[R] hashing
is hashing implemented in R regards søren __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problem with number characters
Hi Scott, What's the result of running the linux file command on your input file? Does it give ISO-8859 text or something else? example: [EMAIL PROTECTED] bobby]$ file test2.txt test2.txt: ISO-8859 text Best regards, Bobby On Thu, 14 Oct 2004 11:31:33 -0700, Scott Waichler [EMAIL PROTECTED] wrote: I am trying to process text fields scanned in from a csv file that is output from the Windows database program FileMakerPro. The characters onscreen look like regular text, but R does not like their underlying binary form. For example, one of text fields contains a name and a number, but R recognizes the number as something other than what it appears to be in plain text. The character string Draszt 03 after being read into R using scan and = becomes Draszt 03 where the 3 is displayed in my R session as a superscript. Here is the result pasted into this email I'm composing in emacs: Draszt 0%/1iso8859-15³ Another clue for the knowledgable: when I try to display the vector element causing trouble, I get CHARSXP: Draszt 0%/1iso8859-15³ where again the superscipt part is just 3 in my R session. I'm working in Linux, R version 1.9.1, 2004-06-21. Your help will be much appreciated. Scott Waichler Pacific Northwest National Laboratory [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problem Compiling R-2.0.0 on Linux Alpha
Prasad, Rajiv [EMAIL PROTECTED] writes: Error in names-.default(`*tmp*`, value = c(R, Platform, Date, : names attribute [4] must be the same length as the vector [3] Execution halted make[2]: *** [R] Error 1 make[2]: Leaving directory `/home1/rajiv/software/src-7.2/R-2.0.0/src/library' make[1]: *** [R] Error 1 make[1]: Leaving directory `/home1/rajiv/software/src-7.2/R-2.0.0/src' make: *** [R] Error 1 Does this indicate a problem with R, or something in my system installation? A search on R help list archives did not turn up this particular error. Specific to your architecture at least (not too many Alphas around, but this works elsewhere). It's a pretty odd bug: It would seem to come from inside .split_description (tools/R/admin.R) where it would indicate that the Built field has less than three ; separators, but the paste construct in .install_package_description that creates the field does seem to put them in. One conjecture is that a newline sneaked in somehow... I'd first look for malformed DESCRIPTION files in your build directory, then maybe insert some debugging code into .split_description (e.g. print(sys.status());print(Built) immediately before the names- bit). Are you doing this in a clean directory, btw, or might you possibly be picking up bits of a previous build? -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problem with number characters
Since ISO-8859 has many different encodings, you need something more precise. I am seeing 'iso8859-15' embedded in that email. AFAIK GNU `file' or any other Linux utility does not know the encoding of a file, and it is guessing that it has `ISO-8859 text' when it sees 8-bit files that are not apparently UTF-8 etc. R assumes that text files are encoded in the current locale (and does not support multibyte locales). That is planned to change one day. On Sat, 16 Oct 2004, Bobby Corpus wrote: Hi Scott, What's the result of running the linux file command on your input file? Does it give ISO-8859 text or something else? example: [EMAIL PROTECTED] bobby]$ file test2.txt test2.txt: ISO-8859 text Best regards, Bobby On Thu, 14 Oct 2004 11:31:33 -0700, Scott Waichler [EMAIL PROTECTED] wrote: I am trying to process text fields scanned in from a csv file that is output from the Windows database program FileMakerPro. The characters onscreen look like regular text, but R does not like their underlying binary form. For example, one of text fields contains a name and a number, but R recognizes the number as something other than what it appears to be in plain text. The character string Draszt 03 after being read into R using scan and = becomes Draszt 03 where the 3 is displayed in my R session as a superscript. Here is the result pasted into this email I'm composing in emacs: Draszt 0%/1iso8859-15³ Another clue for the knowledgable: when I try to display the vector element causing trouble, I get CHARSXP: Draszt 0%/1iso8859-15³ where again the superscipt part is just 3 in my R session. I'm working in Linux, R version 1.9.1, 2004-06-21. Your help will be much appreciated. -- 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 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Testing for normality of residuals in a regression model
Prof Brian Ripley wrote: However, stats 901 or some such tells you that if the distributions have even slightly longer tails than the normal you can get much better estimates than OLS, and this happens even before a test of normality rejects on a sample size of thousands. Robustness of efficiency is much more important than robustness of distribution, and I believe robustness concepts should be in stats 101. (I was teaching them yesterday in the third lecture of a basic course, albeit a graduate course.) This is a very interesting discussion. So you are basically saying that it's better to use robust regression methods, without having to worry too much about the distribution of residuals, instead of using standard methods and doing a lot of tests to check for normality? Did I get your point? Cheers, Federico __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] tree version 1.0-16
