doubt, the solution is something like this for pair i-country is AUT or j-country is BEL output ~ I(i-country==AUT|j-country==BEL) -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Dummies: Call function or script
Hello all, I am new to R and therefore this question is sort of very basic. Could you kindly let me know how can I call a function or script from R GUI(Windows), my directory where these functions/scripts are located is suppose xyz and I have tried setwd(xyz) but still it gives me the Error: object abc not found where abc is the script/function name... Would appreciate any guidance, Thanks, Ali -- View this message in context: http://www.nabble.com/Dummies%3A-Call-function-or-script-tp23110931p23110931.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tinn-R (Version 2.2.0.2) dose not support R 2.9.0 very well
LI Qi schrieb: I found that Tinn-R (Version 2.2.0.2) dose not support R 2.9.0 very well. Maybe, somebody who have solved this problem can help me. What do you mean with doeas not support very well. I have no problems with Vista + R 2.9.0 + Tinn-R 2.2.0.2 Try to start Rterm instead of Rgui. This opens the R terminal inside Tinn-R, causes less problems and has the advantage of a little syntac highlightning in the terminal. However you have to get used to the error messages being sent to another window. Have a look at the Tinn-R sourceforge page. There is an rsite of jose: http://sourceforge.net/forum/forum.php?forum_id=886959 manipulate it to for your convienience. there is also a line that resolves the .trPath problem which is maybe what you stumbled upon. Cheers Stefan PS Notepad++ installs for me but the installation on npptor does not work for me. Syntax highlightning does not work and commands are not passed to Rgui neither in sdi mode nor in normal mode. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Dummies: Call function or script
xyz does not look to me like a valid directory for Windows. Try a complete directory string: setwd(C://username/My Documents/r-work/) # might be something more likely Note the forward slashes. Backslashes are special characters in R. The interpreter would need to see each backslash doubled in ourder to succeed. (You might reconsider whether you want to address the group as Dummies.) On Apr 18, 2009, at 5:28 AM, Ali Faisal wrote: Hello all, I am new to R and therefore this question is sort of very basic. Could you kindly let me know how can I call a function or script from R GUI(Windows), my directory where these functions/scripts are located is suppose xyz and I have tried setwd(xyz) but still it gives me the Error: object abc not found where abc is the script/function name... Would appreciate any guidance, Thanks, Ali David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how can i program this???
Hi all; i want to programm the no-parametric regression estimator with R, any help for this ??? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Rd: how to put a prime in a code fragment?
I have written a function for my 'oce' package that creates a data.frame
containing a variable name with a prime in it. (I use prime to indicate
coordinate rotation, a reasonably standard convention that motivates the odd
variable name.)
How can I name that in an Rd file? I tried \code{u'} but R-2.9.0 doesn't
like to build my package then, since it thinks I'm opening a string. So I
tried \code{u\'} but then the backslash appears in the documentation.
--
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Re: [R] Rd: how to put a prime in a code fragment?
On 18/04/2009 8:49 AM, Dan Kelley wrote:
I have written a function for my 'oce' package that creates a data.frame
containing a variable name with a prime in it. (I use prime to indicate
coordinate rotation, a reasonably standard convention that motivates the odd
variable name.)
How can I name that in an Rd file? I tried \code{u'} but R-2.9.0 doesn't
like to build my package then, since it thinks I'm opening a string. So I
tried \code{u\'} but then the backslash appears in the documentation.
The new convention is that \code{} contains legal R code (with a few
exceptions for markup). You can use \preformatted{} or \samp{} for
verbatim text. So \preformatted{u'} should display the way you like.
If you are using u' as a variable in R, you'd have to enter it as `u'`,
and that should also be accepted in \code{}.
This stuff is in transition in 2.9.x; 2.10.x should make it all a bit
cleaner, when the old somewhat inconsistent rules are no longer the default.
Duncan Murdoch
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Re: [R] how can i program this???
This list is very good with _specific_ questions, but yours is rather vague. Perhaps you should read the posting guide, then ask us about what you are specifically having a problem with. Sarah On Sat, Apr 18, 2009 at 8:41 AM, A.B dahma...@yahoo.fr wrote: Hi all; i want to programm the no-parametric regression estimator with R, any help for this ??? -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Neural Networks in R - Query
Dear R users, I'd like to ask your guidance regarding the following two questions: (i) I just finished reading Chris Bishop's book Neural Networks for Pattern Recognition. Although the book gave me good theoretical foundation about NN, I'm now looking for something more practical regarding architecture selection strategies. Is there any good reference about best practices for architecure selection? (ii) Which R package provides a good implementation of NN? Many thanks in advance for your help! Lars. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Neural Networks in R - Query
Hi Lars, I'd like to ask your guidance regarding the following two questions: (i) I just finished reading Chris Bishop's book Neural Networks for Pattern Recognition. Although the book gave me good theoretical foundation about NN, I'm now looking for something more practical regarding architecture selection strategies. Is there any good reference about best practices for architecure selection? (ii) Which R package provides a good implementation of NN? The CRAN Task View on Machine Learning http://cran.r-project.org/web/views/MachineLearning.html opens with the following topical item: o Neural Networks : Single-hidden-layer neural network are implemented in package nnet as part of the VR bundle (shipped with base R). HTH, Tobias __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] matching subvectors in vector sets
that works very well. how do I store the results into a variable instead of
doing a print?
On Fri, Apr 17, 2009 at 5:51 PM, jim holtman jholt...@gmail.com wrote:
How about this:
x - A00096:A00096:A00096:A00096:A02178:A02178:A07776
x.s - unlist(strsplit(x, :))
for (i in 2:length(x.s)){
+ x.seq - embed(length(x.s):1, i)
+ print(table(apply(x.seq, 1, function(z){
+ paste(x.s[z], collapse=:)
+ })))
+ }
A00096:A00096 A00096:A02178 A02178:A02178 A02178:A07776
3 1 1 1
A00096:A00096:A00096 A00096:A00096:A02178 A00096:A02178:A02178
A02178:A02178:A07776
211
1
A00096:A00096:A00096:A00096 A00096:A00096:A00096:A02178
A00096:A00096:A02178:A02178 A00096:A02178:A02178:A07776
1 1
1 1
A00096:A00096:A00096:A00096:A02178 A00096:A00096:A00096:A02178:A02178
A00096:A00096:A02178:A02178:A07776
1 1
1
A00096:A00096:A00096:A00096:A02178:A02178
A00096:A00096:A00096:A02178:A02178:A07776
1
1
A00096:A00096:A00096:A00096:A02178:A02178:A07776
1
On Fri, Apr 17, 2009 at 9:33 AM, Albert Vilella avile...@gmail.com
wrote:
Starting by the first entry:
A00096:A00096:A00096:A00096:A02178:A02178:A07776
and supposing there aren't any other subvectors identical in the set, the
algorithm will slide through the vector, first in pairs, then in trios,
then
in sets of four, etc, and count the occurrences:
A00096:A00096
3
A00096:A02178
1
A02178:A02178
1
A02178:A07776
1
A00096:A00096:A00096
2
A00096:A00096:A02178
1
A00096:A02178:A02178
1
A02178:A02178:A07776
1
A00096:A00096:A00096:A00096
1
A00096:A00096:A00096:A02178
1
A00096:A00096:A02178:A02178
1
A00096:A02178:A02178:A07776
1
A00096:A00096:A00096:A00096:A02178
1
A00096:A00096:A00096:A02178:A02178
1
A00096:A00096:A02178:A02178:A07776
1
A00096:A00096:A00096:A00096:A02178:A02178
1
A00096:A00096:A00096:A02178:A02178:A07776
1
A00096:A00096:A00096:A00096:A02178:A02178:A07776
1
On Fri, Apr 17, 2009 at 1:04 PM, jim holtman jholt...@gmail.com wrote:
Can you provide the output that you would expect from the data you
gave. I am not sure what you mean by a 'subvector'.
