[R] 3d plot
Hi, I would like to have 3d plot, and I found that there is a command scatterplot3d. x is V, y=sigmaV, and z=TSE (note: After some calculation, I update TSE). How could I do to have 3d plot I want? V-seq(1279,1280,,100) sigmaV-seq(0.28,0.29,,100) TSE-matrix(0, length(V),length(sigmaV)) ps: TSE[i,j] corresponds to V[i] and sigmaV[j]. Thanks, Kate [[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] solver in R
Dear R-users, Is there any function in R that works similarly like solver in Excel. I have a set of daily rainfall data and I would like to estimate alpha and beta for the gamma function. Here is my daily rainfall data: [1] 0.2 1.2 0.0 0.0 0.0 0.0 0.0 0.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.8 1.6 0.0 1.8 1.8 0.0 2.6 [24] 33.0 19.0 0.0 0.0 0.6 0.0 0.0 6.8 0.2 0.0 0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.0 0.0 0.0 1.6 5.4 0.0 [47] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.0 0.0 6.0 6.6 0.0 0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.0 [70] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 12.8 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.0 3.8 2.2 1.2 12.2 3.0 0.0 0.0 [93] 0.0 0.0 5.0 3.6 0.0 0.0 0.0 1.8 1.8 1.2 1.2 1.6 3.6 0.2 0.8 0.0 0.0 0.0 0.0 0.0 1.4 16.6 0.0 [116] 1.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 5.4 [139] 1.6 5.8 0.4 34.2 0.0 0.4 0.4 1.4 1.8 0.0 0.0 0.2 0.0 2.4 24.2 0.4 0.0 0.0 0.0 1.6 0.0 0.0 0.0 [162] 0.0 0.2 0.0 0.0 0.0 4.4 0.0 4.4 21.2 3.0 0.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 [185] 0.0 0.8 0.8 3.6 0.0 1.2 0.0 0.0 0.0 0.0 0.0 0.0 13.8 1.6 6.6 5.0 0.0 1.2 0.0 1.0 1.2 2.4 6.0 [208] 2.8 7.0 0.2 0.6 0.0 0.0 0.0 0.0 0.0 3.2 0.2 2.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.0 [231] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 6.6 2.4 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.6 0.0 0.0 0.0 [254] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.4 2.6 0.0 0.6 4.8 0.0 0.2 3.8 17.0 0.2 0.0 0.0 0.0 [277] 43.6 2.6 0.0 0.0 0.0 0.4 8.2 14.4 0.2 0.0 0.0 0.0 1.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 [300] 0.0 Thank you in advance for your help. Looking for last minute shopping deals? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Finding a mean value of a variable holding a dummyvariablefixed
(I don't recall any Prime Minister called Johnson or Nixon, by the way...) That's why the variable was called president. prime.minister is four letters and a dot longer. That was too complicated ;) - cuncta stricte discussurus - -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Gesendet: Sunday, March 30, 2008 11:50 PM An: [EMAIL PROTECTED]; [EMAIL PROTECTED]; r-help@r-project.org Betreff: RE: [R] Finding a mean value of a variable holding a dummyvariablefixed Here is another way, starting with similar dummy data: PMData - data.frame(PM = c(Thatcher, Thatcher, Thatcher, Thatcher, Thatcher, Thatcher,Major, Major, Major, Major, Major, Major), approval = c(3, 4, 5, 6, 7, 8, 6, 5, 4, 3, 2, 1)) PMData - transform(PMData, Month = 1:nrow(PMData)) ## add the time variable PM_average - with(PMData, tapply(approval, PM, mean)) PM_span - with(PMData, sapply(tapply(Month, PM, range), function(x) structure(approval[Month[x]], names = c(First, Last rbind(mean = PM_average, PM_span) Major Thatcher mean3.5 5.5 First 6.0 3.0 Last1.0 8.0 (I don't recall any Prime Minister called Johnson or Nixon, by the way...) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Daniel Malter Sent: Monday, 31 March 2008 1:22 PM To: 'Alexander Ovodenko'; r-help@r-project.org Subject: Re: [R] Finding a mean value of a variable holding a dummyvariablefixed I found a solution. It's probably not the easiest one, but it works. It assumes that your data frame is ordered from earliest to latest record for each president, but it can be easily adjusted if you want to make it dependent on a third column. The final vector index gives you the line indices for the first record for each president. If you replace min by max you get the last instead of the first record. You can then find the values by ##Sample data president=c(Johnson,Johnson,Johnson,Johnson,Johnson,Johnson,Nix on,Nixon,Nixon,Nixon,Nixon,Nixon) approval=c(3,4,5,6,7,8,6,5,4,3,2,1) tapply(approval,president,mean) ##Find index for first row of each president; assumes ascending order of observations; change min to max to find last record index=NULL for(i in 1:length(unique(president))) index[i]=min(which((president==unique(president)[i])==TRUE)) index ##Generate table with first approvals first.approval=data.frame(cbind(index,president[index],approval[index])) names(first.approval)=c(Index,President,Approval) first.approval Cheers, Daniel - cuncta stricte discussurus - -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Alexander Ovodenko Gesendet: Sunday, March 30, 2008 9:47 PM An: r-help@r-project.org Betreff: [R] Finding a mean value of a variable holding a dummy variablefixed I have time-series data on approval ratings of British Prime Ministers. The prime ministers dating from MacMillan onward till today are coded as dummy variables and the approval ratings are entered for each month. I want to know the mean value of the approval rating of each Prime Minister in the dataset and the approval rating during his/her first month and last month as PM. What R code should I enter for these data? In other words, I want hold the dummy corresponding to each Prime Minister fixed at value one and know the first rating that PM has, the last rating s/he has, and the mean rating s/he has. Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] solver in R
library(MASS) check out fitdistr( ) ... ?fitdistr This isn't 'solver', but then again R isn't MS Excel! Roslina Zakaria wrote: Dear R-users, Is there any function in R that works similarly like solver in Excel. I have a set of daily rainfall data and I would like to estimate alpha and beta for the gamma function. Here is my daily rainfall data: [1] 0.2 1.2 0.0 0.0 0.0 0.0 0.0 0.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.8 1.6 0.0 1.8 1.8 0.0 2.6 [24] 33.0 19.0 0.0 0.0 0.6 0.0 0.0 6.8 0.2 0.0 0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.0 0.0 0.0 1.6 5.4 0.0 [47] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.0 0.0 6.0 6.6 0.0 0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.0 [70] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 12.8 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.0 3.8 2.2 1.2 12.2 3.0 0.0 0.0 [93] 0.0 0.0 5.0 3.6 0.0 0.0 0.0 1.8 1.8 1.2 1.2 1.6 3.6 0.2 0.8 0.0 0.0 0.0 0.0 0.0 1.4 16.6 0.0 [116] 1.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.8 5.4 [139] 1.6 5.8 0.4 34.2 0.0 0.4 0.4 1.4 1.8 0.0 0.0 0.2 0.0 2.4 24.2 0.4 0.0 0.0 0.0 1.6 0.0 0.0 0.0 [162] 0.0 0.2 0.0 0.0 0.0 4.4 0.0 4.4 21.2 3.0 0.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 [185] 0.0 0.8 0.8 3.6 0.0 1.2 0.0 0.0 0.0 0.0 0.0 0.0 13.8 1.6 6.6 5.0 0.0 1.2 0.0 1.0 1.2 2.4 6.0 [208] 2.8 7.0 0.2 0.6 0.0 0.0 0.0 0.0 0.0 3.2 0.2 2.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.0 [231] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 6.6 2.4 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.6 0.0 0.0 0.0 [254] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.4 2.6 0.0 0.6 4.8 0.0 0.2 3.8 17.0 0.2 0.0 0.0 0.0 [277] 43.6 2.6 0.0 0.0 0.0 0.4 8.2 14.4 0.2 0.0 0.0 0.0 1.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 [300] 0.0 Thank you in advance for your help. Looking for last minute shopping deals? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr Richard Rowe Zoology Tropical Ecology School of Marine Tropical Biology James Cook University Townsville 4811 AUSTRALIA ph +61 7 47 81 4851 fax +61 7 47 25 1570 JCU has CRICOS Provider Code 00117J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Comparing Time Series
Dear All, I need to compare hundreds (about 200-300) of time series. Would anyone tell me how to do this in R? If R has no package for doing this, can I get some insight what method I should use? best regards, Nathanael Gratias -- View this message in context: http://www.nabble.com/Comparing-Time-Series-tp16392632p16392632.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] MANOVA, SCC and multiple comparisons
Bonjour, we wanted to fit a manova as descripted in Marieta /et al./ 2003, convergent habitat segregation of /Aedes aegyptii/ and /Aedes albopictus/ (Diptera: /Culicidae/) in Southern Brazil and Florida, /J. Med. Entomol./, *40* (6), 785-794. They did their analysis with SAS software. We have the same kind of data with two insect species and city, habitat and season factors. We used the function manova() and it was OK for multivariate analysis. We have two questions (we have searched, with the help tools we know, on the site and on many packages without success ) : 1/ The next step of our analysis was to calculate the Standardized Canonical Coefficients (SCC); we have no idea of how to do. Is there some link with a PCA on Instrumental Variables ? The explaination in the article was not clear for us. 2/ As in a ANOVA, we wanted to realize mean comparisons. Marieta et al. used bivariate pairwise contrasts of a factor within each level of the other factor (for significant interactions) using Dunn-Sidak method and univariate contrasts of pairs of main effects (for significant main effects) using Dunn-Sidak method. We tried to use the glht() function (from multcomp package) without success because of the nature of manova() object. Has someone an helpful advice to give ? Merci beaucoup, (Thanks a lot) Bien cordialement, (Sincerely yours) Frédéric. -- Dr. Frédéric Chiroleu Biométricien CIRAD-Systèmes Biologiques (Cirad-Bios) UMR 53 PVBMT (Peuplements Végétaux et Bio-agresseurs en Milieu Tropical) Laboratoire d'Ecologie Terrestre et de Lutte Intégrée (LETLI) Pôle de Protection des Plantes (3P) 7, chemin de l'IRAT - Ligne Paradis 97410 Saint-Pierre Île de la Réunion - France Tél. : +262 (0)262 499 230 Standard : +262 (0)262 499 200 Fax : +262 (0)262 499 293 Courriel : [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Number of variables in lrm (Design)
Hi I'm sure this is straightforward, but I can't find an answer. I'm trying to find the function that returns the number of dependent variables/rank of the regression/degrees of freedom/something similar. Any ideas? Thanks! Paul [[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] S4 : the list of all the object
CG == Christophe Genolini [EMAIL PROTECTED]
on Sat, 29 Mar 2008 16:20:41 +0100 writes:
CG Hi the list, Is it possible to get the list of all the
CG S4 user define classes?
do you mean those in .GlobalEnv?
If you have written a package using S4 classes, these are also
user defined ..
CG S4 user define classes? I would like to set up a
CG package.skeleton.S4 but for that, I need the list of
CG the classes...
But (I thought you knew ??) package.skeleton does work in
R-devel and R-alpha,
so we'd rather be very glad if you used R-alpha and told use
about deficiencies you'll find in package.skeleton() there..