ZKala == Kalaylioglu, Zeynep (IMS) [EMAIL PROTECTED] on Fri, 15 Oct 2004 15:52:11 -0400 writes: ZKala Hi There is something weird going on with the tree ZKala package version 1.0-16. So I want to download an ZKala updated tree package. I found that R website has ZKala version 1.0-18. I want to try this one and see how ZKala the tree algorithm is performing. What is the best ZKala way to download tree package version 1.0-18 which can ZKala be found in index of src/contrib at ZKala www.r-project.org? So far I tried two different ways ZKala none of which has worked for me: 1) I tried to ZKala download it by running the command ZKala install.packages(tree,C:\Program ZKala Files\R\rw1081\library,http://cran.r-project.org/src/contrib,destdir= ZKala C:\Program ZKala http://cran.r-project.org/src/contrib,destdir=C:/Program ZKala Files\R\rw1081\library,installWithVers=TRUE) But it ZKala tries to connect to ZKala http://cran.r-project.org/src/contrib/bin/windows/contrib/1.8/PACKAGES ZKala http://cran.r-project.org/src/contrib/bin/windows/contrib/1.8/PACKAGES ZKala instead of the URL I enter above. yes, I see that your R version is 'R (for windows) 1.8.1' and that's probably why it tries to get package for that version. If you want to upgrade packages, I strongly recommend to first upgrade R to the current version 2.0.0 (or maybe 2.0.0 patched) -- and then follow the advice from James MacDonald. Martin Maechler, ETH Zurich ZKala 2) Also I tried to download it by copying the zip ZKala file first tree_1.0-18.tar.gz ZKala http://cran.r-project.org/src/contrib/tree_1.0-18.tar.gz ZKala in my directory and then using the Packages-Install ZKala package from local -zip files in the R menu. ZKala How can I download this version of tree? Thanks. ZKala Zeynep ZKala [[alternative HTML version deleted]] ZKala __ ZKala [EMAIL PROTECTED] mailing list ZKala https://stat.ethz.ch/mailman/listinfo/r-help PLEASE ZKala do read the posting guide! ZKala http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] combine many .csv files into a single file/data frame
On 15-Oct-04 bogdan romocea wrote: I have a few hundred .csv files which I need to put together (full outer joins on a common variable) to do a factor analysis. Each file may contain anywhere from a few hundred to a few thousand rows. What would be the most efficient way to do this in R? Please include some sample code if applicable. Sorry, the example I posted last night was written too hastily and does not illustrate your precise query well. Here is a better one, with explanation. $ cat j1 1,a 1,b 2,c 2,d 2,e 3,f 3,g 4,h 4,i 5,j $ cat j2 1,A 1,B 1,C 2,D 2,E 4,F 4,G 5,H 6,I 6,J $ join -t , -a 1 -a 2 -o 0,1.2,2.2 j1 j2 1,a,A 1,a,B 1,a,C 1,b,A 1,b,B 1,b,C 2,c,D 2,c,E 2,d,D 2,d,E 2,e,D 2,e,E 3,f, 3,g, 4,h,F 4,h,G 4,i,F 4,i,G 5,j,H 6,,I 6,,J Explanation of options: -t ,Input and output field separator is , (for CSV) -a 1Output a line for every line of j1 not matched in j2 -a 2Output a line for every line of j2 not matched in j1 -o 0,1.2,2.2 Output field format specification: 0 denotes the match (join) field (needed when using -a) 1.2 denotes field 2 from file 1 (j1) 2.2 denotes field 2 from file 2 (j2) j1 and j2 are of course the two files to be joined. Using the -a option gives you the full outer join which you want. This command only works for two files at a time (and you must give two). To join several files you would have to loop through them on the lines of $ join -t , -a 1 -a 2 -o 0,1.2,2.2 j1 j2 J which creates a file J which is the full outer join of j1, j2. Then $ join -t , -a 1 -a 2 -o 0,1.2,2.2 J j3 J and so on through j4, j5, ... For your few hundred files this is best done with a loop like $ for i in * ; do join -t , -a 1 -a 2 -o 0,1.2,2.2 J $i J ; done having first done it for j1, j2 as above and put these two file out of sight in a different directory. J itself also needs to be out of sight otherwise it will get joined to itself at some stage. E.g. where J is written above it could in fact be written as ../joins/J and similarly for ../joins/j1 and ../joins/j2. E.g. first move j1 j2 to ../joins and then do $ join -t , -a 1 -a 2 -o 0,1.2,2.2 ../joins/j1 ../joins/j2 ../joins/J and then $ for i in * ; do join -t , -a 1 -a 2 -o 0,1.2,2.2 ../joins/J $i ../joins/J done Sorry for the previous sloppy response, and hoping the above helps. Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 [NB: New number!] Date: 16-Oct-04 Time: 10:39:05 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] barplot
I have a table with two columns, one with types of blood (A, B, AB or 0) and one with the factor (negative = -1 or positive = 1). How can I combine those two columns so that 7 bars are plotted (A, B, AB, 0, -A, -B and -0)? -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] sapply and loop
Zhen, how to count the exact time ? system.time(base) Returns CPU (and other) times that expr used. If you only need seconds, you can also do date();zz-sapply(ma, myfunction);date() I do not know about how to reduce the time. For very comlex iterations, I use for( ) myself, which maybe inneficient. Mayeul KAUFFMANN Université Pierre Mendès France Grenoble France __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Cox PH Warning Message
Hi, Can anybody tell me what the message below means and how to overcome it. Thanks, Neil Warning message: X matrix deemed to be singular; variable 2 in: coxph(Surv(age_at_death, death) ~ project$pluralgp + project$yrborn + . __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] 20,000 * 6 data values