On Fri, Apr 17, 2009 at 5:25 AM, Albert Vilella avile...@gmail.com
wrote:
Hi,
I've got a list of ~2 elements that look like this:
[1]
A00096:A00096:A00096:A00096:A02178:A02178:A07776
[2]
A00046:A00076:A01101:A04146:A05671:A07169
[3]
A00038:A00932:A02185:A02370:A02818:A02818:A02818:A02818:A04732:A07142:A07142
[4]
A00096:A01352:A01352:A02023:A05001:A05001:A07776
[5]
A00036:A00047:A00059:A00503:A00904:A00904:A00904:A01023:A01023:A01399:A02029:A03941:A07679
[6]
A00041:A00533:A00855:A02178:A02178:A02178:A05671:A05671:A05671:A05671:A05671:A05671:A05671
...
And I would like to have a table with the frequency of occurrences for
matching subvectors in all elements, i.e., not
only the number of times a vector is found but also how many times a
subvector (of at least 2 ids) is found.
How can I do that?
Thanks in advance,
Albert.
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] Create histogram from data matrix
On Fri, Apr 17, 2009 at 2:07 PM, Paul Warren Simonin paul.simo...@uvm.edu wrote: Thank you all for your advice. I have received some good tips, but it was suggested I write back with a small simulated data set to better illustrate my needs. So, currently my data frame looks something like: ID (date) Temperature Number of fish 200706183 5 456 200706183 5 765 200706183 4 567 200706183 3 876 200706183 3 888 200706183 2 111 200706184 8 2345 200706184 8 654 200706184 8 7786 200706184 7 345 200706184 6 234 200706184 6 123 I need to create a plots for each ID (date) of the number of fish observed at each temperature. Obviously my data frame is much larger. These plots do not have to be in a specific histogram format, but it seems this may be appropriate. Thanks for any additional advice as to how this may be done, either using plot commands or reformatting my data. It seemed the ggplot2 options may be good but so far I have tried qplot with no success: my most recent code looks like: qplot(temp,number of fish, geom=histogram,binwidth=1) I have tried various tweaks of this, but no success. The problem is number of fish is not a valid R variable name because it has spaces in it (and you didn't specify the data frame to look for those variable in). Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Binomial simulation
This sounds like a potential homework problem. You don't quite need to simulate anything if your question is all you have been asked to do. dbinom(x = 1:10, size = 10, prob = 0.25) Perhaps you have been asked to simulate 1000 realizations and compare the relative frequencies with these probabilities: use rbinom(n = 1000, size = 10, prob = 0.25) in that case and compare the relative frequencies. Btw, there is a small chance of getting a 0. Are you sure the instructor (or whoever has issued the orders) wants only from 1:10? HTH! Ranjan On Fri, 17 Apr 2009 22:23:11 -0700 (PDT) beetle2 samandbren...@aapt.net.au wrote: Not being entirely sure what you mean, I think rbinom(1000, 10, .25) may be what you want. Hi, Thanks for your reply. It is close to that but I need to know the probabilty of how many judges pick a certain brand. Just say x= 6 judges pick brand A which has P=0.25. Using R it would be: dbinom(6,10,.25) [1] 0.016222 Probability of six judges choosing brand A. Hence not very likely. I have been asked to do this for all values of x = 1 to 10. But the question says to simulate 1000 trials for each x value. I'm not sure how to construct the simulation. regards Brendan -- View this message in context: http://www.nabble.com/Binomial-simulation-tp23106347p23109522.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] matching subvectors in vector sets
xlist -list()
for (i in 2:length(x.s)){
x.seq - embed(length(x.s):1, i)
xlist[[i]] - table(apply(x.seq, 1, function(z){
paste(x.s[z], collapse=:)
}))
}
xlist
--
David Winsemius
On Apr 18, 2009, at 11:46 AM, Albert Vilella wrote:
that works very well. how do I store the results into a variable
instead of
doing a print?
On Fri, Apr 17, 2009 at 5:51 PM, jim holtman jholt...@gmail.com
wrote:
How about this:
x - A00096:A00096:A00096:A00096:A02178:A02178:A07776
x.s - unlist(strsplit(x, :))
for (i in 2:length(x.s)){
+ x.seq - embed(length(x.s):1, i)
+ print(table(apply(x.seq, 1, function(z){
+ paste(x.s[z], collapse=:)
+ })))
+ }
A00096:A00096 A00096:A02178 A02178:A02178 A02178:A07776
3 1 1 1
A00096:A00096:A00096 A00096:A00096:A02178 A00096:A02178:A02178
A02178:A02178:A07776
211
1
A00096:A00096:A00096:A00096 A00096:A00096:A00096:A02178
A00096:A00096:A02178:A02178 A00096:A02178:A02178:A07776
1 1
1 1
A00096:A00096:A00096:A00096:A02178 A00096:A00096:A00096:A02178:A02178
A00096:A00096:A02178:A02178:A07776
1 1
1
A00096:A00096:A00096:A00096:A02178:A02178
A00096:A00096:A00096:A02178:A02178:A07776
1
1
A00096:A00096:A00096:A00096:A02178:A02178:A07776
1
On Fri, Apr 17, 2009 at 9:33 AM, Albert Vilella avile...@gmail.com
wrote:
Starting by the first entry:
A00096:A00096:A00096:A00096:A02178:A02178:A07776
and supposing there aren't any other subvectors identical in the
set, the
algorithm will slide through the vector, first in pairs, then in
trios,
then
in sets of four, etc, and count the occurrences:
A00096:A00096
3
A00096:A02178
1
A02178:A02178
1
A02178:A07776
1
A00096:A00096:A00096
2
A00096:A00096:A02178
1
A00096:A02178:A02178
1
A02178:A02178:A07776
1
A00096:A00096:A00096:A00096
1
A00096:A00096:A00096:A02178
1
A00096:A00096:A02178:A02178
1
A00096:A02178:A02178:A07776
1
A00096:A00096:A00096:A00096:A02178
1
A00096:A00096:A00096:A02178:A02178
1
A00096:A00096:A02178:A02178:A07776
1
A00096:A00096:A00096:A00096:A02178:A02178
1
A00096:A00096:A00096:A02178:A02178:A07776
1
A00096:A00096:A00096:A00096:A02178:A02178:A07776
1
On Fri, Apr 17, 2009 at 1:04 PM, jim holtman jholt...@gmail.com
wrote:
Can you provide the output that you would expect from the data you
gave. I am not sure what you mean by a 'subvector'.
On Fri, Apr 17, 2009 at 5:25 AM, Albert Vilella
avile...@gmail.com
wrote:
Hi,
I've got a list of ~2 elements that look like this:
[1]
A00096:A00096:A00096:A00096:A02178:A02178:A07776
[2]
A00046:A00076:A01101:A04146:A05671:A07169
[3]
A00038
:A00932
:A02185:A02370:A02818:A02818:A02818:A02818:A04732:A07142:A07142
[4]
A00096:A01352:A01352:A02023:A05001:A05001:A07776
[5]
A00036
:A00047
:A00059
:A00503
:A00904:A00904:A00904:A01023:A01023:A01399:A02029:A03941:A07679
[6]
A00041
:A00533
:A00855
:A02178
:A02178:A02178:A05671:A05671:A05671:A05671:A05671:A05671:A05671
...
And I would like to have a table with the frequency of
occurrences for
matching subvectors in all elements, i.e., not
only the number of times a vector is found but also how many
times a
subvector (of at least 2 ids) is found.