{I'm sure there still are some, so we'd be very interested in
well-tested patches against
https://svn.r-project.org/R/trunk/src/library/utils/R/package.skeleton.R
}
Bonnes salutations,
Martin
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Re: [R] Comparing Time Series
Surely you need the insight before choosing a package? What is the problem you are trying to solve? There are many different aspects of time series which could be of interest, and we have no idea which are relevant to your problem. On Sun, 30 Mar 2008, nathan3073 wrote: Dear All, I need to compare hundreds (about 200-300) of time series. Would anyone tell me how to do this in R? If R has no package for doing this, can I get some insight what method I should use? best regards, Nathanael Gratias -- View this message in context: http://www.nabble.com/Comparing-Time-Series-tp16392632p16392632.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. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplot2 argument handling odd
Hi Hadley, I tried today your solution to this problem. However, the finally working code was like eval(substitute(ggplot( aes_string(...) )) The qplot code below did not work at all, neither works a ggplot(aes_string(...)) BTW: Since I am always writing scripts to produce my plots, how can I split accross multiple lines one command in a R-script? These ggplot commands can become rather long and the syntax pl - ggplot( ... long expression . ) pl - pl + some_layer( ... ) is not very intuituve. So, how do I break lines in R-Scripts. These questions must sound like beginner questions - but I have searched archives and documentation without success. As we speak of it - is your book finally out? Greetings, Sebastian Weber On Thu, 2008-03-27 at 07:57 -0500, hadley wickham wrote: Ok, I will try that, thanks. BTW, where is this aes_string option documented, sounds useful? How could I do the same thing with facetting? If I want to save something like . ~ groupVar as a string in a variable, could I pass it with facet_string to ggplot? It's a see also from ?aes (or at least it is in the development version). facet_grid should already accept a string instead of a function, so it should just work. But anyway, I would be curious how to do it with qplot. The basic question is: How to pass the string of a variable to a function which is supposed to interpret it. Aka: var - magicVar fun(doSomeMagic(var)) doSomeMagic should then write magicVar at the place. In general, you can use as.name, substitute and eval: x - as.name(mpg) y - as.name(wt) eval(substitute(qplot(x, y, data=mtcars), list(x=x, y=y))) Hadley __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] WinXP exhibits sluggish graphics window movement mouse tracking (repainting?) over windows graphics devices
I don't see this on a much older and slower machine, so suspect a problem with your Windows. It looks like something is set up to ask R to repaint after the mouse pointer, whereas Windows ought to be doing that. Even then, R uses double buffering, so the repaint should be fast (provided graphics acceleration is turned on). If you want to try updated versions of R, we suggest you use 2.7.0 alpha and not R-patched. But R updates will not solve Windows problems. On Mon, 31 Mar 2008, Richard Yeh wrote: I just noticed when using 2.6.1, 2.6.2 (2008-02-08), 2.6.2pat (2008-02-21 r44582), and 2.6.2pat (2008-03-24 r44975) on my poor old Celeron D330 (2.6 GHz; 3.5 years old) running Windows XP, that the mouse cursor appears to be redrawn more sluggishly when the pointer is over R windows graphics devices than over the R console window or other applications' windows. The slowdown seems to start only after I plotting something in the window (running windows() to open the device does not cause any slowdown), but the behavior starts after a plot(rnorm(100)). The slowdown seems to depend on the area of the R graphics window that is visible. The slowdown also occurs when moving the plot window. For example, if the graphics window is frontmost, and I want to move it until it is mostly offscreen, then the movement is jerky. However, once the window is mostly offscreen, dragging it back onto the screen is smooth and fast. (When I drag windows, I only see the frame, not the contents.) This seems to affect SDI and MDI modes. The Windows task manager confirms that when the plot window is frontmost, Rgui.exe takes most of the CPU time. My graphics card is based on an ATI Radeon 7000, with 32 MB of RAM. I am surprised that I never noticed this before. Nobody else seems to have reported it to the r-help list. Is this problem restricted to my system? (My work machine is much newer, and I do not notice the problem there.) -- 607-351-4838 / [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] hsv2rgb in R
I noted that there is a hsv2rgb in the C-API, but no corresponding function in R, while rgb2hsv is available in R. Does the functionality of hsv2rgb hide under some other name? Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R-packages installation problems
Dear, I've installed the following R version on HP compaq 6710 b under Fedora 8: version _ platform i386-redhat-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 6.2 year 2008 month 02 day 08 svn rev 44383 language R version.string R version 2.6.2 (2008-02-08) When I try to install some R packages I receive the following message: [EMAIL PROTECTED] R-packages]# R CMD INSTALL gtools_2.4.0.tar.gz * Installing to library '/usr/lib/R/library' * Installing *source* package 'gtools' ... ** libs gcc -m32 -std=gnu99 -I/usr/include/R -I/usr/include/R -I/usr/local/include -fpic -O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector --param=ssp-buffer-size=4 -m32 -march=i386 -mtune=generic -fasynchronous-unwind-tables -c setTCPNoDelay.c -o setTCPNoDelay.o setTCPNoDelay.c:1:15: error: R.h: No such file or directory setTCPNoDelay.c:2:24: error: Rinternals.h: No such file or directory setTCPNoDelay.c: In function ‘checkStatus’: setTCPNoDelay.c:66: warning: implicit declaration of function ‘strncpy’ setTCPNoDelay.c:66: warning: incompatible implicit declaration of built-in function ‘strncpy’ setTCPNoDelay.c:72: warning: implicit declaration of function ‘strerror’ setTCPNoDelay.c:72: warning: passing argument 2 of ‘strncpy’ makes pointer from integer without a cast make: *** [setTCPNoDelay.o] Error 1 ERROR: compilation failed for package 'gtools' ** Removing '/usr/lib/R/library/gtools' Indeed R headers (R.h and Rinternals.h) don' t exist on my filesystem. Where can I find them? How can I install them? Thanks for help, Best regards, Roberto -- ARPA Piemonte SC05 - Area Previsione e Monitoraggio Ambientale Roberto Cremonini via Pio VII, 9 10135 TORINO ITALIA Tel. +39 011 1968 0282 Fax. +39 011 1968 1341 Riservatezza/Confidentiality In ottemperanza al D.Lgs. n. 196 del 30.06.2003 in materia di protezione dei dati personali, le informazioni contenute in questo messaggio sono strettamente riservate ed esclusivamente indirizzate al destinatario indicato (oppure alla persona responsabile di rimetterlo al destinatario). Vogliate tener presente che qualsiasi uso, riproduzione o divulgazione di questo messaggio è vietato. Nel caso in cui aveste ricevuto questo messaggio per errore, vogliate cortesemente avvertire il mittente e distruggere il presente messaggio. According to Italian law D.Lgs. 196/2003 concerning privacy, if you are not the addressee (or responsible for delivery of the message to such person) you are hereby notified that any disclosure, reproduction, distribution or other dissemination or use of this communication is strictly prohibited. If you have received this message in error, please destroy it and notify us by email. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] hsv2rgb in R
DM == Dieter Menne [EMAIL PROTECTED] on Mon, 31 Mar 2008 08:06:39 + (UTC) writes: DM I noted that there is a hsv2rgb in the C-API, but no DM corresponding function in R, while rgb2hsv is available DM in R. DM Does the functionality of hsv2rgb hide under some other DM name? yes: hsv() __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: I need really help
Hi 1. Install R 2. Read some intro documents (doc folder) 3. Try to reproduce simple examples from let say R-intro.pdf and/or R-data.pdf 4. Try to follow examples with your data 5. When there is an error read a message and try to get some explanation in help page associated with particular command 6. If this does not lead to success, make a simple reproducible example and post it together with your question. From what you write, I deduct that ?read.table and maybe ?reshape could be your first choice. Regards Petr [EMAIL PROTECTED] [EMAIL PROTECTED] napsal dne 30.03.2008 23:26:42: Dear , Madam/Sir I'm student doing master in statistics, I'd like to know how i can set data in R and apply meta-analysis. i have tries many ways but I'm really stuck. i really appreciate if you can help me in this matter. Mohammed Owlowa - [[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] R-packages installation problems
How did you install R? If from the F8 RPMs, it looks like you missed the R-devel RPM (which was always recommended to install packages, and has become necessary in the F8 RPM for 2.6.2). Note that Fedora are responsible for making clear what you need to install from their re-packaged distributions (the R developers do not recommend splitting up R), so if this was not clear to you, please file a Fedora bug report. See also https://stat.ethz.ch/pipermail/r-help/2008-March/158262.html On Mon, 31 Mar 2008, Roberto Cremonini wrote: Dear, I've installed the following R version on HP compaq 6710 b under Fedora 8: version _ platform i386-redhat-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 6.2 year 2008 month 02 day 08 svn rev 44383 language R version.string R version 2.6.2 (2008-02-08) When I try to install some R packages I receive the following message: [EMAIL PROTECTED] R-packages]# R CMD INSTALL gtools_2.4.0.tar.gz * Installing to library '/usr/lib/R/library' * Installing *source* package 'gtools' ... ** libs gcc -m32 -std=gnu99 -I/usr/include/R -I/usr/include/R -I/usr/local/include -fpic -O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector --param=ssp-buffer-size=4 -m32 -march=i386 -mtune=generic -fasynchronous-unwind-tables -c setTCPNoDelay.c -o setTCPNoDelay.o setTCPNoDelay.c:1:15: error: R.h: No such file or directory setTCPNoDelay.c:2:24: error: Rinternals.h: No such file or directory setTCPNoDelay.c: In function ‘checkStatus’: setTCPNoDelay.c:66: warning: implicit declaration of function ‘strncpy’ setTCPNoDelay.c:66: warning: incompatible implicit declaration of built-in function ‘strncpy’ setTCPNoDelay.c:72: warning: implicit declaration of function ‘strerror’ setTCPNoDelay.c:72: warning: passing argument 2 of ‘strncpy’ makes pointer from integer without a cast make: *** [setTCPNoDelay.o] Error 1 ERROR: compilation failed for package 'gtools' ** Removing '/usr/lib/R/library/gtools' Indeed R headers (R.h and Rinternals.h) don' t exist on my filesystem. Where can I find them? How can I install them? Thanks for help, Best regards, Roberto -- ARPA Piemonte SC05 - Area Previsione e Monitoraggio Ambientale Roberto Cremonini via Pio VII, 9 10135 TORINO ITALIA Tel. +39 011 1968 0282 Fax. +39 011 1968 1341 Riservatezza/Confidentiality In ottemperanza al D.Lgs. n. 196 del 30.06.2003 in materia di protezione dei dati personali, le informazioni contenute in questo messaggio sono strettamente riservate ed esclusivamente indirizzate al destinatario indicato (oppure alla persona responsabile di rimetterlo al destinatario). Vogliate tener presente che qualsiasi uso, riproduzione o divulgazione di questo messaggio è vietato. Nel caso in cui aveste ricevuto questo messaggio per errore, vogliate cortesemente avvertire il mittente e distruggere il presente messaggio. According to Italian law D.Lgs. 196/2003 concerning privacy, if you are not the addressee (or responsible for delivery of the message to such person) you are hereby notified that any disclosure, reproduction, distribution or other dissemination or use of this communication is strictly prohibited. If you have received this message in error, please destroy it and notify us by email. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] hsv2rgb in R
I noted that there is a hsv2rgb in the C-API, but no corresponding
function in
R, while rgb2hsv is available in R.
Does the functionality of hsv2rgb hide under some other name?
I agree that it does seem like such a function ought to exist. You can
use col2rgb(hsv(...)) to get what you want.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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Re: [R] 3d plot
If you want to plot some surface, use persp(); or wireframe() from package lattice; or persp3d() from package rgl or therelike. Uwe Ligges kate wrote: Hi, I would like to have 3d plot, and I found that there is a command scatterplot3d. x is V, y=sigmaV, and z=TSE (note: After some calculation, I update TSE). How could I do to have 3d plot I want? V-seq(1279,1280,,100) sigmaV-seq(0.28,0.29,,100) TSE-matrix(0, length(V),length(sigmaV)) ps: TSE[i,j] corresponds to V[i] and sigmaV[j]. Thanks, Kate [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] png file in batch mode
Dear R users, I have an R script which produce me a plot. I run it in batch mode. The outputed file is .ps How ca I ouput the plot in .png in stead of .ps ? I found in R FAQ 7.19 the solution but it is for Linux, and I'm using Windows. Thanks for your help, Ciao, Adel -- View this message in context: http://www.nabble.com/png-file-in-batch-mode-tp16395299p16395299.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] Reorder the x-axis using lattice
Dear list, Is there a way to reorder the xaxis using lattice. Using the following data, the x-axis is ordered as BP GH MH PF RE RP SF VT but I would like the x-axis to be ordered as PF RP BP GH VT SF RE MH. Kön Skalor Tillfälle Medelvärde 1 Kvinnor BP 1-inskrivning 36.45283 2 Kvinnor GH 1-inskrivning 38.62255 3 Kvinnor MH 1-inskrivning 62.9 4 Kvinnor PF 1-inskrivning 39.80710 5 Kvinnor RE 1-inskrivning 41.50943 6 Kvinnor RP 1-inskrivning 22.2 7 Kvinnor SF 1-inskrivning 59.19811 8 Kvinnor VT 1-inskrivning 34.84568 9 Kvinnor BP 2-utskrivning 43.14815 10 Kvinnor GH 2-utskrivning 44.11321 11 Kvinnor MH 2-utskrivning 77.2 12 Kvinnor PF 2-utskrivning 44.74280 13 Kvinnor RE 2-utskrivning 68.95425 14 Kvinnor RP 2-utskrivning 39.90385 15 Kvinnor SF 2-utskrivning 64.62264 16 Kvinnor VT 2-utskrivning 51.97531 bwplot(Medelvärde ~ Skalor| Kön , kt, panel = panel.superpose, groups = Tillfälle,scales = list(x = list(rot = 45),cex=0.7,alternating=2), panel.groups = panel.linejoin,lty=c(1:3),lwd=3,col=c(steelblue,grey50,green4), ylab = list(label = skalpoäng (0-100), cex = 0.8), xlab = list(label = skalor, cex = 0.8), key = list(lines = Rows(list(col=c(steelblue,grey50,green4),lty=c(1:3)), c(1:3, 0)),cex=0.8,text = list(lab = as.character(unique(kt$Tillfälle))), columns = 2, title = SF-36: Skalpoäng för respektive kön vid 3 mättillfälle , cex.title=0.9)) Thanks in advance, Tom - Jämför pris på flygbiljetter och hotellrum: http://shopping.yahoo.se/c-169901-resor-biljetter.html [[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] Reverse seq
On 3/31/2008 6:31 AM, Mario Maiworm wrote: Hi all, I thought I was not SUCH a nooby:) How can I reverse a sequence/ vector? i.e., turn X - 3 5 4 2 6 5 4 3 6 Into X - 6 3 4 5 6 2 4 5 3 I mean without looping and indexing. There should be a very easy solution, shouldn't it? rev(X) [1] 6 3 4 5 6 2 4 5 3 ?rev Mario __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Phone: +49 40 42838 3515 Fax: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.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. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] aov and errors used for interaction effects
Dear list-members, I just switched to R, so here probably a newbie question, I couldn't find a answer for yet: Assume a design with three repeated measures factors (F1-F3), I tried to run the following: aov(rt ~ F1*F2*F3 + Error(subject / (F1*F2*F3))) Comparing the output from here and SPSS (General Linear Model - Repeated measures) I found, that R does all calculations right for the main effects of F1-F3, but seems to do something different for the interactions: Whereas SPSS gets the F for an interaction by dividing the MS of the interaction term (e.g. F1*F2) by the corresponding MS of the interaction with subject (then F1*F2*subject), R just divides all interaction MS by the residual MS. Could somebody point out to me, what I'm doing wrong or how I can get aov to use the appropriate error terms also for the higher order interactions? Thanks for helping! -- View this message in context: http://www.nabble.com/aov-and-errors-used-for-interaction-effects-tp16395617p16395617.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] plot 3dimensional points with log axes
Dear R users I am looking for functions that can plot 3 dimensional figures *with logarithmic axes*. I tried scatterplot3d (under package scatterplot3d) but the log parameter is not implemented (although log appears in the parameter list). Does anyone has some suggestions? Thanks a lot Austin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Reverse seq
Hi all, I thought I was not SUCH a nooby:) How can I reverse a sequence/ vector? i.e., turn X - 3 5 4 2 6 5 4 3 6 Into X - 6 3 4 5 6 2 4 5 3 I mean without looping and indexing. There should be a very easy solution, shouldn't it? Mario __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Phone: +49 40 42838 3515 Fax: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.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] L-BFGS-B needs finite values of 'fn'
Dear All,
I am trying to solve the optimization problem below, but I am always
getting the following error:
Error in optim(rep(20, nvar), f, gr, method = L-BFGS-B, lower = rep(0, :
L-BFGS-B needs finite values of 'fn'
Any ideas?