Sun, There are hundreds using R for microarray analysis. You probably want to avail yourself of the bioconductor tools (http://www.bioconductor.org). 20,000 rows with 6 data values is a small data set, by microarray standards, and is easily handled by R and the bioconductor tools. I routinely analyze a couple of hundred experiments with 40-50k rows, and most of the time, that doesn't push R very hard (on my 2 processor, 4Gb machine, at least). Sean - Original Message - From: Sun [EMAIL PROTECTED] To: R-help mailing list [EMAIL PROTECTED] Sent: Saturday, October 16, 2004 12:24 AM Subject: [R] 20,000 * 6 data values Hello, Rusers: What is the maximum number of data R can handle? Or I have to use SAS? I am trying to do some microarray data analysis. But I am totally new. Did anyone use R to do microarray analysis? Many thanks, Sun __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Cox PH Warning Message
Hi, Can anybody tell me what the message below means and how to overcome it. Thanks, Neil Warning message: X matrix deemed to be singular; variable 2 in: coxph(Surv(age_at_death, death) ~ project$pluralgp + project$yrborn + . Your 2nd covariate (yrborn) is colinear to the other covariates (or nearly colinear). R drops it, but warns you. The results are OK. Mayeul KAUFFMANN Université Pierre Mendès France Grenoble France __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Cox PH Warning Message
This is the output. I think it means that there is a problem with project$pluralgpTriplet??? Neil fm.CPH Call: coxph(formula = Surv(age_at_death, death) ~ project$pluralgp + project$yrborn + project$private + project$pcdead + project$mheight + project$ga2 + project$apgcode + project$apgar5 + project$VLBWgp + project$PEBW + project$LBWgp + project$LBWcat + project$totadm + project$totlos + project$sex + teen + project$DS + project$pcsborn + project$disadv + project$ecores + project$edoccu) coef exp(coef) se(coef) z p project$pluralgpTwin -0.509203 0.601 0.74415 -0.6843 4.9e-01 project$pluralgpTriplet NANA 0.0 NA NA project$yrborn1988-1992-0.083348 0.920 0.22553 -0.3696 7.1e-01 project$private 0.260465 1.298 0.12384 2.1032 3.5e-02 project$pcdead -0.058200 0.943 0.49418 -0.1178 9.1e-01 project$mheight 0.016361 1.016 0.01745 0.9378 3.5e-01 project$ga2 0.104246 1.110 0.10162 1.0258 3.0e-01 project$apgcode4-7 -0.266885 0.766 0.50211 -0.5315 6.0e-01 project$apgcode1-3 -1.704620 0.182 1.12545 -1.5146 1.3e-01 project$apgar5 -0.427139 0.652 0.14099 -3.0296 2.4e-03 project$VLBWgp=1500 grams 0.046203 1.047 0.65494 0.0705 9.4e-01 project$PEBW0.015297 1.015 0.01356 1.1281 2.6e-01 project$LBWgp=2500 grams -0.257472 0.773 0.42496 -0.6059 5.4e-01 project$LBWcat -0.222823 0.800 0.23938 -0.9308 3.5e-01 project$totadm 0.005836 1.006 0.01536 0.3800 7.0e-01 project$totlos 0.007342 1.007 0.00176 4.1664 3.1e-05 project$sexFemale -0.016431 0.984 0.22803 -0.0721 9.4e-01 teen0.286282 1.331 0.34807 0.8225 4.1e-01 project$DSDown syndrome 1.193727 3.299 0.27227 4.3844 1.2e-05 project$pcsborn-0.704743 0.494 0.75943 -0.9280 3.5e-01 project$disadv -0.000573 0.999 0.00250 -0.2296 8.2e-01 project$ecores -0.001588 0.998 0.00249 -0.6392 5.2e-01 project$edoccu 0.000590 1.001 0.00193 0.3054 7.6e-01 Likelihood ratio test=102 on 22 df, p=2.87e-12 n=2126 (1396 observations deleted due to missing) On 16/10/2004, at 6:32 PM, Mayeul KAUFFMANN wrote: Hi, Can anybody tell me what the message below means and how to overcome it. Thanks, Neil Warning message: X matrix deemed to be singular; variable 2 in: coxph(Surv(age_at_death, death) ~ project$pluralgp + project$yrborn + . Your 2nd covariate (yrborn) is colinear to the other covariates (or nearly colinear). R drops it, but warns you. The results are OK. Mayeul KAUFFMANN Université Pierre Mendès France Grenoble France __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Testing for normality of residuals in a regression model
Prof Brian Ripley wrote: However, stats 901 or some such tells you that if the distributions have even slightly longer tails than the normal you can get much better estimates than OLS, and this happens even before a test of normality rejects on a sample size of thousands. Robustness of efficiency is much more important than robustness of distribution, and I believe robustness concepts should be in stats 101. (I was teaching them yesterday in the third lecture of a basic course, albeit a graduate course.) Federico Gherardini answered: This is a very interesting discussion. So you are basically saying that it's better to use robust regression methods, without having to worry too much about the distribution of residuals, instead of using standard methods and doing a lot of tests to check for normality? Did I get your point? My feeling is that symmetry is more important than, let's say kurtosis 0 in the error. Is this correct? Now the problem is: the lower number of observations, the more severe an effect of non-normality (at least, asymmetry?) could be on the regression AND at the same time, power of tests to detect non normality drops. So, I can imagine easily situations where non-normality is not detected, yet asymmetry is such that regression is significantly biased... It is mainly a question of sample size from this point of view... But not only: Andy Liaw wrote: Also, I was told by someone very smart that fitting OLS to data with heteroscedastic errors can make the residuals look `more normal' than they really are... Don't know how true that is, though. That very smart person is not me, but it happens that I experimented also a little bit on this a while ago! Just experiment with artificial data, and you will see what happens: residuals look often more normal that the error distribution you introduced in your artificial data... Another consequence, is a biased estimate of parameters. Indeed, both come together: parameters are biased in a direction that lowers residuals sum of square, obviously, but also in some circumstances, in a direction that make residuals looking more normal... And that is not (how can it be?) taken into account in the test of normality. That is, I believe, a second reason why non-normality of error could not be detected, yet it has a major impact on the OLS regression. And I am pretty sure there are other reasons, like distribution of error both in the dependent and in the independent variables, another violation of the assumptions made for OLS... Best regards, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 6, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: + 32.65.37.34.97, fax: + 32.65.37.33.12 ( ( ( ( (email: [EMAIL PROTECTED] ) ) ) ) ) ( ( ( ( (web: http://www.umh.ac.be/~econum ) ) ) ) ) .. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Testing for normality of residuals in a regression model
I am assuming everyone is on R-help and doesn't want two copies so have trimmed the Cc: list to R-help. On Sat, 16 Oct 2004, Philippe Grosjean wrote: Prof Brian Ripley wrote: [ Other contributions previously excised here without comment. ] However, stats 901 or some such tells you that if the distributions have even slightly longer tails than the normal you can get much better estimates than OLS, and this happens even before a test of normality rejects on a sample size of thousands. Robustness of efficiency is much more important than robustness of distribution, and I believe robustness concepts should be in stats 101. (I was teaching them yesterday in the third lecture of a basic course, albeit a graduate course.) Federico Gherardini answered: This is a very interesting discussion. So you are basically saying that it's better to use robust regression methods, without having to worry too much about the distribution of residuals, instead of using standard methods and doing a lot of tests to check for normality? Did I get your point? My feeling is that symmetry is more important than, let's say kurtosis 0 in the error. Is this correct? Now the problem is: the lower number of observations, the more severe an effect of non-normality (at least, asymmetry?) could be on the regression AND at the same time, power of tests to detect non normality drops. So, I can imagine easily situations where non-normality is not detected, yet asymmetry is such that regression is significantly biased... Before you can even talk about bias you have to agree what it is you are trying to estimate. For asymmetric error distributions it is unlikely to be the population mean, but if it is then least-squares linear regression is unbiased provided only that the error distribution has a finite first moment. (Part of the so-called Gauss-Markov Theorem. This seems to suggest that Philippe's `easy imagination' is of impossible things.) For contaminated normal distributions it is possibly the mean of the uncontaminated normal component, and the latter seems the commonest aim of mainstream robust methods, which do often assume symmetry. (This may not affect interpretation of coefficients other than the intercept.) The (non-linear) robust regression estimators may be biased for the population mean but have a (much) smaller variability for long-tailed distributions. There is a lot of careful discussion about this in the statistical literature, and I don't believe that it is profitable for people to be discussing this without knowing the literature, and probably not _here_ even then. -- 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 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hashing
In a sense, yes. Elements of named vectors can be accessed via their names and that (internally) uses hashing, I believe. -roger Søren Merser wrote: is hashing implemented in R regards søren __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Roger D. Peng http://www.biostat.jhsph.edu/~rpeng/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] sapply and loop
You can use system.time() to time your procedure. There's no guarantee that sapply() will be faster than a for() loop, especially if you preallocate the matrices. -roger Zhen Pang wrote: Dear all, I am doing 200 times simulation. For each time, I generate a matrix and define some function on this matrix to get a 6 dimension vector as my results. As the loop should be slow, I generate 200 matrice first, and save them into a list named ma, then I define zz-sapply(ma, myfunction) To my surprise, It almost costs me the same time to get my results if I directly use a loop from 1 to 200. Is it common? Can I improve any further? Ps, how to count the exact time to finish my code? Thanks. Zhen __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Roger D. Peng http://www.biostat.jhsph.edu/~rpeng/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] sapply and loop
Without seeing what myfunction is, it's almost impossible to tell. In addition to system.time(), you might want to profile your code. e.g., Rprof() zz - sapply(ma, myfunction) Rprof(NULL) summaryRprof() HTH, Andy From: Zhen Pang Dear all, I am doing 200 times simulation. For each time, I generate a matrix and define some function on this matrix to get a 6 dimension vector as my results. As the loop should be slow, I generate 200 matrice first, and save them into a list named ma, then I define zz-sapply(ma, myfunction) To my surprise, It almost costs me the same time to get my results if I directly use a loop from 1 to 200. Is it common? Can I improve any further? Ps, how to count the exact time to finish my code? Thanks. Zhen __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] sapply and loop
Below is my code. myfunction is the myfunction I mentioned in my last email.