How can I do that?
Thanks in advance,
Albert.
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What is the problem that you are trying to solve?
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Heritage Laboratories
West Hartford, CT
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Re: [R] Neural Networks in R - Query
More specifically, it is nnet package. Best 2009/4/18 Tobias Verbeke tobias.verb...@telenet.be: Hi Lars, I'd like to ask your guidance regarding the following two questions: (i) I just finished reading Chris Bishop's book Neural Networks for Pattern Recognition. Although the book gave me good theoretical foundation about NN, I'm now looking for something more practical regarding architecture selection strategies. Is there any good reference about best practices for architecure selection? (ii) Which R package provides a good implementation of NN? The CRAN Task View on Machine Learning http://cran.r-project.org/web/views/MachineLearning.html opens with the following topical item: o Neural Networks : Single-hidden-layer neural network are implemented in package nnet as part of the VR bundle (shipped with base R). HTH, Tobias __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Monotone Transformation
FJ == Feng Jingyu feng...@battelle.org on Fri, 17 Apr 2009 11:22:01 -0700 (PDT) writes: FJ Hi, I am trying to use R to mimic what I did in SAS. FJ proc transreg data=x ; FJ model identity(GSI)=monotone(group1); FJ output out=d2 pprefix=M; FJ run; FJ Accroding to SAS documentation, the MONOTONE transfomation algorithm comes FJ from (Kruskal 1964, secondary approach to ties). I have tried ace. it does FJ provide some kind of monotone transformation, but it is not what I expected. FJ Here is how sas output look like: FJ Obs GSITGSI group1Tgroup1 FJ 1 0.813010.81301 1 1.55594 FJ 2 0.793590.79359 2 1.55594 FJ 3 1.269001.26900 3 2.59702 FJ 4 2.026802.02680 4 4.29111 FJ group1 is the inital value. Tgroup1 is the monotone transformed value. FJ Here is how Transformed value output from ace: FJ $tx FJ [,1] FJ [1,] -0.5698602 FJ [2,] -0.1899534 FJ [3,] 0.1899534 FJ [4,] 0.5698602 FJ Does anybody have any idea whether I could do the similar thing in R? If I FJ can do it, which function I should use? I don't know what exactly you want, and I'd never want to read SAS code to understand your question, but it could be that isoreg() can solve your problem, notably rr - isoreg(x,y) rr $yf ## your transformed y Regards, Martin Maechler, ETH Zurich FJ Thanks, FJ Jingyu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Monotone Transformation
An alternative to isoreg is smooth.monotone in the fda package, though it would doubtless require more work to use smooth.monotone. Hope this helps. Spencer Martin Maechler wrote: FJ == Feng Jingyu feng...@battelle.org on Fri, 17 Apr 2009 11:22:01 -0700 (PDT) writes: FJ Hi, I am trying to use R to mimic what I did in SAS. FJ proc transreg data=x ; FJ model identity(GSI)=monotone(group1); FJ output out=d2 pprefix=M; FJ run; FJ Accroding to SAS documentation, the MONOTONE transfomation algorithm comes FJ from (Kruskal 1964, secondary approach to ties). I have tried ace. it does FJ provide some kind of monotone transformation, but it is not what I expected. FJ Here is how sas output look like: FJ Obs GSITGSI group1Tgroup1 FJ 1 0.813010.81301 1 1.55594 FJ 2 0.793590.79359 2 1.55594 FJ 3 1.269001.26900 3 2.59702 FJ 4 2.026802.02680 4 4.29111 FJ group1 is the inital value. Tgroup1 is the monotone transformed value. FJ Here is how Transformed value output from ace: FJ $tx FJ [,1] FJ [1,] -0.5698602 FJ [2,] -0.1899534 FJ [3,] 0.1899534 FJ [4,] 0.5698602 FJ Does anybody have any idea whether I could do the similar thing in R? If I FJ can do it, which function I should use? I don't know what exactly you want, and I'd never want to read SAS code to understand your question, but it could be that isoreg() can solve your problem, notably rr - isoreg(x,y) rr $yf ## your transformed y Regards, Martin Maechler, ETH Zurich FJ Thanks, FJ Jingyu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Welcome to the R-help mailing list
Hi all,
I'm a newbie R developer, am trying to dotplot a few graphs using a for
loop.
The following code works fine but once I wanna plot inside a loop, nothing
happens.
for(i in 1:1){dotplot(y~x)}
y - c(1,2,3)
x - c('a','b','c')
dotplot(y~x)
for (i in 1:3) {dotplot(y~x)} (y and x depends on I in actual case)
Nothing happens.
I appreciate your advice on what is going wrong? Thanks.
Best,
Tony
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Loop question
Brendan,
Matrix is atomic. Once you define t1 in matrix, t1[1]=0 rather than the
whole column. I would just convert t1 to a data frame, which is a special
list, by adding t1- data.frame(t1). Now t1[1] represents the whole column.
Then you can use your loop to add more columns.
Jun
On Fri, Apr 17, 2009 at 9:12 PM, Brendan Morse morse.bren...@gmail.comwrote:
Hi everyone, I am trying to accomplish a small task that is giving me quite
a headache. I would like to automatically generate a series of matrices and
give them successive names. Here is what I thought at first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices with a
single column of 250 values randomly selected from a normal distribution,
and that they would be labeled t11, t12, t13, t14 etc.
Can anyone steer me in the right direction with this one?
Thanks!
Brendan
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--
Jun Shen PhD
PK/PD Scientist
BioPharma Services
Millipore Corporation
15 Research Park Dr.
St Charles, MO 63304
Direct: 636-720-1589
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[R] Modelling an incomplete Poisson distribution ?
Dear list, I have the following problem : I want to model a series of observations of a given hospital activity on various days under various conditions. among my outcomes (dependent variables) is the number of patients for which a certain procedure is done. The problem is that, when no relevant patient is hospitalized on said day, there is no observation (for which the number of patients item would be 0). My goal is to model this number of patients as a function of the various conditions described by my independant variables, mosty of them observed but uncontrolled, some of them unobservable (random effects). I am tempted to model them along the lines of : glm(NoP~X+Y+..., data=MyData, family=poisson(link=log)) or (accounting for some random effects) : lmer(NoP~X+Y+(X|Center)), data=Mydata, family=poisson(link=log)) While the preliminary analysis suggest that (the right part of) a Poisson distribution might be reasonable for all real observations, the lack of observations with count==0 bothers me. Is there a way to cajole glm (and lmer, by the way) into modelling these data to an incomplete Poisson model, i. e. with unobserved 0 values ? Sincerely, Emmanuel Charpentier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Welcome to the R-help mailing list
When plotting in loops, you need to wrap your plot call in print().
On Sat, Apr 18, 2009 at 2:14 PM, Qifei Zhu zhu_qi...@yahoo.com.sg wrote:
Hi all,
I'm a newbie R developer, am trying to dotplot a few graphs using a for
loop.
The following code works fine but once I wanna plot inside a loop, nothing
happens.
for(i in 1:1){dotplot(y~x)}
y - c(1,2,3)
x - c('a','b','c')
dotplot(y~x)
for (i in 1:3) {dotplot(y~x)} (y and x depends on I in actual case)
Nothing happens.
I appreciate your advice on what is going wrong? Thanks.
Best,
Tony
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student
Department of Psychology
Dalhousie University
Looking to arrange a meeting? Check my public calendar:
http://tr.im/mikes_public_calendar
~ Certainty is folly... I think. ~
__
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Re: [R] Welcome to the R-help mailing list
It works. Thanks!