Thanks in advance,
Paul
---
k - 1
b - 0.3
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
return(r)
}
gr - function(x) {
n - length(x)
r - (b^(0:(n-1)))*(1/x) - 400*(sum(x)-k)
return(r)
}
nvar - 10
(sols -
optim(rep(20,nvar),f,gr,method=L-BFGS-B,lower=rep(0,nvar),upper=rep(k,nvar),control=list(fnscale=-1,parscale=rep(2000,nvar),factr=1e-300,pgtol=1e-300)))
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Re: [R] L-BFGS-B needs finite values of 'fn'
Your function gives -Inf at the lower bound on the parameters, so you are
minimizing a function without a lower bound.
Using the trace facilities of optim() would have got you thereeasily
enough.
On Mon, 31 Mar 2008, Paul Smith wrote:
Dear All,
I am trying to solve the optimization problem below, but I am always
getting the following error:
Error in optim(rep(20, nvar), f, gr, method = L-BFGS-B, lower = rep(0, :
L-BFGS-B needs finite values of 'fn'
Any ideas?
Thanks in advance,
Paul
---
k - 1
b - 0.3
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
return(r)
}
gr - function(x) {
n - length(x)
r - (b^(0:(n-1)))*(1/x) - 400*(sum(x)-k)
return(r)
}
nvar - 10
(sols -
optim(rep(20,nvar),f,gr,method=L-BFGS-B,lower=rep(0,nvar),upper=rep(k,nvar),control=list(fnscale=-1,parscale=rep(2000,nvar),factr=1e-300,pgtol=1e-300)))
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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
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Re: [R] Clash between alr3 and AlgDesign. (Was: Re: Second subsequent calls to function fails. Please help debug.)
Dear all, thanks for the advice. The latest version of crossdes (1.0-8) on CRAN now has namespaces. Hope this helps. Regards Oliver Sailer Duncan Murdoch schrieb: On 30/03/2008 8:43 AM, Michael Kubovy wrote: Thanks, Duncan, I started a new session under the suspicion that packages were clashing. ... # # BAD INTERACTION BETWEEN alr3 and AlgDesign # What now? There's not much you can do; this is up to the package authors to fix. Some advice for them: alr3, crossdes, faraway, and HH don't use a namespace. This makes them very vulnerable to this sort of interaction, because they don't get to choose where the functions they use come from. They should add one. In my opinion, all packages should have namespaces, and I'd like to create a default one if the author doesn't. (The default would be: import what is listed in the Depends clause, export everything.) I don't think this will happen for 2.7.0, though package.skeleton might start creating one. Duncan Murdoch -- -- Dipl.-Stat. Oliver Sailer Fakultät Statistik Technische Universität Dortmund D-44221 Dortmund Tel.: ++49 (0)231 755 - 5409 Fax: ++49 (0)231 755 - 3454 E-Mail: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lmer function.
I am using the lmer function from the lme4 package. I wrote the following statement, specifying the method to be adaptive Gaussian quadrature. I am getting an error saying method = AGQ not yet implemented for supernodal representation. Please help. fit - lmer(response~beta1+(1|patient),family=binomial,method=AGQ,data=data.2) Boikanyo. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplots - how to increase the number of levels
Hi, Thanks to Hadley Wickham, I am now able to plot some nice weather graphs (see http://www.nabble.com/how-to-plot-image%28%29-without-painting-a-map-%28the-background%29-td15546906.html#a15584405 How to plot a graph on a map ). For some parameters, I would like to increase the number of levels (the number of grey tinges in fact (nuances in french - not sure tinges is the correct english translation)). I tried to add nlevels=10 to the following command : p - qplot(data=lonlat1, longitude, latitude, geom=tile, fill=TCDC, xlab=, ylab=) + scale_fill_gradient(low=white, high=grey) + geom_contour(aes(z = TCDC)) + france_map ... but nothings changes. I have always 5 levels for TCDC : 20 / 40 / 60 / 80 /100. http://www.nabble.com/file/p16396028/tcdc_map.jpg Could you please tell me how to solve this easy (except for me) problem ? Have a nice week, Ptit Bleu. -- View this message in context: http://www.nabble.com/ggplots---how-to-increase-the-number-of-levels-tp16396028p16396028.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot 3dimensional points with log axes
On 3/31/2008 7:05 AM, Hung-Hsuan Chen (Sean) wrote: Dear R users I am looking for functions that can plot 3 dimensional figures *with logarithmic axes*. I tried scatterplot3d (under package scatterplot3d) but the log parameter is not implemented (although log appears in the parameter list). Does anyone has some suggestions? Thanks a lot A general way to do this sort of thing is to take the logs yourself, and get the general purpose function to draw without axes. You can then add ticks, etc. after the fact. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] unexpected GAM result - at least for me!
Hi I am afraid i am not understanding something very fundamental and does not matter how much i am looking into the book Generalized Additive Models of S. Wood i still don't understand my result. I am trying to model presence / absence (presence = 1, absence = 0) of a species using some lidar metrics (i have 4 of these). I am using different models and such and when i used gam i got this very weird (for me) result which i thought it is not possible - or i have no idea how to interpret it. can3.gam - gam(can0~s(be)+s(crr)+s(ch)+s(home), family = 'binomial') summary(can3.gam) Family: binomial Link function: logit Formula: can 0 ~ s(be) + s(crr) + s(ch) + s(home) Parametric coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)85.39 162.88 0.524 0.6 Approximate significance of smooth terms: edf Est.rank Chi.sq p-value s(be) 1.0001 0.100 0.751 s(crr) 3.9298 0.380 1.000 s(ch) 6.8209 0.396 1.000 s(home) 1.0001 0.314 0.575 R-sq.(adj) = 1 Deviance explained = 100% UBRE score = -0.81413 Scale est. = 1 n = 148 Is this a perfect fit with no statistical significance, an over-estimating or what It seems that the significance of the smooths terms is null. Of course with such a model i predict perfectly presence / absence of species. Again, i hope you don't mind i'm asking you this. Any explanation will be very much appreciated. Thanks, Monica PS. I've contacted the author of the book who is the package maintainer as well but until now i didn't get a reply. _ esh_realtime_042008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 maps in R
Aleksandr Andreev aleksandr.andreev at duke.edu writes: Roger Bivand Roger.Bivand at nhh.no writes: Merge using: FS1 - spCbind(FS, agg2) This call fails, because: Error in spCbind(FS, agg2) : different numbers of rows This was exactly why I emphasised care. One way to try to do this is to extract the FS data slot: FSd - as(FS, data.frame) and then merge() FSd and agg2, using - untried - something like: FS1d - merge(FSd, agg2, by=row.names, all=TRUE) and check str(FS1d) for sanity, and possibly eyeball: match(row.names(FS1d), sapply(slot(FS, polygons), function(x) slot(x, ID))) before moving on to: FS1 - SpatialPolygonsDataFrame(as(FS, SpatialPolygons), data=FSD1) The alternative is to subset FS to remove the geometries for which there is no data, but using merge to insert NAs ought to work too. Roger The reason is because I have data in a001ter for 79 Federal Subjects, but russia.shp contains 193 labeled objects (93 Federal Subjects plus 100 Arctic Ocean Islands and other things). Is there a way I can merge without an equal number of rows, labeling the data for the Arctic islands as missing? Thanks, Aleks Andreev Duke 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] unexpected GAM result - at least for me!
On 3/31/2008 8:34 AM, Monica Pisica wrote: Hi I am afraid i am not understanding something very fundamental and does not matter how much i am looking into the book Generalized Additive Models of S. Wood i still don't understand my result. I am trying to model presence / absence (presence = 1, absence = 0) of a species using some lidar metrics (i have 4 of these). I am using different models and such and when i used gam i got this very weird (for me) result which i thought it is not possible - or i have no idea how to interpret it. can3.gam - gam(can0~s(be)+s(crr)+s(ch)+s(home), family = 'binomial') summary(can3.gam) Family: binomial Link function: logit Formula: can 0 ~ s(be) + s(crr) + s(ch) + s(home) Parametric coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)85.39 162.88 0.524 0.6 Approximate significance of smooth terms: edf Est.rank Chi.sq p-value s(be) 1.0001 0.100 0.751 s(crr) 3.9298 0.380 1.000 s(ch) 6.8209 0.396 1.000 s(home) 1.0001 0.314 0.575 R-sq.(adj) = 1 Deviance explained = 100% UBRE score = -0.81413 Scale est. = 1 n = 148 Is this a perfect fit with no statistical significance, an over-estimating or what It seems that the significance of the smooths terms is null. Of course with such a model i predict perfectly presence / absence of species. Again, i hope you don't mind i'm asking you this. Any explanation will be very much appreciated. Look at the data. You can get a perfect fit to a logistic regression model fairly easily, and it looks as though you've got one. (In fact, the huge intercept suggests that all predictions will be 1. Do you actually have any variation in the data?) Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] L-BFGS-B needs finite values of 'fn'
Thanks for you help and for having called my attention to the trace
facilities of optim(), which are really helpful.
Paul
On Mon, Mar 31, 2008 at 12:43 PM, Prof Brian Ripley
[EMAIL PROTECTED] wrote:
Your function gives -Inf at the lower bound on the parameters, so you are
minimizing a function without a lower bound.
Using the trace facilities of optim() would have got you thereeasily
enough.
On Mon, 31 Mar 2008, Paul Smith wrote:
Dear All,
I am trying to solve the optimization problem below, but I am always
getting the following error:
Error in optim(rep(20, nvar), f, gr, method = L-BFGS-B, lower = rep(0, :
L-BFGS-B needs finite values of 'fn'
Any ideas?
Thanks in advance,
Paul
---
k - 1
b - 0.3
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
return(r)
}
gr - function(x) {
n - length(x)
r - (b^(0:(n-1)))*(1/x) - 400*(sum(x)-k)
return(r)
}
nvar - 10
(sols -
optim(rep(20,nvar),f,gr,method=L-BFGS-B,lower=rep(0,nvar),upper=rep(k,nvar),control=list(fnscale=-1,parscale=rep(2000,nvar),factr=1e-300,pgtol=1e-300)))
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PLEASE do read the posting guide
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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
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing Time Series
It is a reaction time experiment. My program will display 25 circles, arranged in a 5x5 square. The program will light one circle at a time, which the subject should click ASAP. The time interval between two lighting is 1.5s. If the subject fails to click in approriate time, the data will be considered missing. The one I want to explore is the relationship between intellegence and the ability to recognize visual stimuli regularity. I arrange the experiment so there will be 4 repeated pattern. At last, after all pattern have been displayed, my program will light the circle randomly. I want to check the time series plots, in order to clarify my hypothesis that the 'smart' and the 'average' deal with stimuli in a different way. For example, because of 'smart' people is known to be able to recognize regularity faster, we may expect their plot to decline quite rapidly. But, because of their automation, we may also expect that they will find some difficulties to adapt when the pattern has changed. The adaptation difficulty may be represented (hopefully) by an increasing reaction time. I think to check every subject's plot, and determine their ARIMA(p,d,q) model. Then, if they all follow same ARIMA model, obviously my hypothesis is wrong, and I may conclude that there are no differences in the way subject deal with stimuli. But, if let's say 20 subject follow ARIMA(1,0,1) and the others follow ARIMA(0,0,2) I can say that there are differences in the subject's cognitive ability for dealing with stimuli. For now, I set aside what the differences is and its psychological explanation. The problem is, it will be very frustating to have 200 subjects and determine approriate ARIMA model for each plot. So, there're only 2 ways left. One, I reduce the number of subjects to manageable quantity, let's say 30 subjects. Or two, I know method(s) that make me possible to analyze multiple time series plot, and say whether they follow same or different model. Thank you, regards, Nathanael PS: please forgive grammatical error Prof Brian Ripley wrote: Surely you need the insight before choosing a package? What is the problem you are trying to solve? There are many different aspects of time series which could be of interest, and we have no idea which are relevant to your problem. On Sun, 30 Mar 2008, nathan3073 wrote: Dear All, I need to compare hundreds (about 200-300) of time series. Would anyone tell me how to do this in R? If R has no package for doing this, can I get some insight what method I should use? best regards, Nathanael Gratias -- View this message in context: http://www.nabble.com/Comparing-Time-Series-tp16392632p16392632.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. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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/Comparing-Time-Series-tp16392632p16396039.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] unexpected GAM result - at least for me!