p0-.2
rho0-.2
nl-200
simu-200
set.seed(135)
setwd(d:/r)
options(warn=1)
ns-rep(1,nl)
configuration-runif(nl)
frequency-c(0.0046,0.0057,0.0099,0.0139,0.0147,0.0148,0.0225,0.0321,0.0475,0.0766,0.1179,0.1529,0.1605,0.1424,0.0975,0.0542,0.0207,0.0086,0.0030)
for (i in 1:nl)
{if (configuration[i]=frequency[1]) {ns[i]-1
} else {for (j in 2:length(frequency))
{if (sum(frequency[1:(j-1)])configuration[i]
configuration[i]=sum(frequency[1:j])) {ns[i]-j}
}
}
}
nu-unique(ns)
k-max(nu)
a-vector(simu,mode=list)
for (si in 1:simu)
{
print (si)
print (si)
data-c(0,0,0)
for (iu in 1:length(nu))
{fr-length(ns[ns==nu[iu]])
y-rep(0,fr)
for (ju in 1:fr)
{y[ju]-rbinom(1,nu[iu],rbeta(1,(1-rho0)*p0/rho0,(1-rho0)*(1-p0)/rho0))
}
yu-sort(unique(y))
yy-rep(0,length(yu))
for (ku in 1:length(yu))
{yy[ku]-length(y[y==yu[ku]])
}
ma-cbind(rep(nu[iu],length(yu)),yu,yy)
data-rbind(data,ma)
}
data-data[-1,]
a[[si]]-data
}
myfunction-function(data)
{
llb-function(theta)
{
s - apply(data, 1, function(data) {
n-data[1]; y-data[2] ; re-data[3]
p-1/(1+exp(-theta[1]))
t - exp(theta[2])
s - log(choose(n,y))
r-c(0:(n-1))
s - s-sum(log(1+r*t))
if (n-y-1=0)
{
r-c(0:(n-y-1))
s - s+sum(log(1-p+r*t))
}
if (y-1=0)
{r-c(0:(y-1))
s- s+sum(log(p+r*t))
}
s*re
})
-sum(s)
}
est2-optim(c(log(p0/(1-p0)),log(rho0/(1-rho0))),llb,hessian=T,control =
list(maxit=500))
est2$par
}
zz-sapply(a,myfunction)
If we move the myfucntion to the for(si in 1:simu) loop, results are the
same and there are no time spare. Can you improve a little? Thanks.
Zhen
From: Liaw, Andy [EMAIL PROTECTED]
To: 'Zhen Pang' [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: RE: [R] sapply and loop
Date: Sat, 16 Oct 2004 08:23:54 -0400
Without seeing what myfunction is, it's almost impossible to tell.
In addition to system.time(), you might want to profile your code. e.g.,
Rprof()
zz - sapply(ma, myfunction)
Rprof(NULL)
summaryRprof()
HTH,
Andy
From: Zhen Pang
Dear all,
I am doing 200 times simulation. For each time, I generate a
matrix and
define some function on this matrix to get a 6 dimension vector as my
results.
As the loop should be slow, I generate 200 matrice first, and
save them into
a list named ma,
then I define zz-sapply(ma, myfunction)
To my surprise, It almost costs me the same time to get my
results if I
directly use a loop from 1 to 200. Is it common? Can I
improve any further?
Ps, how to count the exact time to finish my code?
Thanks.