-Original Message-
From: mike.lw...@gmail.com [mailto:mike.lw...@gmail.com] On Behalf Of Mike
Lawrence
Sent: Saturday, April 18, 2009 1:30 PM
To: Qifei Zhu
Cc: r-help@r-project.org
Subject: Re: [R] Welcome to the R-help mailing list
When plotting in loops, you need to wrap your plot call in print().
On Sat, Apr 18, 2009 at 2:14 PM, Qifei Zhu zhu_qi...@yahoo.com.sg wrote:
Hi all,
I'm a newbie R developer, am trying to dotplot a few graphs using a for
loop.
The following code works fine but once I wanna plot inside a loop, nothing
happens.
for(i in 1:1){dotplot(y~x)}
y - c(1,2,3)
x - c('a','b','c')
dotplot(y~x)
for (i in 1:3) {dotplot(y~x)} (y and x depends on I in actual case)
Nothing happens.
I appreciate your advice on what is going wrong? Thanks.
Best,
Tony
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student
Department of Psychology
Dalhousie University
Looking to arrange a meeting? Check my public calendar:
http://tr.im/mikes_public_calendar
~ Certainty is folly... I think. ~
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Welcome to the R-help mailing list
Qifei Zhu wrote:
Hi all,
I'm a newbie R developer, am trying to dotplot a few graphs using a for
loop.
The following code works fine but once I wanna plot inside a loop, nothing
happens.
for(i in 1:1){dotplot(y~x)}
y - c(1,2,3)
x - c('a','b','c')
dotplot(y~x)
for (i in 1:3) {dotplot(y~x)} (y and x depends on I in actual case)
Nothing happens.
dotplot is a lattice function that need to be print()-ed which happened
before automatically (as you'd type x in it is printed).
Uwe Ligges
I appreciate your advice on what is going wrong? Thanks.
Best,
Tony
__
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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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Re: [R] From daily series to monthly and viceversa
Well Gabor, this is actually a really good help, thanks so much. There is only one problem, I'm getting what you say, but in the code there are a couple of errors time(z2) - as.Date(time(zz)) #probably z2 z3 - na.locf( cbind(zz, zoo(, dd), #dd object not found day.of.month = as.numeric(format(time(zz), %d)) )) z3$day.of.month - as.numeric(format(time(zz2), %d)) #z2? And why did you call partial this solution? Thanks again for you great help Take care Gabor Grothendieck wrote: Here is a partial solution: library(zoo) # z is CPI. Just use 1, 2, 3, ... for example. z - zooreg(1:12, as.yearmon(2008-01), freq = 12) days.in.month - as.numeric(as.Date(time(z), frac = 1) - as.Date(time(z)) + 1) z2 - cbind(z, z1 = lag(z, -1), z2 = lag(z, -2), z3 = lag(z, -3), days.in.month) time(z2) - as.Date(time(zz)) z3 - na.locf( cbind(zz, zoo(, dd), day.of.month = as.numeric(format(time(zz), %d)) )) z3$day.of.month - as.numeric(format(time(zz2), %d)) # now each row of z3 has all the data you need so apply(z3, 1, your.function) On Fri, Apr 17, 2009 at 2:13 PM, manta mantin...@libero.it wrote: any update anybody? I'm really stucked! -- View this message in context: http://www.nabble.com/From-daily-series-to-monthly-and-viceversa-tp23064454p23103052.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/From-daily-series-to-monthly-and-viceversa-tp23064454p23115847.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Modelling an incomplete Poisson distribution ?
I forgot to add that yes, I've done my homework, and that it seems to me that answers pointing to zero-inflated Poisson (and negative binomial) are irrelevant ; I do not have a mixture of distributions but only part of one distribution, or, if you'll have it, a zero-deflated Poisson. An answer by Brian Ripley (http://finzi.psych.upenn.edu/R/Rhelp02/archive/11029.html) to a similar question leaves me a bit dismayed : if it is easy to compute the probability function of this zero-deflated RV (off the top of my head, Pr(X=x)=e^-lambda.lambda^x/(x!.(1-e^-lambda))), and if I think that I'm (still) able to use optim to ML-estimate lambda, using this to (correctly) model my problem set and test it amounts to re-writing some (large) part of glm. Furthermore, I'd be a bit embarrassed to test it cleanly (i. e. justifiably) : out of the top of my head, only the likelihood ration test seems readily applicable to my problem. Testing contrasts in my covariates ... hum ! So if someone has written something to that effect, I'd be awfully glad to use it. A not-so-cursory look at the existing packages did not ring any bells to my (admittedly untrained) ears... Of course, I could also bootstrap the damn thing and study the distribution of my contrasts. I'd still been hard pressed to formally test hypotheses on factors... Any ideas ? Emmanuel Charpentier Le samedi 18 avril 2009 à 19:28 +0200, Emmanuel Charpentier a écrit : Dear list, I have the following problem : I want to model a series of observations of a given hospital activity on various days under various conditions. among my outcomes (dependent variables) is the number of patients for which a certain procedure is done. The problem is that, when no relevant patient is hospitalized on said day, there is no observation (for which the number of patients item would be 0). My goal is to model this number of patients as a function of the various conditions described by my independant variables, mosty of them observed but uncontrolled, some of them unobservable (random effects). I am tempted to model them along the lines of : glm(NoP~X+Y+..., data=MyData, family=poisson(link=log)) or (accounting for some random effects) : lmer(NoP~X+Y+(X|Center)), data=Mydata, family=poisson(link=log)) While the preliminary analysis suggest that (the right part of) a Poisson distribution might be reasonable for all real observations, the lack of observations with count==0 bothers me. Is there a way to cajole glm (and lmer, by the way) into modelling these data to an incomplete Poisson model, i. e. with unobserved 0 values ? Sincerely, Emmanuel Charpentier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] From daily series to monthly and viceversa