Thanks Duncan. Yes i do have variation in the lidar metrics (be, ch, crr, and home) although i have a quite high correlation between ch and home. But even if i eliminate one metric (either ch or home) i end up with a deviation of 99.99. The species has values of 0 and 1 since i try to predict presence / absence. Do you think it is still a valid result? Thanks again, Monica Date: Mon, 31 Mar 2008 08:47:48 -0400 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] CC: r-help@r-project.org Subject: Re: [R] unexpected GAM result - at least for me! On 3/31/2008 8:34 AM, Monica Pisica wrote: Hi I am afraid i am not understanding something very fundamental and does not matter how much i am looking into the book Generalized Additive Models of S. Wood i still don't understand my result.I am trying to model presence / absence (presence = 1, absence = 0) of a species using some lidar metrics (i have 4 of these). I am using different models and such and when i used gam i got this very weird (for me) result which i thought it is not possible - or i have no idea how to interpret it.can3.gam - gam(can0~s(be)+s(crr)+s(ch)+s(home), family = 'binomial') summary(can3.gam) Family: binomial Link function: logit Formula: can 0 ~ s(be) + s(crr) + s(ch) + s(home) Parametric coefficien! ts: Estimate Std. Error z value Pr(|z|) (Intercept) 85.39 162.88 0.524 0.6 Approximate significance of smooth terms: edf Est.rank Chi.sq p-value s(be) 1.000 1 0.100 0.751 s(crr) 3.929 8 0.380 1.000 s(ch) 6.820 9 0.396 1.000 s(home) 1.000 1 0.314 0.575 R-sq.(adj) = 1 Deviance explained = 100% UBRE score = -0.81413 Scale est. = 1 n = 148Is this a perfect fit with no statistical significance, an over-estimating or what It seems that the significance of the smooths terms is null. Of course with such a model i predict perfectly presence / absence of species.Again, i hope you don't mind i'm asking you this. Any explanation will be very much appreciated. Look at the data. You can get a perfect fit to a logistic regression model fairly easily, and it looks as though you've got one. (In fact, the huge intercept suggests that all predictions will be 1. Do you actually have any variation in the data?) Duncan Murdoch _ esh_realtime_042008 [[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] Reorder the x-axis using lattice
Is there a way to reorder the xaxis using lattice. Using the
following data, the x-axis is ordered as BP GH MH PF RE RP SF VT but
I would like the x-axis to be ordered as PF RP BP GH VT SF RE MH.
Kön Skalor Tillfälle Medelvärde
1 Kvinnor BP 1-inskrivning 36.45283
2 Kvinnor GH 1-inskrivning 38.62255
3 Kvinnor MH 1-inskrivning 62.9
4 Kvinnor PF 1-inskrivning 39.80710
bwplot(Medelvärde ~ Skalor| Kön , kt, panel = panel.superpose,
groups = Tillfälle,scales = list(x = list(rot = 45),cex=0.7,
alternating=2),
...
When the x-axis is a factor, bwplot plots in order of the factor levels.
You'll need to explicitly set the levels in your data frame:
Skalor - factor(c(BP, GH, MH, PF, RE, RP, SF, VT, BP,
GH, MH, PF, RE, RP, SF, VT), levels=c(PF, RP, BP,
GH, VT, SF, RE, MH))
kt - data.frame(Kon=Kon, Skalor=Skalor, ...)
bwplot(...) #as before
Regards,
Richie.
Mathematical Sciences Unit
HSL
[EMAIL PROTECTED] wrote on 31/03/2008 11:05:18:
Dear list,
5 Kvinnor RE 1-inskrivning 41.50943
6 Kvinnor RP 1-inskrivning 22.2
7 Kvinnor SF 1-inskrivning 59.19811
8 Kvinnor VT 1-inskrivning 34.84568
9 Kvinnor BP 2-utskrivning 43.14815
10 Kvinnor GH 2-utskrivning 44.11321
11 Kvinnor MH 2-utskrivning 77.2
12 Kvinnor PF 2-utskrivning 44.74280
13 Kvinnor RE 2-utskrivning 68.95425
14 Kvinnor RP 2-utskrivning 39.90385
15 Kvinnor SF 2-utskrivning 64.62264
16 Kvinnor VT 2-utskrivning 51.97531
panel.groups = panel.linejoin,lty=c(1:3),lwd=3,
col=c(steelblue,grey50,green4),
ylab = list(label = skalpoäng (0-100), cex = 0.8),
xlab = list(label = skalor, cex = 0.8),
key = list(lines =
Rows(list(col=c(steelblue,grey50,green4),lty=c(1:3)),
c(1:3, 0)),cex=0.8,text = list(lab = as.
character(unique(kt$Tillfälle))),
columns = 2, title = SF-36: Skalpoäng för
respektive kön vid 3 mättillfälle ,
cex.title=0.9))
Thanks in advance,
Tom
-
Jämför pris på flygbiljetter och hotellrum: http://shopping.yahoo.
se/c-169901-resor-biljetter.html
[[alternative HTML version deleted]]
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Re: [R] Comparing Time Series
You could look at the dtw package. If x and y are two time series, dtw(x, y)$distance returns the distance between them in the sense discussed in the package. On Mon, Mar 31, 2008 at 8:37 AM, nathan3073 [EMAIL PROTECTED] wrote: It is a reaction time experiment. My program will display 25 circles, arranged in a 5x5 square. The program will light one circle at a time, which the subject should click ASAP. The time interval between two lighting is 1.5s. If the subject fails to click in approriate time, the data will be considered missing. The one I want to explore is the relationship between intellegence and the ability to recognize visual stimuli regularity. I arrange the experiment so there will be 4 repeated pattern. At last, after all pattern have been displayed, my program will light the circle randomly. I want to check the time series plots, in order to clarify my hypothesis that the 'smart' and the 'average' deal with stimuli in a different way. For example, because of 'smart' people is known to be able to recognize regularity faster, we may expect their plot to decline quite rapidly. But, because of their automation, we may also expect that they will find some difficulties to adapt when the pattern has changed. The adaptation difficulty may be represented (hopefully) by an increasing reaction time. I think to check every subject's plot, and determine their ARIMA(p,d,q) model. Then, if they all follow same ARIMA model, obviously my hypothesis is wrong, and I may conclude that there are no differences in the way subject deal with stimuli. But, if let's say 20 subject follow ARIMA(1,0,1) and the others follow ARIMA(0,0,2) I can say that there are differences in the subject's cognitive ability for dealing with stimuli. For now, I set aside what the differences is and its psychological explanation. The problem is, it will be very frustating to have 200 subjects and determine approriate ARIMA model for each plot. So, there're only 2 ways left. One, I reduce the number of subjects to manageable quantity, let's say 30 subjects. Or two, I know method(s) that make me possible to analyze multiple time series plot, and say whether they follow same or different model. Thank you, regards, Nathanael PS: please forgive grammatical error Prof Brian Ripley wrote: Surely you need the insight before choosing a package? What is the problem you are trying to solve? There are many different aspects of time series which could be of interest, and we have no idea which are relevant to your problem. On Sun, 30 Mar 2008, nathan3073 wrote: Dear All, I need to compare hundreds (about 200-300) of time series. Would anyone tell me how to do this in R? If R has no package for doing this, can I get some insight what method I should use? best regards, Nathanael Gratias -- View this message in context: http://www.nabble.com/Comparing-Time-Series-tp16392632p16392632.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. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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/Comparing-Time-Series-tp16392632p16396039.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] Comparing Time Series
Nathanael, On 31 Mar 2008, at 14:37, nathan3073 wrote: It is a reaction time experiment. My program will display 25 circles, arranged in a 5x5 square. The program will light one circle at a time, which the subject should click ASAP. The time interval between two lighting is 1.5s. If the subject fails to click in approriate time, the data will be considered missing. The one I want to explore is the relationship between intellegence and the ability to recognize visual stimuli regularity. I arrange the experiment so there will be 4 repeated pattern. At last, after all pattern have been displayed, my program will light the circle randomly. I want to check the time series plots, in order to clarify my hypothesis that the 'smart' and the 'average' deal with stimuli in a different way. For example, because of 'smart' people is known to be able to recognize regularity faster, we may expect their plot to decline quite rapidly. But, because of their automation, we may also expect that they will find some difficulties to adapt when the pattern has changed. The adaptation difficulty may be represented (hopefully) by an increasing reaction time. How important is the time series aspect here? Why not just do some anova on the difference between these blocks of trials? If the time information is essential, repeated measures anova could be applied. I think to check every subject's plot, and determine their ARIMA (p,d,q) model. Then, if they all follow same ARIMA model, obviously my hypothesis is wrong, and I may conclude that there are no differences in the way subject deal with stimuli. But, if let's say 20 subject follow ARIMA(1,0,1) and the others follow ARIMA(0,0,2) I can say that there are differences in the subject's cognitive ability for dealing with stimuli. For now, I set aside what the differences is and its psychological explanation. This sounds like you need a mixture of ARIMA models. Is that indeed the hypothesis that you want to test: whether there are different strategies that result in either an AR learning process or in an MA learning process? The problem is, it will be very frustating to have 200 subjects and Who said doing research was going to be fun (-; determine approriate ARIMA model for each plot. So, there're only 2 ways left. One, I reduce the number of subjects to manageable quantity, let's say 30 subjects. Or two, I know method(s) that make me possible to analyze multiple time series plot, and say whether they follow same or different model. Using a mixture of ARIMA models would allow you to analyze these data simultaneously. hth, Ingmar Thank you, regards, Nathanael PS: please forgive grammatical error Prof Brian Ripley wrote: Surely you need the insight before choosing a package? What is the problem you are trying to solve? There are many different aspects of time series which could be of interest, and we have no idea which are relevant to your problem. On Sun, 30 Mar 2008, nathan3073 wrote: Dear All, I need to compare hundreds (about 200-300) of time series. Would anyone tell me how to do this in R? If R has no package for doing this, can I get some insight what method I should use? best regards, Nathanael Gratias -- View this message in context: http://www.nabble.com/Comparing-Time-Series-tp16392632p16392632.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. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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/Comparing-Time- Series-tp16392632p16396039.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. Ingmar Visser Department of Psychology, University of Amsterdam Roetersstraat 15 1018 WB Amsterdam The Netherlands t: +31-20-5256723 [[alternative HTML version deleted]]
Re: [R] lmer function.
On Mon, Mar 31, 2008 at 7:13 AM, Boikanyo Makubate [EMAIL PROTECTED] wrote: I am using the lmer function from the lme4 package. I wrote the following statement, specifying the method to be adaptive Gaussian quadrature. I am getting an error saying method = AGQ not yet implemented for supernodal representation. Please help. fit - lmer(response~beta1+(1|patient),family=binomial,method=AGQ,data=data.2) As the message indicates, the AGQ method has not yet been implemented. It is very tricky to implement the AGQ method. For one thing it would be infeasible for more than one grouping factor for the random effects if these were non-nested. The Laplacian method is the default method in the development version of the lme4 package, available as install.packages(lme4, repos = http://r-forge.r-project.org;) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] unexpected GAM result - at least for me!
On 3/31/2008 9:01 AM, Monica Pisica wrote: Thanks Duncan. Yes i do have variation in the lidar metrics (be, ch, crr, and home) although i have a quite high correlation between ch and home. But even if i eliminate one metric (either ch or home) i end up with a deviation of 99.99. The species has values of 0 and 1 since i try to predict presence / absence. Do you think it is still a valid result? I repeat: look at the data. Compare the observed and predicted. That's the only way to know whether this is reasonable or not. If you're getting reasonable predictions, then it's a valid fit. (The tests and approximations used in the reported p-values may not be at all valid. I don't know what the requirements are for those in a GAM, but if you're getting a perfect fit, then they probably aren't being met.) Duncan Murdoch Thanks again, Monica Date: Mon, 31 Mar 2008 08:47:48 -0400 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] CC: r-help@r-project.org Subject: Re: [R] unexpected GAM result - at least for me! On 3/31/2008 8:34 AM, Monica Pisica wrote: Hi I am afraid i am not understanding something very fundamental and does not matter how much i am looking into the book Generalized Additive Models of S. Wood i still don't understand my result. I am trying to model presence / absence (presence = 1, absence = 0) of a species using some lidar metrics (i have 4 of these). I am using different models and such and when i used gam i got this very weird (for me) result which i thought it is not possible - or i have no idea how to interpret it. can3.gam - gam(can0~s(be)+s(crr)+s(ch)+s(home), family = 'binomial') summary(can3.gam) Family: binomial Link function: logit Formula: can 0 ~ s(be) + s(crr) + s(ch) + s(home) Parametric coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 85.39 162.88 0.524 0.6 Approximate significance of smooth terms: edf Est.rank Chi.sq p-value s(be) 1.000 1 0.100 0.751 s(crr) 3.929 8 0.380 1.000 s(ch) 6.820 9 0.396 1.000 s(home) 1.000 1 0.314 0.575 R-sq.(adj) = 1 Deviance explained = 100% UBRE score = -0.81413 Scale est. = 1 n = 148 Is this a perfect fit with no statistical significance, an over-estimating or what It seems that the significance of the smooths terms is null. Of course with such a model i predict perfectly presence / absence of species. Again, i hope you don't mind i'm asking you this. Any explanation will be very much appreciated. Look at the data. You can get a perfect fit to a logistic regression model fairly easily, and it looks as though you've got one. (In fact, the huge intercept suggests that all predictions will be 1. Do you actually have any variation in the data?) Duncan Murdoch In a rush? Get real-time answers with Windows Live Messenger. http://www.windowslive.com/messenger/overview.html?ocid=TXT_TAGLM_WL_Refresh_realtime_042008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] convert delimited string to vector
Hi R Users, Simple question: How might I convert the text a, b, c (or for that matter a b c with any delimiter - space, comma, etc.) into a 3-element character vector? [1] a b c Thanks, Brad __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Number of variables in lrm (Design)
Paul Sweeting wrote: Hi I'm sure this is straightforward, but I can't find an answer. I'm trying to find the function that returns the number of dependent variables/rank of the regression/degrees of freedom/something similar. Any ideas? Thanks! Paul Design does not allow for singularities. To get the regression d.f. you can use length(coef(fit))-num.intercepts(fit) or just fit$stats['d.f.'] Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Retrieving locations of element of a vector or matrix
Hello all, I have a problem of getting the locations of elements in a vector or matrix. I know one can use the command which.min or which.max to get the location of minimum or maximum values in a vector. But is there any command to get the first k minimum values or maximum values in a vector. Thanks - You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost. [[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] L-BFGS-B needs finite values of 'fn'
try something like this before wrapping up your function else i guess u'll
have to stick to Prof Brian Ripley suggestion his suggestions are usually
best bet .