Zhen
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Re: [R] tree version 1.0-16
Martin Maechler wrote: ZKala == Kalaylioglu, Zeynep (IMS) [EMAIL PROTECTED] on Fri, 15 Oct 2004 15:52:11 -0400 writes: ZKala Hi There is something weird going on with the tree ZKala package version 1.0-16. So I want to download an ZKala updated tree package. I found that R website has ZKala version 1.0-18. I want to try this one and see how ZKala the tree algorithm is performing. What is the best ZKala way to download tree package version 1.0-18 which can ZKala be found in index of src/contrib at ZKala www.r-project.org? So far I tried two different ways ZKala none of which has worked for me: 1) I tried to ZKala download it by running the command ZKala install.packages(tree,C:\Program ZKala Files\R\rw1081\library,http://cran.r-project.org/src/contrib,destdir= ZKala C:\Program ZKala http://cran.r-project.org/src/contrib,destdir=C:/Program ZKala Files\R\rw1081\library,installWithVers=TRUE) But it ZKala tries to connect to ZKala http://cran.r-project.org/src/contrib/bin/windows/contrib/1.8/PACKAGES ZKala http://cran.r-project.org/src/contrib/bin/windows/contrib/1.8/PACKAGES ZKala instead of the URL I enter above. yes, I see that your R version is 'R (for windows) 1.8.1' and that's probably why it tries to get package for that version. If you want to upgrade packages, I strongly recommend to first upgrade R to the current version 2.0.0 (or maybe 2.0.0 patched) -- and then follow the advice from James MacDonald. Yes! Let me add, that if you are going to stay with R-1.8.1 for some unbelievable reason, you can compile from sources (the one that ends with .tar.gz) yourself (follow .../src/gnuwin32/readme.packages). That might be much harder than upgrading, though - and we cannot promise that a recent package version works under outdated versions of R. Uwe Ligges Martin Maechler, ETH Zurich ZKala 2) Also I tried to download it by copying the zip ZKala file first tree_1.0-18.tar.gz ZKala http://cran.r-project.org/src/contrib/tree_1.0-18.tar.gz ZKala in my directory and then using the Packages-Install ZKala package from local -zip files in the R menu. ZKala How can I download this version of tree? Thanks. ZKala Zeynep ZKala [[alternative HTML version deleted]] ZKala __ ZKala [EMAIL PROTECTED] mailing list ZKala https://stat.ethz.ch/mailman/listinfo/r-help PLEASE ZKala do read the posting guide! ZKala http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] barplot
Sepp Gurgel wrote: I have a table Do you mean a data.frame? with two columns, one with types of blood (A, B, AB or 0) and one with the factor (negative = -1 or positive = 1). And from these you made a table? my.table - table(my.data.frame) How can I combine those two columns so that 7 bars are plotted (A, B, AB, 0, -A, -B and -0)? Now you can say blood.id - outer(rownames(my.table), colnames(m.ytable), paste, sep=) barplot(as.vector(my.table), names = blood.id) Uwe Ligges -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Lazy loading... advices
Hello, I am looking for more information about lazy loading introduced in R 2.0.0. Doing ?lazyLoad I got some and there is a 'see also' section that points to 'makeLazyLoading'... But I cannot reach this page. My problem is: I recompiled a library that uses a lot of functions from other libraries (of course I can give details if needed). I load it in my computer: library(svGUI), and it takes something like 20 seconds to load. In R 1.9.1 it took 3-4 seconds on the same machine (Windows XP). So, I try now to understand the mechanism and to find a way to lower the loading time of this library with lazy loading (its goal is to load faster, isn't, so I probably do something wrong). Any help or advice would be appreciated. Here is a Rprof of library(svGUI) on my machine: » summaryRprof() $by.self self.time self.pct total.time total.pct file.exists 7.42 24.9 8.46 28.4 list.files 7.24 24.3 7.32 24.6 file 6.78 22.8 6.88 23.1 read.dcf 1.42 4.8 9.24 31.0 file.info0.54 1.8 0.82 2.8 lapply 0.52 1.7 8.96 30.1 inherits 0.34 1.1 28.76 96.5 names0.34 1.1 0.38 1.3 names- 0.30 1.0 0.42 1.4 paste0.30 1.0 0.66 2.2 close.connection 0.24 0.8 0.24 0.8 .Call0.20 0.7 0.20 0.7 apply0.20 0.7 0.58 1.9 .find.package0.18 0.6 18.14 60.9 [... More here] $by.total total.time total.pct self.time self.pct library 29.70 99.7 0.00 0.0 try 29.64 99.5 0.10 0.3 f29.36 98.5 0.00 0.0 firstlib 29.36 98.5 0.00 0.0 Require 28.96 97.2 0.00 0.0 match28.80 96.6 0.08 0.3 inherits 28.76 96.5 0.34 1.1 is.factor28.76 96.5 0.00 0.0 %in% 28.60 96.0 0.00 0.0 installed.packages 28.60 96.0 0.00 0.0 unlist 21.46 72.0 0.08 0.3 packageDescription 21.22 71.2 0.10 0.3 system.file 19.18 64.4 0.08 0.3 .find.package18.14 60.9 0.18 0.6 guiInstall 11.82 39.7 0.00 0.0 read.dcf 9.24 31.0 1.42 4.8 lapply8.96 30.1 0.52 1.7 file.exists 8.46 28.4 7.42 24.9 FUN 7.82 26.2 0.02 0.1 list.files7.32 24.6 7.24 24.3 .packages 7.20 24.2 