Try this: library(zoo) # z is CPI. Just use 1, 2, 3, ... for example. z - zooreg(1:12, as.yearmon(2008-01), freq = 12) # z2 contains z and lags of z and days.in.month # It has a Date class time index, rather than yearmon. days.in.month - as.numeric(as.Date(time(z), frac = 1) - as.Date(time(z)) + 1) z2 - cbind(z, z1 = lag(z, -1), z2 = lag(z, -2), z3 = lag(z, -3), days.in.month) time(z2) - as.Date(time(z2)) # z3 is z2 expanded to include each day of each month and # day of the month that each row represents dd - seq(time(z2)[1], as.Date(as.yearmon(tail(time(z2), 1)), frac = 1), day) z3 - na.locf(cbind(z2, zoo(, dd))) z3$day.of.month - as.numeric(format(time(z3), %d)) # now each row of z3 has all the data you need so apply(z3, 1, your.function) On Sat, Apr 18, 2009 at 2:28 PM, manta mantin...@libero.it wrote: Well Gabor, this is actually a really good help, thanks so much. There is only one problem, I'm getting what you say, but in the code there are a couple of errors time(z2) - as.Date(time(zz)) #probably z2 z3 - na.locf( cbind(zz, zoo(, dd), #dd object not found day.of.month = as.numeric(format(time(zz), %d)) )) z3$day.of.month - as.numeric(format(time(zz2), %d)) #z2? And why did you call partial this solution? Thanks again for you great help Take care Gabor Grothendieck wrote: Here is a partial solution: library(zoo) # z is CPI. Just use 1, 2, 3, ... for example. z - zooreg(1:12, as.yearmon(2008-01), freq = 12) days.in.month - as.numeric(as.Date(time(z), frac = 1) - as.Date(time(z)) + 1) z2 - cbind(z, z1 = lag(z, -1), z2 = lag(z, -2), z3 = lag(z, -3), days.in.month) time(z2) - as.Date(time(zz)) z3 - na.locf( cbind(zz, zoo(, dd), day.of.month = as.numeric(format(time(zz), %d)) )) z3$day.of.month - as.numeric(format(time(zz2), %d)) # now each row of z3 has all the data you need so apply(z3, 1, your.function) On Fri, Apr 17, 2009 at 2:13 PM, manta mantin...@libero.it wrote: any update anybody? I'm really stucked! -- View this message in context: http://www.nabble.com/From-daily-series-to-monthly-and-viceversa-tp23064454p23103052.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/From-daily-series-to-monthly-and-viceversa-tp23064454p23115847.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dummy (factor) based on a pair of variables
Bernardo: this is not quite what I am looking for, Let the data be: y,i,j 1,AUT,BEL 2,AUT,GER 3,BEL,GER then the dummies sould look like: y,i,j,d_AUT,d_BEL,d_GER 1,AUT,BEL,1,1,0 2,AUT,GER,1,0,1 3,BEL,GER,0,1,1 I can generate the above dummies but can this design be imputed in a reg. model directly? Serguei __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] (no subject)
Hi, as a newcomer to R i have what probably is a very simple issue but i have been to solve it for hours: all i want to know is how to arrange a set of numbers in size order without putting them in a table. just arranging them from for e.g. 2,1,3,5,4 into 1,2,3,4,5 - it must be simple but i cant find how to do it anywhere thanksdan _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Can't read table encoded in Unicode (R-2.8.1)
Hi all, I have problems reading Unicode (UTF-16) coded tables in R 2.8.1 under Windows Vista. Imagine the following table: abcd X1,21,31,4 Y2,22,32,4 Z3,23,33,4 Usually I would use the following code to read the table: t = read.table(test.txt, header=T, sep=\t,dec=,) This works well if I create the table using Notepad (the text will be in UTF-8 or ASCII, then). However, If I use e.g. OpenOffice scalc to create a spreadsheet holding the same data and save this data as text (using tabs as separators, no quotes and using Unicode encoding) the command above gives this: t = read.table(test.csv, header=T, sep=\t,dec=,) t ÿþa 1 NA 2 NA 3 NA I tried to play with the encoding parameter but that would not change anything. The file from OpenOffice is in UTF-16, as shown by hexdump: $ hexdump test.csv 000 feff 0061 0009 0062 0009 0063 0009 0064 010 000d 000a 0058 0009 0031 002c 0032 0009 020 0031 002c 0033 0009 0031 002c 0034 000d 030 000a 0059 0009 0032 002c 0032 0009 0032 040 002c 0033 0009 0032 002c 0034 000d 000a 050 005a 0009 0033 002c 0032 0009 0033 002c 060 0033 0009 0033 002c 0034 000d 000a 06e I tried to read the file using file/readLines, which seemed to work after specifying the encoding: a = file(test.csv,open=r, encoding=UTF-16) b = readLines(a) b [1] a\tb\tc\td X\t1,2\t1,3\t1,4 Y\t2,2\t2,3\t2,4 Z\t3,2\t3,3\t3,4 Looking at the code of readtable.R in R-2.8.1. and R-2.9.0 it seems that the encoding does not get passed through in the second call to scan() appearing in the code. I'm not sure if this is a bug or if I'm doing something wrong here. Regards, Hilmar -- My system and R settings are: sessionInfo() R version 2.8.1 (2008-12-22) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252;LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.8.1 Sys.info() sysname release version nodename Windows Vista build 6001, Service Pack 1 PC machine login user x86 options(encoding) $encoding [1] native.enc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (no subject)
Welcome to R! I don't quite know what you mean by size order. But if you have five numbers: a - c(2, 1, 3, 5, 4) then sort(a) will do what you want? HTH! Ranjan On Sat, 18 Apr 2009 14:46:04 + Dan Cary daniel_c...@hotmail.co.uk wrote: Hi, as a newcomer to R i have what probably is a very simple issue but i have been to solve it for hours: all i want to know is how to arrange a set of numbers in size order without putting them in a table. just arranging them from for e.g. 2,1,3,5,4 into 1,2,3,4,5 - it must be simple but i cant find how to do it anywhere thanksdan _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (no subject)
mylist - c( 2,1,3,5,4 ) make a vector of numbers sort(mylist) [1] 1 2 3 4 5in sorted order mylist - c( this, is, a, test) sort(mylist) [1] ais test this in sorted order order(mylist) [1] 3 2 4 1 original positions, e.g. mylist[3] is a On Sat, Apr 18, 2009 at 10:46 AM, Dan Cary daniel_c...@hotmail.co.uk wrote: ...all i want to know is how to arrange a set of numbers in size order without putting them in a table. just arranging them from for e.g. 2,1,3,5,4 into 1,2,3,4,5 - it must be simple but i cant find how to do it anywhere __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Loop question
On Fri, Apr 17, 2009 at 10:12 PM, Brendan Morse morse.bren...@gmail.com wrote:
...I would like to automatically generate a series of matrices and
give them successive names. Here is what I thought at first:
t1-matrix(0, nrow=250, ncol=1)
for(i in 1:10){
t1[i]-rnorm(250)
}
What I intended was that the loop would create 10 different matrices with a
single column of 250 values randomly selected from a normal distribution,
and that they would be labeled t11, t12, t13, t14 etc.
Very close. But since you've started out with a *matrix* t1, your
assignments to t1[i] will assign to parts of the matrix. To correct
this, all you need to do is initialize t1 as a *list of matrices* or
(even better) as an *empty list*, like this:
t1 - list()
and then assign to *elements* of the list (using [[ ]] notation), not
to *sublists* of the list (which is what [ ] notation means in R),
like this:
for(i in 1:10){
t1[[i]] - rnorm(250)
}
Is that what you had in mind?