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
if(!is.finite(r))
r-1e+20 return(r)
}
have a nice day.
On Mon, 31 Mar 2008 12:24:09 +0100 Paul Smith wrote:
Dear All,
I am trying to solve the optimization problem below, but I am always
getting the following error:
Error in optim(rep(20, nvar), f, gr, method = L-BFGS-B, lower = rep(0, :
L-BFGS-B needs finite values of 'fn'
Any ideas?
Thanks in advance,
Paul
-! --
k - 1
b - 0.3
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
return(r)
}
gr - function(x) {
n - length(x)
r - (b^(0:(n-1)))*(1/x) - 400*(sum(x)-k)
return(r)
}
nvar - 10
(sols -
optim(rep(20,nvar),f,gr,method=L-BFGS-B,lower=rep(0,nvar),upper=rep(k,nvar
),control=list(fnscale=-1,parscale=rep(2000,nvar),factr=1e-300,pgtol=1e-300)
))
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert delimited string to vector
On 3/31/2008 9:43 AM, Brad Christoffersen wrote: Hi R Users, Simple question: How might I convert the text a, b, c (or for that matter a b c with any delimiter - space, comma, etc.) into a 3-element character vector? [1] a b c Thanks, Brad unlist(strsplit(a b c, split= )) [1] a b c unlist(strsplit(a, b, c, split=, )) [1] a b c ?strsplit __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. (www.ndri.org) 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Defining reference category for a cph model summary inside of a for loop
Frank,
Thanks again, I didn't realize that continuous variables could be
manipulated that way inside of the summary function.
I realize that my code was kind of confusing.
The variables A...F are all categorical variables. They each have
four levels named 1st Quartile4th Quartile
I tried the code below with the same result.
print(summary(f, eval(parse(text=paste(i,='1st Quartile', sep='')
In the output, the reference category is different for each of the
variables.
Brian
-Original Message-
From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED]
Sent: Sunday, March 30, 2008 9:14 AM
To: Wells, Brian
Cc: r-help@r-project.org
Subject: Re: [R] Defining reference category for a cph model summary
inside of a for loop
Wells, Brian wrote:
Dr. Harrell,
Thanks for you help.
I tried:
print(summary(f,parse(text=paste(i,'=1st Quartile', sep=''
Same result. No error, the reference category simply doesn't change.
That's good, because the default in summary is to compare the outer
quartiles for a continuous variable. And as I said before the string
'1st Quartile' has no special meaning for R or Design.
Get what you are trying to do to work without parse (and you'll need
eval() with parse) first. When you want total control over a setting,
say getting a hazard ratio for the .2 to the .8 quantile, do something
like
summary(f, age=quantile(age,c(.2,.8),na.rm=TRUE))
Frank
Brian
-Original Message-
From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2008 8:34 PM
To: Wells, Brian
Cc: r-help@r-project.org
Subject: Re: [R] Defining reference category for a cph model summary
inside of a for loop
Wells, Brian wrote:
I have the following code.
f - cph(formula = Surv(TimeToDeath, Dead == Yes)
~1,data=single.dat, x=T, y=T, surv=T)
for(i in c('A', 'B', 'C', 'D', 'E', 'F')){
f -update(f,as.formula(paste('Surv(TimeToDeath, Dead ==
Yes)~',i,sep='')))
print(summary(f, paste(i,=1st Quartile, sep='')))
There is no error message generated in R, but R ignores the reference
category defined with paste in the summary function for the cph
model.
The output uses the 1st Quartile as the reference category to
calculate hazards for some of the variables defined by i, but not all
of
them.
Your code is confusing. What is to the right of ~ in a formula is a
predictor variable name, not a value. If your variables are named A,
B,
C, ... you are OK.
'1st Quartile' has no special meaning to R or Design, and you can't
pass
a character string as a second argument to summary and expect it to
work.
You will need parse(text=paste(...)) to create an appropriate
expression.
But Design gives you inter-quartile range hazard ratios by default
anyway.
Beware of getting hazard ratios that are not adjusted for other
variables needed in the model.
Frank Harrell
Any help would be greatly appreciated.
thanks
Brian J. Wells, MD, MS
Research Associate
Quantitative Health Sciences
Cleveland Clinic
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt
University
P Please consider the environment before printing this e-mail
Cleveland Clinic is ranked one of the top hospitals
in America by U.S. News World Report (2007).
Visit us online at http://www.clevelandclinic.org for
a complete listing of our services, staff and
locations.
Confidentiality Note: This message is intended for use\...{{dropped:13}}
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert delimited string to vector
Or: scan(textConnection(a, b, c), what=character, sep=,) On 31/03/2008, Brad Christoffersen [EMAIL PROTECTED] wrote: Hi R Users, Simple question: How might I convert the text a, b, c (or for that matter a b c with any delimiter - space, comma, etc.) into a 3-element character vector? [1] a b c Thanks, Brad __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Retrieving locations of element of a vector or matrix
I think that you can use: x - sample(1:20, 10) head(sort(x), 5) #First 5 minimum values head(sort(x, decreasing = TRUE), 5) #First 5 maximum values On 31/03/2008, Yemi Oyeyemi [EMAIL PROTECTED] wrote: Hello all, I have a problem of getting the locations of elements in a vector or matrix. I know one can use the command which.min or which.max to get the location of minimum or maximum values in a vector. But is there any command to get the first k minimum values or maximum values in a vector. Thanks - You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost. [[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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Retrieving locations of element of a vector or matrix
For the position use order indeed of sort On 31/03/2008, Henrique Dallazuanna [EMAIL PROTECTED] wrote: I think that you can use: x - sample(1:20, 10) head(sort(x), 5) #First 5 minimum values head(sort(x, decreasing = TRUE), 5) #First 5 maximum values On 31/03/2008, Yemi Oyeyemi [EMAIL PROTECTED] wrote: Hello all, I have a problem of getting the locations of elements in a vector or matrix. I know one can use the command which.min or which.max to get the location of minimum or maximum values in a vector. But is there any command to get the first k minimum values or maximum values in a vector. Thanks - You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost. [[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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Defining reference category for a cph model summary inside of a for loop
Wells, Brian wrote:
Frank,
Thanks again, I didn't realize that continuous variables could be
manipulated that way inside of the summary function.
I realize that my code was kind of confusing.
The variables A...F are all categorical variables. They each have
four levels named 1st Quartile4th Quartile
I tried the code below with the same result.
print(summary(f, eval(parse(text=paste(i,='1st Quartile', sep='')
In the output, the reference category is different for each of the
variables.
Brian
Thanks for clarifying. That approach will NOT provide estimates at the
quartiles. For example a hazard ratio for the upper quartile category
to the lower quartile category will estimate the ratio of hazards when
XQ3 to when XQ1 where outer quartiles are Q1 and Q3. This represents
a hazard ratio of an unknown mixture of distributions and will not
transport to another sample with a different mixture.
In addition you will have serious residual confounding with that
approach by not adjusting for all the information in continuous predictors.
Frank
-Original Message-
From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED]
Sent: Sunday, March 30, 2008 9:14 AM
To: Wells, Brian
Cc: r-help@r-project.org
Subject: Re: [R] Defining reference category for a cph model summary
inside of a for loop
Wells, Brian wrote:
Dr. Harrell,
Thanks for you help.
I tried:
print(summary(f,parse(text=paste(i,'=1st Quartile', sep=''
Same result. No error, the reference category simply doesn't change.
That's good, because the default in summary is to compare the outer
quartiles for a continuous variable. And as I said before the string
'1st Quartile' has no special meaning for R or Design.
Get what you are trying to do to work without parse (and you'll need
eval() with parse) first. When you want total control over a setting,
say getting a hazard ratio for the .2 to the .8 quantile, do something
like
summary(f, age=quantile(age,c(.2,.8),na.rm=TRUE))
Frank
Brian
-Original Message-
From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2008 8:34 PM
To: Wells, Brian
Cc: r-help@r-project.org
Subject: Re: [R] Defining reference category for a cph model summary
inside of a for loop
Wells, Brian wrote:
I have the following code.
f - cph(formula = Surv(TimeToDeath, Dead == Yes)
~1,data=single.dat, x=T, y=T, surv=T)
for(i in c('A', 'B', 'C', 'D', 'E', 'F')){
f -update(f,as.formula(paste('Surv(TimeToDeath, Dead ==
Yes)~',i,sep='')))
print(summary(f, paste(i,=1st Quartile, sep='')))
There is no error message generated in R, but R ignores the reference
category defined with paste in the summary function for the cph
model.
The output uses the 1st Quartile as the reference category to
calculate hazards for some of the variables defined by i, but not all
of
them.
Your code is confusing. What is to the right of ~ in a formula is a
predictor variable name, not a value. If your variables are named A,
B,
C, ... you are OK.
'1st Quartile' has no special meaning to R or Design, and you can't
pass
a character string as a second argument to summary and expect it to
work.
You will need parse(text=paste(...)) to create an appropriate
expression.
But Design gives you inter-quartile range hazard ratios by default
anyway.
Beware of getting hazard ratios that are not adjusted for other
variables needed in the model.
Frank Harrell
Any help would be greatly appreciated.
thanks
Brian J. Wells, MD, MS
Research Associate
Quantitative Health Sciences
Cleveland Clinic
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
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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] Memory in R.