0.02 0.1 file 6.88 23.1 6.78 22.8 require 3.42 11.5 0.00 0.0 [... More here] This is the description of my package (in the bundle SciViews): Package: svGUI Title: SciViews GUI API - Main GUI features Description: Functions to communicate with a GUI client, to implement an object browser, etc... Bundle: SciViews Version: 0.7-0 Date: 2004-10-10 Depends: utils, grDevices, graphics, stats, methods, tcltk, R2HTML, svMisc Suggests: Hmisc, MASS, wxPython Author: Philippe Grosjean Eric Lecoutre Maintainer: Philippe Grosjean [EMAIL PROTECTED] BundleDescription: SciViews GUI API A series of packages to implement a full reusable GUI API for R. License: GPL 2 or above URL: http://www.sciviews.org/SciViews-R Thank you. Best, Philippe Grosjean ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 6, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: + 32.65.37.34.97, fax: + 32.65.37.33.12 ( ( ( ( (email: [EMAIL PROTECTED] ) ) ) ) ) ( ( ( ( (web: http://www.umh.ac.be/~econum ) ) ) ) ) .. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] moving functions between namespaces
Hi all, I'm a newbie wrt R but that's okay, since I don't do any programming in R. What I do need to do is to administrate the activites of an R programmer, so my questions will be related to adminning R as opposed to statistcal programming. I hope that's okay with y'all. :-) I've been through most of the docs that I could find (the PDFs, StatsRUs, etc.) and I've found the answers to most of my questions, but not to an important one: how do I move a function from one namespace to another. On the chance that I'm not using the right terminlogoy, here's my situation: Programmer A has written a function foobar() which resides in his namespace /home/progA/.RData. I want to move foobar(), and only foobar(), into the QA enviroment which means replicating foobar() inside of /home/qa/.RData. I've found that I can use sink() to copy foobar() to a file but 1) I haven't found how to read it back in and 2) I figure you guys have a more elegant way of doing this. Any advice? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Lazy loading... advices
Philippe Grosjean wrote: Hello, I am looking for more information about lazy loading introduced in R 2.0.0. Doing ?lazyLoad I got some and there is a 'see also' section that points to 'makeLazyLoading'... But I cannot reach this page. My problem is: I recompiled a library that uses a lot of functions from other libraries (of course I can give details if needed). I load it in my computer: library(svGUI), and it takes something like 20 seconds to load. In R 1.9.1 it took 3-4 seconds on the same machine (Windows XP). So, I try now to understand the mechanism and to find a way to lower the loading time of this library with lazy loading (its goal is to load faster, isn't, so I probably do something wrong). Any help or advice would be appreciated. Have you read the article about lazyLoad mechanism that Brian Ripley wrote for the latest R News issue? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] moving functions between namespaces
On Sat, 16 Oct 2004, Faber Fedor wrote: Hi all, I'm a newbie wrt R but that's okay, since I don't do any programming in R. What I do need to do is to administrate the activites of an R programmer, so my questions will be related to adminning R as opposed to statistcal programming. I hope that's okay with y'all. :-) I've been through most of the docs that I could find (the PDFs, StatsRUs, etc.) and I've found the answers to most of my questions, but not to an important one: how do I move a function from one namespace to another. On the chance that I'm not using the right terminlogoy, here's my I think you mean `workspace', not `namespace', so the explanation was very helpful. situation: Programmer A has written a function foobar() which resides in his namespace /home/progA/.RData. I want to move foobar(), and only foobar(), into the QA enviroment which means replicating foobar() inside of /home/qa/.RData. I've found that I can use sink() to copy foobar() to a file but 1) I haven't found how to read it back in and 2) I figure you guys have a more elegant way of doing this. In /home/progA start R and run save(foobar, file=foobar.rda). End. In /home/qa start R and run load(/home/progA/foobar.rda). If this is more than once off, your programmer should write a package containing the useful functions: see `Writing R Extensions'. -- 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 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] moving functions between namespaces