-s
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can't read table encoded in Unicode (R-2.8.1)
On 18/04/2009 1:18 PM, Hilmar Berger wrote: Hi all, I have problems reading Unicode (UTF-16) coded tables in R 2.8.1 under Windows Vista. Imagine the following table: abcd X1,21,31,4 Y2,22,32,4 Z3,23,33,4 Usually I would use the following code to read the table: t = read.table(test.txt, header=T, sep=\t,dec=,) This works well if I create the table using Notepad (the text will be in UTF-8 or ASCII, then). I haven't tried 2.8.1 (which is obsolete, since yesterday :-), but in 2.9.0 it works fine if I use the fileEncoding argument to read.table. Duncan Murdoch However, If I use e.g. OpenOffice scalc to create a spreadsheet holding the same data and save this data as text (using tabs as separators, no quotes and using Unicode encoding) the command above gives this: t = read.table(test.csv, header=T, sep=\t,dec=,) t ÿþa 1 NA 2 NA 3 NA I tried to play with the encoding parameter but that would not change anything. The file from OpenOffice is in UTF-16, as shown by hexdump: $ hexdump test.csv 000 feff 0061 0009 0062 0009 0063 0009 0064 010 000d 000a 0058 0009 0031 002c 0032 0009 020 0031 002c 0033 0009 0031 002c 0034 000d 030 000a 0059 0009 0032 002c 0032 0009 0032 040 002c 0033 0009 0032 002c 0034 000d 000a 050 005a 0009 0033 002c 0032 0009 0033 002c 060 0033 0009 0033 002c 0034 000d 000a 06e I tried to read the file using file/readLines, which seemed to work after specifying the encoding: a = file(test.csv,open=r, encoding=UTF-16) b = readLines(a) b [1] a\tb\tc\td X\t1,2\t1,3\t1,4 Y\t2,2\t2,3\t2,4 Z\t3,2\t3,3\t3,4 Looking at the code of readtable.R in R-2.8.1. and R-2.9.0 it seems that the encoding does not get passed through in the second call to scan() appearing in the code. I'm not sure if this is a bug or if I'm doing something wrong here. Regards, Hilmar -- My system and R settings are: sessionInfo() R version 2.8.1 (2008-12-22) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252;LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.8.1 Sys.info() sysname release version nodename Windows Vista build 6001, Service Pack 1 PC machine login user x86 options(encoding) $encoding [1] native.enc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dummy (factor) based on a pair of variables
On 2009.04.18 13:52:35, Serguei Kaniovski wrote: Bernardo: this is not quite what I am looking for, Let the data be: y,i,j 1,AUT,BEL 2,AUT,GER 3,BEL,GER then the dummies sould look like: y,i,j,d_AUT,d_BEL,d_GER 1,AUT,BEL,1,1,0 2,AUT,GER,1,0,1 3,BEL,GER,0,1,1 I can generate the above dummies but can this design be imputed in a reg. model directly? Serguei Hello Serguei, I am sure there is a better way to do this, but the following seems to work: # Create sample data.frame() i - c(AUT, AUT, BEL) j - c(BEL, GER, GER) df - data.frame(i=i, j=j) # Create dummy vectors df$d.aut - ifelse(df$i==AUT|df$j==AUT, 1, 0) df$d.bel - ifelse(df$i==BEL|df$j==BEL, 1, 0) df$d.ger - ifelse(df$i==GER|df$j==GER, 1, 0) # Print results df HTH, ~Jason -- Jason W. Morgan Graduate Student, Political Science *The Ohio State University* __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dummy (factor) based on a pair of variables
On 2009.04.18 15:58:30, Jason Morgan wrote: On 2009.04.18 13:52:35, Serguei Kaniovski wrote: I can generate the above dummies but can this design be imputed in a reg. model directly? Oops, I apologize for not reading the whole question. Can you do the following: lm(y ~ I(ifelse(df$i==AUT|df$j==AUT, 1, 0)) + I(ifelse(df$i==BEL|df$j==BEL, 1, 0)) + I(ifelse(df$i==GER|df$j==GER, 1, 0)), data=df) If you exclude the ifelse(), you will get a vector of TRUE/FALSE, which may or may not work. ~Jason Hello Serguei, I am sure there is a better way to do this, but the following seems to work: # Create sample data.frame() i - c(AUT, AUT, BEL) j - c(BEL, GER, GER) df - data.frame(i=i, j=j) # Create dummy vectors df$d.aut - ifelse(df$i==AUT|df$j==AUT, 1, 0) df$d.bel - ifelse(df$i==BEL|df$j==BEL, 1, 0) df$d.ger - ifelse(df$i==GER|df$j==GER, 1, 0) # Print results df HTH, ~Jason -- Jason W. Morgan Graduate Student, Political Science *The Ohio State University* __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] From daily series to monthly and viceversa
Thanks so much -- View this message in context: http://www.nabble.com/From-daily-series-to-monthly-and-viceversa-tp23064454p23116875.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dummy (factor) based on a pair of variables
df - read.table(textConnection(y,i,j + 1,AUT,BEL + 2,AUT,GER + 3,BEL,GER), header=T,sep=,, as.is=T) df y i j 1 1 AUT BEL 2 2 AUT GER 3 3 BEL GER countries - unique(c(df$i,df$j)) countries [1] AUT BEL GER df[countries] - sapply(countries, function(x) df[x] - df$i == x | df$j == x) df y i j AUT BEL GER 1 1 AUT BEL TRUE TRUE FALSE 2 2 AUT GER TRUE FALSE TRUE 3 3 BEL GER FALSE TRUE TRUE Obviously it would not be possible to test this arrangement with lm. So I tried scaling it up and testing on: dft - data.frame(y=rnorm(100), i = sample(countries, 100, replace=T), j= sample(countries, 100, replace=T)) #Removed all the duplicates with: dft - dft(dft$i != dft$j, ] #and it did not give proper answers. This seems to give correct answers dft[countries] - sapply(countries, function(y) apply(dft, 1, function(x) x[2] == y | x[3] == y)) And application of those variables is handles in a reasonable manner by the R formula parser: lm(y ~ AUT + BEL + GER, data=dft) Call: lm(formula = y ~ AUT + BEL + GER, data = dft) Coefficients: (Intercept) AUTTRUE BELTRUE GERTRUE 0.09192 0.15130 -0.29274 NA - David Winsemius On Apr 18, 2009, at 4:09 PM, Jason Morgan wrote: On 2009.04.18 15:58:30, Jason Morgan wrote: On 2009.04.18 13:52:35, Serguei Kaniovski wrote: I can generate the above dummies but can this design be imputed in a reg. model directly? Oops, I apologize for not reading the whole question. Can you do the following: lm(y ~ I(ifelse(df$i==AUT|df$j==AUT, 1, 0)) + I(ifelse(df$i==BEL|df$j==BEL, 1, 0)) + I(ifelse(df$i==GER|df$j==GER, 1, 0)), data=df) If you exclude the ifelse(), you will get a vector of TRUE/FALSE, which may or may not work. ~Jason Hello Serguei, I am sure there is a better way to do this, but the following seems to work: # Create sample data.frame() i - c(AUT, AUT, BEL) j - c(BEL, GER, GER) df - data.frame(i=i, j=j) # Create dummy vectors df$d.aut - ifelse(df$i==AUT|df$j==AUT, 1, 0) df$d.bel - ifelse(df$i==BEL|df$j==BEL, 1, 0) df$d.ger - ifelse(df$i==GER|df$j==GER, 1, 0) # Print results df HTH, ~Jason -- Jason W. Morgan Graduate Student, Political Science *The Ohio State University* __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] igraph 0.5.2
igraph is a package for graphs/networks. It has a C core and uses a simple and fast graph representation allowing millions of vertices and edges. LINKS Release notes for the 0.5.2 version: http://igraph.sourceforge.net/relnotes-0.5.2.html Release notes for the 0.5.1 version: http://igraph.sourceforge.net/relnotes-0.5.1.html Complete list of changes: http://igraph.sourceforge.net/news.html The igraph homepage: http://igraph.sf.net Since there was no announcement about igraph version 0.5.1 on R-pkgs, here is a list of the more important new things of both version 0.5.2 and 0.5.1. NEW FEATURES IN THE 0.5.2 VERSION - We have some support for bipartite (two-mode) graphs now. See ?graph.incidence, ?get.incidence for dealing with incidence matrices, ?graph.bipartite for a way to create bipartite graphs, ?bipartite.projection for creating one-mode projections. See ?is.bipartite for deciding whether a graph has a bipartite structure. - A new simple and fast community finding algorithm called 'label propagation' was added, see ?label.propagation.community for details. - The DrL layout generator now supports 3d layouts. - A limited Tcl/Tk igraph GUI was added, please note that it supports only a small fraction of the igraph functions. You can start it with the tkigraph() function. - Johnson's algorithm is now supported by the shortest path finder. It allows negative edge weight, but not negative cycles. See ?shortest.paths. - The graph.knn function was added to calculate the average nearest neighbor degree. It supports edge weights as well. - Curved edges are now supported by plot() and tkplot(), see ?igraph.plotting for the details. - Many bugs and memory leaks were fixed, see the News section on the igraph homepage for details. NEW FEATURES IN THE 0.5.1 VERSION - DrL layout generator was added, see ?layout.drl. - Uniform sampling of graphs with a given degree sequence, see ?degree.sequence.game. - Conversion functions to/from graphNEL objects (graph package), see ?igraph.to.graphNEL and ?igraph.from.graphNEL. - Conversion functions to/from sparse matrices (Matrix package), see ?graph.adjacency and the 'sparse' argument of ?get.adjacency. - Creating graphs from adjacency lists, see ?graph.adjlist. - The function graph.data.frame has an argument called 'vertices' and this makes it very easy to create graphs with a lot of vertex and edge metadata. - The Dijkstra and Bellman-Ford shortest path algorithms were added, see ?shortest.paths. - The function is.mutual() was added, this tests edge reciprocity. - The plot() function now supports plottings different vertex shapes. See ?igraph.plotting and ?igraph.vertex.shapes for detaills. - Many bugs were fixed, see the News section on the igraph homepage for details. PACKAGE DESCRIPTION: igraph is originally a C library for graphs, but has interfaces to high level languages like R, Python and Ruby. The R package contains BOTH the C library and its R interface. igraph supports: - graph generators, creating both regular structures like trees, lattices, etc. and various random graphs. - a rich set of functions calculating structural properties of graphs, like vertex centrality (degree, betweenness, closeness, page rank, eigenvector centrality, Burt's constraints, etc.), shortest paths, dyad and triad census, network motifs, girth, K-core decomposition, etc. - attributes can be associated with the vertices/edges of the graph, or the graph itself. The attributes can be arbitrary R objects. - graph visualization using regular R devices, interactive visualization using Tcl/Tk, 3D visualization using RGL. - graph layout generators, the standard Kamada-Kawai and Fruchterman-Reingold algorithms are included, plus many more. - Functions for graph and subgraph isomorphism, the BLISS and the VF2 algorithms are included. - Functions for maximal network flows, minimal cuts, vertex and edge connectivity. - igraph can read and write many popular file formats used for storing graph data: GraphML, Pajek, GML and others. - igraph contains implementations of many community structure detection algorithms proposed recently. -- Gabor Csardi gabor.csa...@unil.ch UNIL DGM ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Modelling an incomplete Poisson distribution ?