Hi, I am pretty new in R. In fact, I started using R last month. So, be indulgent if my questions seem too boring or obvious. I have 2 basic questions: 1 - I want to increase memory, but I can't find the right way. E.g: in stata, you just type set memory 1g. When I try to load huge datasets, R crashes. 2 - I want to clear the memory (no object or data), such as the clear command in stata. Best regards. Amadou DIALLO. PhD student. Cerdi, University of Auvergne France. [[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] concatenating two successive time series
Dear Helpers, I am looking for methods and tools to compare and then to concatenate two successive time series. They are both in the same frequency and they describe one phenomena. There is no time gap between them. The problem is that the method of measurements has changed between both time series and they are no statistically the same. I would like to merge them to receive one homogeneous time series. Here is an example of time series ts1 (1982-1999) and ts2 (2000-2006). Zeros are NA's. ts1 - structure(c(100, 48, 31, 10, 16, 25, 56, 81, 51, 45, 73, 112, 103, 60, 30, 21, 10, 26, 60, 55, 40, 46, 78, 125, 55, 21, 22, 11, 13, 20, 38, 25, 17, 27, 49, 68, 40, 29, 44, 10, 21, 28, 58, 71, 50, 68, 88, 38, 71, 37, 51, 11, 30, 30, 52, 59, 40, 47, 91, 95, 62, 31, 44, 12, 27, 35, 36, 69, 32, 50, 81, 88, 72, 47, 28, 20, 9, 15, 34, 38, 10, 18, 56, 124, 79, 26, 29, 9, 16, 21, 71, 74, 52, 49, 80, 84, 52, 36, 37, 12, 21, 23, 70, 75, 69, 69, 81, 66, 53, 26, 18, 12, 12, 11, 45, 60, 40, 39, 64, 80, 72, 30, 23, 9, 18, 22, 53, 77, 45, 31, 54, 59, 26, 16, 14, 4, 4, 12, 20, 21, 7, 13, 28, 56, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 50, 36, 17, 18, 12, 28, 99, 90, 72, 69, 98, 78, 56, 31, 48, 11, 15, 31, 70, 79, 67, 80, 107, 76, 43, 15, 29, 13, 17, 30, 70, 86, 52, 35, 45, 93, 80, 49, 29, 16, 12, 23, 68, 80, 38, 52, 86, 94, 67, 30, 14, 10, 14, 21, 35, 49, 33, 20, 39, 77), .Tsp = c(1982, 1999.917, 12), class = ts) ts2 - structure(c(0, 0, 0, 220, 72, 45, 45, 56, 60, 80, 81, 99, 97, 92, 220, 141, 85, 40, 74, 115, 88, 97, 127, 107, 90, 86, 185, 193, 100, 43, 45, 73, 74, 95, 87, 107, 77, 63, 163, 123, 106, 35, 102, 122, 103, 96, 104, 140, 135, 87, 215, 173, 56, 38, 80, 90, 113, 72, 86, 95, 88, 93, 198, 128, 40, 28, 76, 133, 80, 89, 72, 108, 102, 86, 199, 204, 84, 60, 49, 103, 121, 98, 108, 165 ), .Tsp = c(2000, 2006.917, 12), class = ts) When we plot them together the difference is visible: ts12 - ts(c(ts1,ts2),start=1982,frequency=12) plot(ts12) but plot(decompose(ts12)) confirm that both describe one phenomena. I have dozens of such pairs and I would like to find a method to recalculate one ts to the other to make one homogeneous ts. Do you have any suggestions how to statistically concatenate those two ts or zoo (I tried this package) objects? Thanks in advance for the answer and the precious time you might waste. Regards, Jedrzej PS. Włącz radio w Twoim komputerze! Muzyka idealna do pracy i do zabawy! - Kliknij: http://klik.wp.pl/?adr=http%3A%2F%2Fcorto.www.wp.pl%2Fas%2Fradiokosci.htmlsid=291 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Cannot find new threads: generic error when require(RODBC) with -d gdb
Hi, We have the error below. Any ideas ? Regards, Matt $ R --vanilla -d gdb GNU gdb 6.7.1-debian Copyright (C) 2007 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later http://gnu.org/licenses/gpl.html This is free software: you are free to change and redistribute it. There is NO WARRANTY, to the extent permitted by law. Type show copying and show warranty for details. This GDB was configured as x86_64-linux-gnu... Using host libthread_db library /lib/libthread_db.so.1. (gdb) run Starting program: /misc/local/R-2.4.0/bin/exec/R --vanilla R version 2.4.0 (2006-10-03) Copyright (C) 2006 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. require(RODBC) Loading required package: RODBC [Thread debugging using libthread_db enabled] Error while reading shared library symbols: Cannot find new threads: generic error Cannot find new threads: generic error (gdb) This message and any files attached are intended solely for the use of the individual or the entity to whom they are addressed and may contain information that is confidential. Please notify the sender by e-mail if you are not the intended recipient and if you are not the intended recipient you are hereby notified that any dissemination, distribution or copying of this message, any attachments or their content is prohibited. To the extent this message contains information regarding any of the investment funds managed by Concordia, you should be aware that such funds may only be offered pursuant to the relevant private placement or information memorandum. Any information contained in this message must be reviewed and considered in connection with, and no one is authorized to make any representations or warranties or provide information which is inconsistent with, or in addition to the relevant private placement or information memorandum. To the extent this message contains any performance information, you should be aware that past performance is not necessarily indicative of future results. [[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] concatenating two successive time series
It might just be a matter of scaling: ts12s - ts(c(scale(ts1), scale(ts2)), start = 1982, frequency = 12) plot(ts12s) plot(decompose(ts12s)) 2008/3/31 Jędrzej Bojanowski [EMAIL PROTECTED]: Dear Helpers, I am looking for methods and tools to compare and then to concatenate two successive time series. They are both in the same frequency and they describe one phenomena. There is no time gap between them. The problem is that the method of measurements has changed between both time series and they are no statistically the same. I would like to merge them to receive one homogeneous time series. Here is an example of time series ts1 (1982-1999) and ts2 (2000-2006). Zeros are NA's. ts1 - structure(c(100, 48, 31, 10, 16, 25, 56, 81, 51, 45, 73, 112, 103, 60, 30, 21, 10, 26, 60, 55, 40, 46, 78, 125, 55, 21, 22, 11, 13, 20, 38, 25, 17, 27, 49, 68, 40, 29, 44, 10, 21, 28, 58, 71, 50, 68, 88, 38, 71, 37, 51, 11, 30, 30, 52, 59, 40, 47, 91, 95, 62, 31, 44, 12, 27, 35, 36, 69, 32, 50, 81, 88, 72, 47, 28, 20, 9, 15, 34, 38, 10, 18, 56, 124, 79, 26, 29, 9, 16, 21, 71, 74, 52, 49, 80, 84, 52, 36, 37, 12, 21, 23, 70, 75, 69, 69, 81, 66, 53, 26, 18, 12, 12, 11, 45, 60, 40, 39, 64, 80, 72, 30, 23, 9, 18, 22, 53, 77, 45, 31, 54, 59, 26, 16, 14, 4, 4, 12, 20, 21, 7, 13, 28, 56, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 50, 36, 17, 18, 12, 28, 99, 90, 72, 69, 98, 78, 56, 31, 48, 11, 15, 31, 70, 79, 67, 80, 107, 76, 43, 15, 29, 13, 17, 30, 70, 86, 52, 35, 45, 93, 80, 49, 29, 16, 12, 23, 68, 80, 38, 52, 86, 94, 67, 30, 14, 10, 14, 21, 35, 49, 33, 20, 39, 77), .Tsp = c(1982, 1999.917, 12), class = ts) ts2 - structure(c(0, 0, 0, 220, 72, 45, 45, 56, 60, 80, 81, 99, 97, 92, 220, 141, 85, 40, 74, 115, 88, 97, 127, 107, 90, 86, 185, 193, 100, 43, 45, 73, 74, 95, 87, 107, 77, 63, 163, 123, 106, 35, 102, 122, 103, 96, 104, 140, 135, 87, 215, 173, 56, 38, 80, 90, 113, 72, 86, 95, 88, 93, 198, 128, 40, 28, 76, 133, 80, 89, 72, 108, 102, 86, 199, 204, 84, 60, 49, 103, 121, 98, 108, 165 ), .Tsp = c(2000, 2006.917, 12), class = ts) When we plot them together the difference is visible: ts12 - ts(c(ts1,ts2),start=1982,frequency=12) plot(ts12) but plot(decompose(ts12)) confirm that both describe one phenomena. I have dozens of such pairs and I would like to find a method to recalculate one ts to the other to make one homogeneous ts. Do you have any suggestions how to statistically concatenate those two ts or zoo (I tried this package) objects? Thanks in advance for the answer and the precious time you might waste. Regards, Jedrzej PS. Włącz radio w Twoim komputerze! Muzyka idealna do pracy i do zabawy! - Kliknij: http://klik.wp.pl/?adr=http%3A%2F%2Fcorto.www.wp.pl%2Fas%2Fradiokosci.htmlsid=291 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to extract terms from aov
I couldn't find clues from the achieve of R help that allow to retrieve the terms of aov. av2- summary(aov(Prevalence~Random*Bednet*Flight)); av2 Df Sum Sq Mean Sq F value Pr(F) Targeted2 2.11e+10 1.05e+10 102.90 2e-16 *** Bednet 2 9.48e+09 4.74e+09 46.35 1.2e-13 *** Flight 1 1.08e+09 1.08e+09 10.55 0.0018 ** Targeted:Bednet 4 6.32e+09 1.58e+09 15.45 3.6e-09 *** Targeted:Flight 2 7.19e+08 3.59e+083.51 0.0350 * Bednet:Flight 2 5.12e+08 2.56e+082.50 0.0889 . Targeted:Bednet:Flight 4 3.41e+08 8.53e+070.83 0.5082 Residuals 72 7.37e+09 1.02e+08 I would like to extract the first unlabeled column: terms like Targeted, Bednet... for producing a table, and I know how to extract the others such as Df, F... Help is appreciated Weidong Gu, Department of Medicine University of Alabama, Birmingham 1900 University Blvd., Birmingham, Alabama 35294 Email: [EMAIL PROTECTED] PH: (205)-975-9053 [[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] how to extract terms from aov
you can access it as a data frame by av3-av2[[1]] then, dimnames(av3)[[1]] gives you the leftmost labels, and also, av3$Df av3$Sum etc, gives the values. str(object) always help you to find out the structure of object and how to access the information. HTH I couldn't find clues from the achieve of R help that allow to retrieve the terms of aov. av2- summary(aov(Prevalence~Random*Bednet*Flight)); av2 Df Sum Sq Mean Sq F value Pr(F) Targeted2 2.11e+10 1.05e+10 102.90 2e-16 *** Bednet 2 9.48e+09 4.74e+09 46.35 1.2e-13 *** Flight 1 1.08e+09 1.08e+09 10.55 0.0018 ** Targeted:Bednet 4 6.32e+09 1.58e+09 15.45 3.6e-09 *** Targeted:Flight 2 7.19e+08 3.59e+083.51 0.0350 * Bednet:Flight 2 5.12e+08 2.56e+082.50 0.0889 . Targeted:Bednet:Flight 4 3.41e+08 8.53e+070.83 0.5082 Residuals 72 7.37e+09 1.02e+08 I would like to extract the first unlabeled column: terms like Targeted, Bednet... for producing a table, and I know how to extract the others such as Df, F... Help is appreciated Weidong Gu, Department of Medicine University of Alabama, Birmingham 1900 University Blvd., Birmingham, Alabama 35294 Email: [EMAIL PROTECTED] PH: (205)-975-9053 [[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] Journal for R
Yes. Try: Tutorials in Quantitative Methods for Psychology http://www.tqmp.org/ (I am on the Editorial Board). On 31-Mar-08, at 4:00 AM, [EMAIL PROTECTED] wrote: Hi the list I made up a new statistical procedure. I will publish it in a medical journal, but there will be only the way of using it, no calculation or algorithme detail. So is there a journal (I mean scientific journal) with selection commity to submit an article describing the detail of a package? Christophe -- Please avoid sending me Word or PowerPoint attachments. See http://www.gnu.org/philosophy/no-word-attachments.html -Dr. John R. Vokey __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] operators %/% bug?
Dear all, I have come across a strange behavior of the %/% operator, and I wasn't sure if this was intended. i.e. a - 1.2 b - 1.0 (a-b) %/% 0.1 results in 1? Whereas 0.2 %/% 0.1 results in 2. Am I missing something obvious here? It seems to be consistent up to R 2.6.0 running in WinXP (haven't upgraded yet). Best, Ken This email message may contain legally privileged and/or confidential information. If you are not the intended recipient(s), or the employee or agent responsible for the delivery of this message to the intended recipient(s), you are hereby notified that any disclosure, copying, distribution, or use of this email message is prohibited. If you have received this message in error, please notify the sender immediately by e-mail and delete this email message from your computer. Thank you. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] operators %/% bug?
FAQ 7.31 a-b != 0.2 a - 1.2 b - 1.0 a-b-.2 [1] -5.551115e-17 On 3/31/08, Lo, Ken [EMAIL PROTECTED] wrote: Dear all, I have come across a strange behavior of the %/% operator, and I wasn't sure if this was intended. i.e. a - 1.2 b - 1.0 (a-b) %/% 0.1 results in 1? Whereas 0.2 %/% 0.1 results in 2. Am I missing something obvious here? It seems to be consistent up to R 2.6.0 running in WinXP (haven't upgraded yet). Best, Ken This email message may contain legally privileged and/or confidential information. If you are not the intended recipient(s), or the employee or agent responsible for the delivery of this message to the intended recipient(s), you are hereby notified that any disclosure, copying, distribution, or use of this email message is prohibited. If you have received this message in error, please notify the sender immediately by e-mail and delete this email message from your computer. Thank you. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] DE optimization with equality constraint
Paul Smith phhs80 at gmail.com writes: The problem with DEoptim approach is that is not guaranteed that it converges to the solution. Moreover, from my experience, it seems to be quite slow when the optimization problem is high-dimensional (i.e., with many variables). Paul There is a difference between local and global optimization: 'optim' realizes *local* optimization using a gradient-based approach. This is fast, but will get stuck in local optima (except method SANN). 'DEoptim' is one of many approaches to *global* optimization, of which each has its advantages and drawbacks. ...not guaranteed that it converges to the solution. As a local optimization routine, also 'optim' does not guarantee to reach a (global) optimum. [DEoptim] seems to be quite slow... This is normal for routines in global optimization as they have to search a quite large space. Hans Werner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Memory in R.
On 3/31/08, R RR [EMAIL PROTECTED] wrote: Hi, I am pretty new in R. In fact, I started using R last month. So, be indulgent if my questions seem too boring or obvious. I have 2 basic questions: 1 - I want to increase memory, but I can't find the right way. E.g: in stata, you just type set memory 1g. When I try to load huge datasets, R crashes. PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. What exactly are you doing? How much memory do you have on your system? What is hugh? All your R objects have to fit in physical memory; you don't want to page. So some more information is needed. 2 - I want to clear the memory (no object or data), such as the clear command in stata. ?rm Best regards. Amadou DIALLO. PhD student. Cerdi, University of Auvergne France. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] operators %/% bug?
Consider also the following: a - 1.2 b - 1.0 (a-b)%%.1 [1] 0.1 (a-b)%/%.1 [1] 1 .2%%.1 [1] 0 .2%/%.1 [1] 2 This performs as described on ?%/%. In particular, the second paragraph under Value: '%%' indicates 'x mod y' and '%/%' indicates integer division. It is guaranteed that 'x == (x %% y) + y * ( x %/% y )' (up to rounding error) unless 'y == 0' where the result is 'NA_integer_' or 'NaN' (depending on the 'typeof' of the arguments). See URL: http://en.wikipedia.org/wiki/Modulo_operation for the rationale. hope this helps. Spencer jim holtman wrote: FAQ 7.31 a-b != 0.2 a - 1.2 b - 1.0 a-b-.2 [1] -5.551115e-17 On 3/31/08, Lo, Ken [EMAIL PROTECTED] wrote: Dear all, I have come across a strange behavior of the %/% operator, and I wasn't sure if this was intended. i.e. a - 1.2 b - 1.0 (a-b) %/% 0.1 results in 1? Whereas 0.2 %/% 0.1 results in 2. Am I missing something obvious here? It seems to be consistent up to R 2.6.0 running in WinXP (haven't upgraded yet). Best, Ken This email message may contain legally privileged and/or confidential information. If you are not the intended recipient(s), or the employee or agent responsible for the delivery of this message to the intended recipient(s), you are hereby notified that any disclosure, copying, distribution, or use of this email message is prohibited. If you have received this message in error, please notify the sender immediately by e-mail and delete this email message from your computer. Thank you. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mean vs sum behavior
Dear all Could someone explain me why lapply(split(x,fac),mean) returns means per levels of fac for each column of x whereas lapply(split(x,fac),sum) does not return sums per level of fac and columns of x, but adds all columns together? Hence, how can I get sum to behave as mean in this instance? Thank you very much in advance E. Castella __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mean vs sum behavior
On Mon, 2008-03-31 at 18:41 +0200, Emmanuel Castella wrote: Dear all Could someone explain me why lapply(split(x,fac),mean) returns means per levels of fac for each column of x whereas lapply(split(x,fac),sum) does not return sums per level of fac and columns of x, but adds all columns together? Hence, how can I get sum to behave as mean in this instance? Thank you very much in advance E. Castella You didn't tell us what x is, but I suspect a data.frame. mean has a method for class data.frame, which returns the mean of each column. Sum doesn't have any methods and hence works by summing all the numbers. If you want to replicate the mean behaviour with sum, the following would suffice: fac - gl(4, 50) dat - data.frame(a = rnorm(200), b = rnorm(200), c = rnorm(200)) sp - split(dat, fac) then lapply(sp, function(x) sapply(x, sum)) or even quicker and easier: lapply(sp, colSums) HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] SiZer plots in R
Hello, I am a graduate student at UNC Chapel Hill, and I am attempting to create a SiZer plot for a nonparametric analysis. I have found the file to use this program in Matlab, however I was hoping to find a package to use this in R. Does anyone know of a package that can create this type of graph? Thanks, Stephanie You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ps or pdf
Hi everyone, I have been making a fair amount of figures in R recently that I've been touching up with Illustrator and I've found a difference between pdf and ps files and I was wondering if someone could enlighten me about them. While the figures look the same, the ps version tends to have truncated strings. The last character of short strings tends to be on a string of its own, located right beside the rest. This makes it a bit awkward to manipulate, especially if scaling is involved. Is there a reason for this differences? There also seems to be somewhat arbitrary grouping of the last column cells in heatmaps in ps files. I used to prefer the ps because they embed more easily in latex documents (although pdf are not difficult and conversions are trivial anyhow), but I'm curious if there are other reasons why one format might be preferred over the other in this context. This is with R 2.6 on linux, and I've seen this behavior with older R version also. Francois sessionInfo() R version 2.6.0 (2007-10-03) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=en_US.UTF-8;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] rcompgen_0.1-15 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] L-BFGS-B needs finite values of 'fn'
On Mon, Mar 31, 2008 at 2:57 PM, Zaihra T [EMAIL PROTECTED] wrote:
try something like this before wrapping up your function else i guess u'll
have to stick to Prof Brian Ripley suggestion his suggestions are usually
best bet .
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
if(!is.finite(r))
r-1e+20 return(r)
}
have a nice day.
On Mon, 31 Mar 2008 12:24:09 +0100 Paul Smith wrote:
Dear All,
I am trying to solve the optimization problem below, but I am always
getting the following error:
Error in optim(rep(20, nvar), f, gr, method = L-BFGS-B, lower = rep(0, :
L-BFGS-B needs finite values of 'fn'
Any ideas?
Thanks in advance,
Paul
-! --
k - 1
b - 0.3
f - function(x) {
n - length(x)
r - sum((b^(0:(n-1)))*log(x)) - 200*(sum(x)-k)^2
return(r)
}
gr - function(x) {
n - length(x)
r - (b^(0:(n-1)))*(1/x) - 400*(sum(x)-k)
return(r)
}
nvar - 10
(sols -
optim(rep(20,nvar),f,gr,method=L-BFGS-B,lower=rep(0,nvar),upper=rep(k,nvar),control=list(fnscale=-1,parscale=rep(2000,nvar),factr=1e-300,pgtol=1e-300)))
Not much progress, Zaihra. Unfortunately! I am wondering whether one
can transform the original problem into an equivalent one and solvable
with optim.
I know the analytical solution; I am just trying to check how far can
R go regarding optimization problems.
Paul
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Re: [R] ps or pdf
On Mon, 31 Mar 2008, Francois Pepin wrote: Hi everyone, I have been making a fair amount of figures in R recently that I've been touching up with Illustrator and I've found a difference between pdf and ps files and I was wondering if someone could enlighten me about them. While the figures look the same, the ps version tends to have truncated strings. The last character of short strings tends to be on a string of its own, located right beside the rest. This makes it a bit awkward to manipulate, especially if scaling is involved. Is there a reason for this differences? Please see the footer of this message. Neither postscript() nor pdf() graphics devices split up strings they are passed (by e.g. text()), so this is being done either by the code used to create the plot (and we have no idea what that is) or by the viewer. I suspect the problem is rather in the viewer, but without the example we asked for it is impossible to know. There also seems to be somewhat arbitrary grouping of the last column cells in heatmaps in ps files. Again, we need an example. I used to prefer the ps because they embed more easily in latex documents (although pdf are not difficult and conversions are trivial anyhow), but I'm curious if there are other reasons why one format might be preferred over the other in this context. The graphics devices are very similar (they share a lot of code). One small difference is that PostScript has an arc primitive, and PDF does not. This is with R 2.6 on linux, and I've seen this behavior with older R version also. Nothing has changed at that level for a long time -- not even in current versions of R (and 2.6.0 is obsolete). Francois sessionInfo() R version 2.6.0 (2007-10-03) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=en_US.UTF-8;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] rcompgen_0.1-15 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding device size-independent y=0 line to a lattice plot
Using the following lattice plot as an example, I would like to add horizontal lines where y=0: library(lattice) library(grid) fac - gl(4,12) x - letters[rep(1:3,16)] y - runif(48,min=0.0) dotplot(y~x|fac) I've tried it with grid.lines using npc and native units, which works fine unless I change the size of the output device - then the lines are in the wrong place. Is there a way to do this that is independent of the output device size? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding device size-independent y=0 line to a lattice plot
On 3/31/08, Levi Waldron [EMAIL PROTECTED] wrote:
Using the following lattice plot as an example, I would like to add
horizontal lines where y=0:
library(lattice)
library(grid)
fac - gl(4,12)
x - letters[rep(1:3,16)]
y - runif(48,min=0.0)
dotplot(y~x|fac)
I've tried it with grid.lines using npc and native units, which works
fine unless I change the size of the output device - then the lines
are in the wrong place. Is there a way to do this that is independent
of the output device size?
The obvious thing to try would be
dotplot(y~x|fac,
panel = function(...) {
panel.abline(h = 0)
panel.dotplot(...)
})
Does this not work?
-Deepayan
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Re: [R] ps or pdf
Prof Brian Ripley wrote:
Please see the footer of this message.
Sorry, here is an example. For some reason, I cannot reproduce it
without using actual gene names.
set.seed(1)
##The row names were originally obtained using the hgug4112a library
##from bioconductor. I set it manually for people who don't have it
##installed.
##library(hgug4112a);row-sample(na.omit(unlist(as.list(hgug4112aSYMBOL))),50)
row-c(BDNF, EMX2, ZNF207, HELLS, PWP1, PDXDC1, BTD,
NETO1, SLCO4C1, FZD7, NICN1, TMSB4Y, PSMB7, CADM2,
SIRT3, ADH6, TM6SF1, AARS, TMEM88, CP110, ADORA2A,
ATAD3A, VAPA, NXPH3, IL27RA, NEBL, FANCF, PTPRG,
HSU79275, CCDC34, EPDR1, FBLN1, PCAF, AP1B1, TXNRD2,
MUC20, MBNL1, STAU2, STK32C, PPIAL4, TGFBR2, DPY19L2P3,
TMEM50B, ENY2, MAN2A2, ZFYVE26, TECTA, CD55, LOC400794,
SLC19A3)
postscript('/tmp/heatmap.ps',paper='letter',horizontal=F)
heatmap(matrix(rnorm(2500),50),labRow=row)
dev.off()
Neither postscript() nor pdf()
graphics devices split up strings they are passed (by e.g. text()), so
this is being done either by the code used to create the plot (and we
have no idea what that is) or by the viewer. I suspect the problem is
rather in the viewer, but without the example we asked for it is
impossible to know.
Example of row names that are truncated in Illustrator (* denoting
truncation):
CCDC3*4 (2nd row)
MUC2*0 (3rd row)
MBNL*1 (8th row)
...
It is likely that Illustrator (CS 3, OS X version) is at fault. I do
not see any truncation if I look at the ps file by hand (lines 4801 and
4802):
540.22 545.88 (MUC20) 0 0 0 t
540.22 553.90 (CCDC34) 0 0 0 t
There also seems to be somewhat arbitrary grouping of the last column
cells in heatmaps in ps files.
Again, we need an example.
The top right cell (26, TXNRD2) is grouped with the cell just below it
(26, CCDC34). It's more of a curiosity than anything else.
I used to prefer the ps because they embed more easily in latex
documents (although pdf are not difficult and conversions are trivial
anyhow), but I'm curious if there are other reasons why one format might
be preferred over the other in this context.
The graphics devices are very similar (they share a lot of code). One
small difference is that PostScript has an arc primitive, and PDF does not.
This is what I thought at first, which is why I found these differences
surprising. I think your idea of blaming the viewer is correct. I
thought that Adobe of all people could deal with Postscript files
properly, but I guess I was overly trusting.
Thanks for the help,
Francois
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[R] 3d line smoothing
Dear R People,
I would like to smooth some 3d lines, which consist of line segments
specified by start and end 3d coordinates. These are experimentally derived
from microscope images of neurons, so there is no sense in which one axis is
any different from any other (predictor vs response).
I have tried doing repeated smoothing interpolation with smooth.splines (see
below), but this doesn't behave that well with some of my data which can be
multivalued with respect to x and is not a directly 3d approach since.
Could anyone suggest a package which might have appropriate functionality or
a link to read up about possible approaches?
With many thanks for your help,
Greg Jefferis.
PS As an aside, wouldn't categorising the 1350 downloadable packages on CRAN
according to their content make a nice exercise in applied statistical
learning/visualisation?
--
LineSmooth3d-function(l){
lxy=smooth.spline(l[,1:2])
lxz=smooth.spline(l[,c(1,3)])
cbind(l[,1],predict(lxy,l[,1])$y,predict(lxz,l[,1])$y)
}
# First 2 segments of a neuron
l=structure(c(2.240391, 5.629143, 8.140456, 9.94432, 12.295891,
13.457473, 15.170222, 17.758942, 18.290545, 20.334337, 22.027239,
23.263636, 24.24404, 26.234032, 27.411226, 29.369003, 30.806803,
32.68042, 35.012493, 37.436077, 40.567665, 46.615963, 49.906017,
57.803772, 62.236923, 63.849625, 131.309967, 129.133469, 127.194168,
125.517822, 123.464272, 121.037727, 118.98951, 116.41317, 114.511192,
111.733353, 108.64267, 106.416183, 103.347008, 100.653694, 98.72525,
96.918602, 95.4189, 94.231995, 92.493828, 91.070656, 88.901505,
86.535431, 86.055229, 86.209984, 86.143341, 86.368622, 9.116494,
10.510942, 12.17625, 14.124721, 17.23107, 21.285524, 24.150547,
26.97097, 29.702291, 32.635288, 35.791851, 39.013748, 41.252228,
42.638962, 44.100445, 45.742134, 47.191525, 49.838848, 51.361511,
52.929546, 55.789406, 60.742428, 61.489536, 60.880463, 61.653023,
61.472942), .Dim = c(26L, 3L), .Dimnames = list(NULL, c(X, Y, Z)))
l2=structure(list(X = c(63.849625, 63.710838, 64.202766, 64.650017,
64.965698, 64.609695, 64.352356), Y = c(86.368622, 83.420357,
80.972359, 79.352409, 77.604141, 76.094627, 74.800072), Z = c(61.472942,
62.381229, 62.161339, 63.138336, 64.076202, 65.080811, 68.487137
)), .Names = c(X, Y, Z), row.names = c(26, 27, 28,
29, 30, 31, 32), class = data.frame)
# install.packages('rgl')
require(rgl)
# First line OK
rgl.linestrips(l)
rgl.linestrips(LineSmooth3d(l),col='red')
# second line not so good
rgl.linestrips(l2)
rgl.linestrips(LineSmooth3d(l2),col='red')
--
Gregory Jefferis, PhD and:
Research Fellow
Department of Zoology St John's College
University of Cambridge Cambridge
Downing Street CB2 1TP
Cambridge, CB2 3EJ
United Kingdom
Lab Tel: +44 (0)1223 336683 Office: +44 (0)1223 339899
Lab Fax: +44 (0)1223 336676
[EMAIL PROTECTED]
http://www.zoo.cam.ac.uk/zoostaff/jefferis.html
http://www.neuroscience.cam.ac.uk/directory/profile.php?gsxej2
http://flybrain.stanford.edu
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[R] Can not load msm package
windows XP
R 2.6.0
I have tried to install the msm package several times. Each time the
installation appears to work. I then go to PACKAGES-LOAD PACKAGE but the msm
package does not appear in the SELECT ONE dialog box. Can someone suggest how I
can get msm to run on my system?