Emmanuel Charpentier charpent at bacbuc.dyndns.org writes: I forgot to add that yes, I've done my homework, and that it seems to me that answers pointing to zero-inflated Poisson (and negative binomial) are irrelevant ; I do not have a mixture of distributions but only part of one distribution, or, if you'll have it, a zero-deflated Poisson. An answer by Brian Ripley (http://finzi.psych.upenn.edu/R/Rhelp02/archive/11029.html) to a similar question leaves me a bit dismayed : if it is easy to compute the probability function of this zero-deflated RV (off the top of my head, Pr(X=x)=e^-lambda.lambda^x/(x!.(1-e^-lambda))), and if I think that I'm (still) able to use optim to ML-estimate lambda, using this to (correctly) model my problem set and test it amounts to re-writing some (large) part of glm. Furthermore, I'd be a bit embarrassed to test it cleanly (i. e. justifiably) : out of the top of my head, only the likelihood ration test seems readily applicable to my problem. Testing contrasts in my covariates ... hum ! So if someone has written something to that effect, I'd be awfully glad to use it. A not-so-cursory look at the existing packages did not ring any bells to my (admittedly untrained) ears... Of course, I could also bootstrap the damn thing and study the distribution of my contrasts. I'd still been hard pressed to formally test hypotheses on factors... I would call this a truncated Poisson distribution, related to hurdle models. You could probably use the hurdle function in the pscl package to do this, by ignoring the fitting to the zero part of the model. On the other hand, it might break if there are no zeros at all (adding some zeros would be a pretty awful hack/workaround). If you defined a dtpoisson() for the distribution of the truncated Poisson model, you could probably also use bbmle with the formula interface and the parameters argument. The likelihood ratio test seems absolutely appropriate for this case. Why not? Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Tow to perform diallel analysis in R?
I want to use a diallel analysis in R, for some of my own data. I've been through the primary literature and textbooks, and remain stumped as to how to implment this in R. I can illustrate the problem using a published example dataset: [Cockerham and Weir (1977) Quadratic Analyses of Reciprocal Crosses. Biometrics, Vol. 33, No. 1 pp. 187-203] In this study, 8 different individuals were crossed in a diallel design with reciprocals but without self-crosses. Two individuals were measured for each combination of sire and dam. The goal is to partition the phenotypic variance based on comparisons among full and half-siblings, and draw inferences about the additive and dominant components of variation. The data look like this (complete dataset is at the end of this message) summary(df) val sire dam block Min. : 9.00 1:14 1:14 1:56 1st Qu.:13.35 2:14 2:14 2:56 Median :15.35 3:14 3:14 Mean :16.31 4:14 4:14 3rd Qu.:18.05 5:14 5:14 Max. :31.80 6:14 6:14 (Other):28 (Other):28 The analysis used in that (and other studies) that I want to reproduce in R identifies general combining ability (GCA, the average performance of a line in hybrid combinations), and specific combining ability (SCA, cases where certain hybrid combinations exceed expectations). The output from this analysis looks like this: Df MS Blocks 1 4.32 General 7 175.04 Specific 20 21.2 Reciprocal General 7 53.43 Reciprocal Specific 21 11.66 Error 55 3.62 Specifically, I am interested in using my data to ask what proportion of the phenotypic variance is additive genetic variance? The example dataset is pasted below. I would greatly appreciate any ideas for how to do this. -- Eli Meyer Postdoctoral Fellow Department of Integrative Biology University of Texas at Austin Austin, TX 78712 office: (512) 475-6424 cell: (310) 618-4483 -- Cockerham-Weir.tab val sire dam block 1 14.4 1 2 1 2 16.2 1 2 2 3 27.2 1 3 1 4 30.8 1 3 2 5 17.2 1 4 1 6 27 1 4 2 7 18.3 1 5 1 8 20.2 1 5 2 9 16.2 1 6 1 10 16.8 1 6 2 11 18.6 1 7 1 12 14.4 1 7 2 13 16.4 1 8 1 14 16 1 8 2 15 15.4 2 1 1 16 16.5 2 1 2 17 14.8 2 3 1 18 14.6 2 3 2 19 18.6 2 4 1 20 18.6 2 4 2 21 15.2 2 5 1 22 15.3 2 5 2 23 17 2 6 1 24 15.2 2 6 2 25 14.4 2 7 1 26 14.8 2 7 2 27 10.8 2 8 1 28 13.2 2 8 2 29 31.8 3 1 1 30 30.4 3 1 2 31 21 3 2 1 32 23 3 2 2 33 24.6 3 4 1 34 25.4 3 4 2 35 19.2 3 5 1 36 20 3 5 2 37 29.8 3 6 1 38 28.4 3 6 2 39 12.8 3 7 1 40 14.2 3 7 2 41 13 3 8 1 42 14.4 3 8 2 43 16.2 4 1 1 44 17.8 4 1 2 45 11.4 4 2 1 46 13 4 2 2 47 16.8 4 3 1 48 16.3 4 3 2 49 12.4 4 5 1 50 14.2 4 5 2 51 16.8 4 6 1 52 14.8 4 6 2 53 12.6 4 7 1 54 12.2 4 7 2 55 9.6 4 8 1 56 11.2 4 8 2 57 14.6 5 1 1 58 18.8 5 1 2 59 12.2 5 2 1 60 13.6 5 2 2 61 15.2 5 3 1 62 15.4 5 3 2 63 15.2 5 4 1 64 13.8 5 4 2 65 18 5 6 1 66 16 5 6 2 67 10.4 5 7 1 68 12.2 5 7 2 69 13.4 5 8 1 70 20 5 8 2 71 20.2 6 1 1 72 23.4 6 1 2 73 14.2 6 2 1 74 14 6 2 2 75 18.6 6 3 1 76 14.8 6 3 2 77 22.2 6 4 1 78 17 6 4 2 79 14.3 6 5 1 80 17.3 6 5 2 81 9 6 7 1 82 10.2 6 7 2 83 11.8 6 8 1 84 12.8 6 8 2 85 14 7 1 1 86 16.6 7 1 2 87 12.2 7 2 1 88 9.2 7 2 2 89 13.6 7 3 1 90 16.2 7 3 2 91 13.8 7 4 1 92 14.4 7 4 2 93 15.6 7 5 1 94 15.6 7 5 2 95 15.6 7 6 1 96 11 7 6 2 97 13 7 8 1 98 9.8 7 8 2 99 15.2 8 1 1 100 17.2 8 1 2 101 10 8 2 1 102 11.6 8 2 2 103 17 8 3 1 104 18.2 8 3 2 105 20.8 8 4 1 106 20.8 8 4 2 107 20 8 5 1 108 17.4 8 5 2 109 17 8 6 1 110 12.6 8 6 2 111 13 8 7 1 112 9.8 8 7 2 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] nonparametric regression
i want just make a program to compute the estimator of nonparametric density with Gaussian kernel! i need to compute the function of regression. any help for this [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] get() versus getAnywhere()
On 17/04/2009, at 10:21 PM, Duncan Murdoch wrote:
Benjamin Tyner wrote:
Many thanks Duncan. Perhaps this merits a more explicit note in the
documentation?