Thanks,
John
Confidentiality Statement:
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and provide commented, minimal, self-contained, reproducible code.
[R] tkconfigure throws an error
Thanks everybody for looking at this. I am trying to assign a script to a button please help: library(tcltk) tt- tktoplevel() tktitle(tt)-the title heading-tklabel(tt,text=Enter date as -MM-DD) l1-tklabel(tt,text=Reporting date) b1=tkbutton(tt,text=Run) d.val-tkentry(tt,width=12) tkgrid(heading,columnspan=2) tkgrid(l1,d.val) tkgrid(b1,columnspan=2) tkconfigure(b1,command=source(./src/f.imm2.R)) # and get inconsistent errors: 1. Error in structure(.External(dotTclObjv, objv, PACKAGE = tcltk), class = tclObj) : [tcl] invalid command name .6.4. ## when the tkconfigure is entered from R buffer 2. tkconfigure(b1,command=source(./src/f.imm2.R)) Error in switch(storage.mode(x), character = .External (RTcl_ObjFromCharVector, : cannot handle object of mode 'list' # when I run the full source buffer (stuff pasted above) the script just needs the date from the tcl window and everything else is conventional R code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] investing in ideas and decision support system experimentation
we're looking for help to analyze some theoretical data-sets, essentially outlining the progress of a given number of ideas over time, by using R in conjunction with mysql databases. let's say there is a game, and the point of the game is to find the most worthwhile solution. let's extend that and say that this solution (or number of solutions) need to be validated by a community. and that community can, and has incentive to, promote ideas they think are good. they can do this in multiple ways: investing $ into an idea, rating, commenting on, and adding to the idea. we need to consider investments, including time of investment relative to life of idea (and life of idea within the game's timeline), and risk taken by the investor at time of investment. it may also be important to look at investment amount trends (and possibly risk trends) in the context of time. the ratio of investment count to unique investment count is also potentially important. investments will not be entirely transparent, that is, a player can see their percentage stake in an idea, but not necessarily how much the other players have invested. let's say we also introduce a rating system and a way for initial ideas to be 'enhanced.' now we have multiple possibly estimable parameters that could tell us a story and allow the 'best ideas' within our construct to bubble to the top. we need help approaching our data sets in a mathematical, algorithmic way. we'd ultimately like help on how to make the decision of *what's the most validated idea* but we don't know quite what distributions or statistical models would be appropriate to use for comparative analyses. Possible values: rating: scale from 1-10 investment: limited to individual's available holdings. holdings can range from nothing to potentially billions (although at the time of this writing hover around a max of 50,000). Rules: • ideas and their enhancements may both be rated, but only initial ideas carry an investment value. • time to submit ideas and vote on or invest in them is limited and that limitation varies widely, but typically ranges from a few hours to many weeks. some things we're concerned about: favoring early adopters of ideas; avoiding players 'gaming the system' to push an idea through; making investment importance relative to actual support by the community, not just to larger investors. -- View this message in context: http://www.nabble.com/investing-in-ideas-and-decision-support-system-experimentation-tp16399082p16399082.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] tkconfigure throws an error
The command argument in the tkconfigure call should be a function, you
have the results of sourcing the file, probably not the correct thing.
Try something like:
tkconfigure(b1,command=function(...){source(./src/f.imm2.R)})
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of stephen bond
Sent: Monday, March 31, 2008 2:45 PM
To: r-help@r-project.org
Subject: [R] tkconfigure throws an error
Thanks everybody for looking at this. I am trying to assign a
script to a button please help:
library(tcltk)
tt- tktoplevel()
tktitle(tt)-the title
heading-tklabel(tt,text=Enter date as -MM-DD)
l1-tklabel(tt,text=Reporting date)
b1=tkbutton(tt,text=Run)
d.val-tkentry(tt,width=12)
tkgrid(heading,columnspan=2)
tkgrid(l1,d.val)
tkgrid(b1,columnspan=2)
tkconfigure(b1,command=source(./src/f.imm2.R))
#
and get inconsistent errors:
1. Error in structure(.External(dotTclObjv, objv, PACKAGE =
tcltk), class = tclObj) :
[tcl] invalid command name .6.4. ## when the
tkconfigure is entered from R buffer 2.
tkconfigure(b1,command=source(./src/f.imm2.R))
Error in switch(storage.mode(x), character = .External
(RTcl_ObjFromCharVector, :
cannot handle object of mode 'list' # when I run the
full source buffer (stuff pasted above)
the script just needs the date from the tcl window and
everything else is conventional R code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Dendrogram orientation
Lassana TOURE wrote: I am using the two functions hclust and plot to draw a dendrogram. The result is a vertical dendrogram , but I prefer a horizontal one. Need help. Take a look at `?dendrogram'. HTH, Henric This is my script: dd1=hclust(d1, method=ward) plot(dd1) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] as.character ()
Hello, I'm trying to tranform a numeric vector into a character vector. x=c(2.00,1.20,5.00,6.56) y= as.character(x) y [1] 21.2 56.56 What I want is : [1] 2.001.20 5.006.56 Does someone know how to do this please ? Benoit Bruneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] as.character ()
Is this what you want: x=c(2.00,1.20,5.00,6.56) sprintf(%.2f, x) [1] 2.00 1.20 5.00 6.56 On Mon, Mar 31, 2008 at 4:45 PM, benlafleche [EMAIL PROTECTED] wrote: Hello, I'm trying to tranform a numeric vector into a character vector. x=c(2.00,1.20,5.00,6.56) y= as.character(x) y [1] 21.2 56.56 What I want is : [1] 2.001.20 5.006.56 Does someone know how to do this please ? Benoit Bruneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] question about levelplot panel arrangement
The default arrangement of multiple panels in a lattice plot is from bottom left to top right. Are there any ways to change that, for example, from top left to bottom right? Thank you. J.P. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Memory in R.
Hello Amadou, On Mon, Mar 31, 2008 at 4:53 PM, R RR [EMAIL PROTECTED] wrote: 1 - I want to increase memory, but I can't find the right way. E.g: in stata, you just type set memory 1g. When I try to load huge datasets, R crashes. You will find this recent thread [1] interesting. You'd also want to check packages filehash, ff and sqldf. Regards, Liviu [1] http://www.nabble.com/How-to-read-HUGE-data-sets--tt15729830.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] question about levelplot panel arrangement
On 3/31/08, Jimin Pei [EMAIL PROTECTED] wrote: The default arrangement of multiple panels in a lattice plot is from bottom left to top right. Are there any ways to change that, for example, from top left to bottom right? Thank you. Add as.table=TRUE to your call. -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] download.file error
Dear all, I am looking for a way to work out if a file on the internet exists before attempting to download it using the function download.file(). For example, using a url that does not exist url - http://finance.yahoo.com/ftse.csv; destfile - tempfile() download.file(url = url, destfile = destfile) # gives the following response ... trying URL 'http://finance.yahoo.com/ftse.csv' Error in download.file(url = url, destfile = destfile) : cannot open URL 'http://finance.yahoo.com/ftse.csv' In addition: Warning message: In download.file(url = url, destfile = destfile) : cannot open: HTTP status was '404 Not Found' When I am using the download.file() function in a loop over multiple URLs, the above error will cause the loop to terminate, so I want to avoid this by checking if the file exists first then wrapping the subsequent functions in an if() statment. The original fault came from the function get.hist.quote() in the tseries package. I was trying to iterate over various stocks, but some stocks listed in the yahoo website do not have any downloadable data associated with them, which causes the loop to terminate. The workhorse function of get.hist.quote() is the download.file() function. Kind Regards Chib _ Win 100s of Virgin Experience days with BigSnapSearch.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] download.file error
?try On Mon, Mar 31, 2008 at 6:26 PM, CHIB CO [EMAIL PROTECTED] wrote: Dear all, I am looking for a way to work out if a file on the internet exists before attempting to download it using the function download.file(). For example, using a url that does not exist url - http://finance.yahoo.com/ftse.csv; destfile - tempfile() download.file(url = url, destfile = destfile) # gives the following response ... trying URL 'http://finance.yahoo.com/ftse.csv' Error in download.file(url = url, destfile = destfile) : cannot open URL 'http://finance.yahoo.com/ftse.csv' In addition: Warning message: In download.file(url = url, destfile = destfile) : cannot open: HTTP status was '404 Not Found' When I am using the download.file() function in a loop over multiple URLs, the above error will cause the loop to terminate, so I want to avoid this by checking if the file exists first then wrapping the subsequent functions in an if() statment. The original fault came from the function get.hist.quote() in the tseries package. I was trying to iterate over various stocks, but some stocks listed in the yahoo website do not have any downloadable data associated with them, which causes the loop to terminate. The workhorse function of get.hist.quote() is the download.file() function. Kind Regards Chib _ Win 100's of Virgin Experience days with BigSnapSearch.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing Time Series
Hai Ingmar, nice to see you again. well, I consider time series as the most important aspect in my experiment. ANOVA can only show whether the mean are different, and combined with Turkey HSD, the greatest they can do is showing the difference for each pair of subject. Time Series in other side, can give whether the smart has real difference with the average, in term of their cognitive process. Their learning time, their capability to adapt their mental representation when dealing with new pattern or regularity. They can all be plotted and we can check it. Of course by the plot alone, we can say something meaningfull. But, the conclusion is more subjective than objective and is not as strong as statistical approach. Anyway, my hypothesis for this is the subject follow different models. I don't care if the model being followed are ARIMA(1,2,0) with AR(2); or AR(2) and MA(1). I just want to know if the models are different. So far, the only way I have is checking their model one by one. I'm curious if there is another way to check the hypothesis best regards, Nathanael Ingmar Visser wrote: Nathanael, How important is the time series aspect here? Why not just do some anova on the difference between these blocks of trials? If the time information is essential, repeated measures anova could be applied. This sounds like you need a mixture of ARIMA models. Is that indeed the hypothesis that you want to test: whether there are different strategies that result in either an AR learning process or in an MA learning process? Using a mixture of ARIMA models would allow you to analyze these data simultaneously. hth, Ingmar -- View this message in context: http://www.nabble.com/Comparing-Time-Series-tp16392632p16400910.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] garch prediction
Hello I want to predict the future values of time series with Garch When I specified my model like this: library(fGarch) ret - diff(log(x))*100 fit = garchFit(~arma(1,0,0)+garch(1, 1), data =ret) predict(fit, n.ahead = 10) meanForecast meanError standardDeviation 10.01371299 0.030863500.03305819 20.01211893 0.030945190.03350248 I know that if I use fit = garchFit(~garch(1, 1), data =ret) I got constant mean, so trherefore I include amra term to move with mean Iam not sure what values are hiding in this output. 1. Does menForecast hold my future predicted values? 2.Or I am able to just compute the confidence intervals for my prediction like meanForecast +-2*standardDeviation ?? 3Or I need to compute the future values like yt=meanForecast+meanError*sqrt(standardDeviation) ??? My return looks like standard return series with plus and minus values, [748,] 0.008184311 [749,] 0.024548914 [750,] -0.008182302 so I hope I would get similar prediction to this return, not just a postive mean constant. thanks?? -- View this message in context: http://www.nabble.com/garch-prediction-tp16400909p16400909.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] Dendrogram orientation
Try: plot(as.dendrogram(dd1), horiz = TRUE, nodePar = list(lab.cex = .4)) On Thu, Mar 27, 2008 at 1:08 PM, Lassana TOURE [EMAIL PROTECTED] wrote: I am using the two functions hclust and plot to draw a dendrogram. The result is a vertical dendrogram , but I prefer a horizontal one. Need help. This is my script: dd1=hclust(d1, method=ward) plot(dd1) [[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] as.character ()
An older alternative uses format() x - c(2.00, 1.20, 5.00, 6.56) format(x) [1] 2.00 1.20 5.00 6.56 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of jim holtman Sent: Tuesday, 1 April 2008 7:48 AM To: benlafleche Cc: R-help@r-project.org Subject: Re: [R] as.character () Is this what you want: x=c(2.00,1.20,5.00,6.56) sprintf(%.2f, x) [1] 2.00 1.20 5.00 6.56 On Mon, Mar 31, 2008 at 4:45 PM, benlafleche [EMAIL PROTECTED] wrote: Hello, I'm trying to tranform a numeric vector into a character vector. x=c(2.00,1.20,5.00,6.56) y= as.character(x) y [1] 21.2 56.56 What I want is : [1] 2.001.20 5.006.56 Does someone know how to do this please ? Benoit Bruneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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.