The quote I gave is from the documentation. How could it be more
explicit?
This is unfortunately typical of the attitude of R-core people toward
the
documentation. ``It's clear.'' they say. ``It's explicit.'' Clear and
explicit once you *know* what it's saying. Not before, but.
In this case the documentation is quite opaque to me, and I would
suspect
to a good many like me. Now that you have made it *genuinely* explicit,
I can understand what the documentation is saying. Prior to that I
wouldn't
have had a prayer of guessing that get() would sometimes find things
that
getAnywhere() would not find.
Moreover, if getAnywhere() does not really mean ``get *anywhere*''
then its
name is misleading. Surely it wouldn't be too tough to modify
getAnywhere()
so that it really got anywhere. E.g. get it to call get() when it can't
find an object with a given name?
cheers,
Rolf
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Re: [R] how to control lattice plot parameters
We can see the plotting options using trellis.par.get(). For example, one
listed parameter is $superpose.line for which we can set col, lty, and lwd.
Load the mlmRev package to obtain the Gcsemv data (used in Lattice book).
One way to set the parameters is
data(Gcsemv, package = mlmRev)
my.fonts - list(font = 2,cex = 2)
my.theme - list(
superpose.symbol = list(col=c(1,3)),
superpose.line=list(col=c(1,3),lwd=c(1,2)),
par.xlab.text = my.fonts,
par.ylab.text = my.fonts
)
xyplot(written ~ course , data = Gcsemv, groups = gender, type = c(p,r),
par.settings = my.theme,
auto.key = list(lines = TRUE, points = FALSE))
On Wed, Apr 15, 2009 at 5:40 AM, Paulo E. Cardoso pecard...@netcabo.ptwrote:
I'm not being able to control all parameters of a xyplot.
dataset example:
tabp[1:10,]
time n X id name
1 1 95 0.00 1 Coral reef
2 1 93 0.00 2 Coral reef
3 1 92 0.00 3 Coral reef
4 1 90 0.00 4 Coral reef
5 1 87 8.994321 5 Coral reef
6 1 86 12.580143 6 Coral reef
7 1 84 17.004030 7 Coral reef
8 1 83 18.469500 8 Coral reef
9 1 82 37.919033 9 Coral reef
101 81 39.059352 10 Coral reef
...
plot code:
xyplot(
X~n,groups=name,data=tabp[tabp$time %in% c(1,4,8),],
ylab=% of target,
xlab=PU selection Frequency,
lty=c(1,3,5), #! not responding
xlim=c(-10,110),
scales=list(cex=0.7,at=seq(0,360,by=20),labels=seq(0,360,by=20)), #!
not responding
panel=function(x,y,groups,subscripts)
{
panel.xyplot(x,y,
subscripts=subscripts,
groups=groups,
type=l)
panel.abline(h=30,lty=2,col=grey50)
},
key=list(
space=top,
columns=1,
text=list(levels(tabp$name)[c(2,5,11)]),
lines=list(col=c(grey,blue,green))
)
)
A few controls are not responding and I don't know how to:
1) control cex for xlab and ylab
2) control position of legend
3) control lty for each group level
4) control plot groups colors
5) match legend colors with graph colors
Any help will be very appreciated.
Paulo E. Cardoso
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Re: [R] get() versus getAnywhere()
I doubt that it will cure all of the R documentation complaints, but
this R-news article by Ligges answered all on my questions on the
topic of accessing source (see page 43):
www.r-project.org/doc/Rnews/Rnews_2006-4.pdf
I learned to use methods() and then to use the full function names to
which calls got dispatched.
--
David Winsemius
On Apr 18, 2009, at 8:47 PM, Rolf Turner wrote:
On 17/04/2009, at 10:21 PM, Duncan Murdoch wrote:
Benjamin Tyner wrote:
Many thanks Duncan. Perhaps this merits a more explicit note in the
documentation?
The quote I gave is from the documentation. How could it be more
explicit?
This is unfortunately typical of the attitude of R-core people
toward the
documentation. ``It's clear.'' they say. ``It's explicit.'' Clear
and
explicit once you *know* what it's saying. Not before, but.
In this case the documentation is quite opaque to me, and I would
suspect
to a good many like me. Now that you have made it *genuinely*
explicit,
I can understand what the documentation is saying. Prior to that I
wouldn't
have had a prayer of guessing that get() would sometimes find things
that
getAnywhere() would not find.
Moreover, if getAnywhere() does not really mean ``get *anywhere*''
then its
name is misleading. Surely it wouldn't be too tough to modify
getAnywhere()
so that it really got anywhere. E.g. get it to call get() when it
can't
find an object with a given name?
cheers,
Rolf
##
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9}}
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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[R] dynlm question: How to predefine formula for call to dynlm(formula) call
I want to set up a model with a formula and then run dynlm(formula)
because I ultimately want to loop over a set of formulas (see end of post)
R form - gas~price
R dynlm(form)
Time series regression with ts data:
Start = 1959(1), End = 1990(4)
snip
Works OK without a Lag term
R dynlm(gas ~ L(gas,1))
Time series regression with ts data:
Start = 1959(2), End = 1990(4)
snip
Works OK with a Lag with this type of call
R form - gas~L(gas,1)
R dynlm(form)
Error in merge.zoo(gas, L(gas, 1), retclass = list, all = FALSE) :
could not find function L
Does not work using a predefined formula with a Lag (This type of call
works using dyn$lm from library(dyn))
How do I make the call (or how do I setup form) so that this works in dynlm?
Thanks for any help,
Ron
To be specific the following is an example of what I was attempting to do:
m1 - gas ~ L(gas,1)
m2 - gas ~ L(gas,1) + price
m3 - gas ~ L(gas,1) + price + d(gas)
m4 - gas ~ L(gas,1) + price + d(gas) + L(d(gas),1)
M - c(m1,m2,m3,m4)
A - array(0,c(4,2))
for(i in 1:4){
g - dynlm(M[[i]]) ## works if use dyn$lm from library(dyn) and use
appropriate m's
A[i,1] - AIC(g,k=2)
A[i,2] - AIC(g,k=log(length(fitted(g
}
colnames(A) - c(AIC,BIC)
rownames(A) - c(m1,m2,m3,m4)
A
--
R. R. Burns
Retired in Oceanside, CA
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