Faber Fedor wrote: Hi all, I'm a newbie wrt R but that's okay, since I don't do any programming in R. What I do need to do is to administrate the activites of an R programmer, so my questions will be related to adminning R as opposed to statistcal programming. I hope that's okay with y'all. :-) I've been through most of the docs that I could find (the PDFs, StatsRUs, etc.) and I've found the answers to most of my questions, but not to an important one: how do I move a function from one namespace to another. On the chance that I'm not using the right terminlogoy, here's my situation: Programmer A has written a function foobar() which resides in his namespace /home/progA/.RData. I want to move foobar(), and only foobar(), into the QA enviroment which means replicating foobar() inside of /home/qa/.RData. I've found that I can use sink() to copy foobar() to a file but 1) I haven't found how to read it back in and 2) I figure you guys have a more elegant way of doing this. Any advice? The term namespace now means something other than the way you use it but that's not important. The simple way to do this is to have Programmer A start up R in /home/progA and save a copy of the function foobar to a file. save(foobar, file = /home/qa/foobar.rda) (Depending on the size of the file Programmer A may wish to add the optional argument compress = TRUE in that call.) Now start R in /home/qa and load the saved copy load(/home/qa/foobar.rda) then exit R. I'm tempted to write more about variations on this mechanism but I think I'll just stay with the simple answer. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Lazy loading... advices
Philippe Grosjean wrote: Hello, I am looking for more information about lazy loading introduced in R 2.0.0. Doing ?lazyLoad I got some and there is a 'see also' section that points to 'makeLazyLoading'... But I cannot reach this page. My problem is: I recompiled a library Philippe, citing Doug Bates: Someone named Martin Maechler will shortly be sending you email regarding the distinction between 'library' and 'package'. ;-) [from: install.packages(fortunes) library(fortunes) fortune(library) ] that uses a lot of functions from other libraries (of course I can give details if needed). I load it in my computer: library(svGUI), and it takes something like 20 seconds to load. In R 1.9.1 it took 3-4 seconds on the same machine (Windows XP). So, I try now to understand the mechanism and to find a way to lower the loading time of this library with lazy loading (its goal is to load faster, isn't, so I probably do something wrong). It might not always be faster. See Brian Ripley's article in the most recent R Newsletter. Probably you are using a lot of functions in the startup directly after the call to library() (in .First.lib() or .onLoad() or Hooks or whatever). I think you want to specify LazyLoad: no in the package's DESCRIPTION file (cp. Writing R Extensions). Uwe Any help or advice would be appreciated. Here is a Rprof of library(svGUI) on my machine: » summaryRprof() $by.self self.time self.pct total.time total.pct file.exists 7.42 24.9 8.46 28.4 list.files 7.24 24.3 7.32 24.6 file 6.78 22.8 6.88 23.1 read.dcf 1.42 4.8 9.24 31.0 file.info0.54 1.8 0.82 2.8 lapply 0.52 1.7 8.96 30.1 inherits 0.34 1.1 28.76 96.5 names0.34 1.1 0.38 1.3 names- 0.30 1.0 0.42 1.4 paste0.30 1.0 0.66 2.2 close.connection 0.24 0.8 0.24 0.8 .Call0.20 0.7 0.20 0.7 apply0.20 0.7 0.58 1.9 .find.package0.18 0.6 18.14 60.9 [... More here] $by.total total.time total.pct self.time self.pct library 29.70 99.7 0.00 0.0 try 29.64 99.5 0.10 0.3 f29.36 98.5 0.00 0.0 firstlib 29.36 98.5 0.00 0.0 Require 28.96 97.2 0.00 0.0 match28.80 96.6 0.08 0.3 inherits 28.76 96.5 0.34 1.1 is.factor28.76 96.5 0.00 0.0 %in% 28.60 96.0 0.00 0.0 installed.packages 28.60 96.0 0.00 0.0 unlist 21.46 72.0 0.08 0.3 packageDescription 21.22 71.2 0.10 0.3 system.file 19.18 64.4 0.08 0.3 .find.package18.14 60.9 0.18 0.6 guiInstall 11.82 39.7 0.00 0.0 read.dcf 9.24 31.0 1.42 4.8 lapply8.96 30.1 0.52 1.7 file.exists 8.46 28.4 7.42 24.9 FUN 7.82 26.2 0.02 0.1 list.files7.32 24.6 7.24 24.3 .packages 7.20 24.2 0.02 0.1 file 6.88 23.1 6.78 22.8 require 3.42 11.5 0.00 0.0 [... More here] This is the description of my package (in the bundle SciViews): Package: svGUI Title: SciViews GUI API - Main GUI features Description: Functions to communicate with a GUI client, to implement an object browser, etc... Bundle: SciViews Version: 0.7-0 Date: 2004-10-10 Depends: utils, grDevices, graphics, stats, methods, tcltk, R2HTML, svMisc Suggests: Hmisc, MASS, wxPython Author: Philippe Grosjean Eric Lecoutre Maintainer: Philippe Grosjean [EMAIL PROTECTED] BundleDescription: SciViews GUI API A series of packages to implement a full reusable GUI API for R. License: GPL 2 or above URL: http://www.sciviews.org/SciViews-R Thank you. Best, Philippe Grosjean ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 6, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: +
RE: [R] power in a specific frequency band
Unfortunately the documentation seems somewhat sparse but you can figure out what it is doing by looking at the code of spec.pgram or just pumping a sine wave through it and comparing that to a manually generated periodogram, which we do below: tt - 1:32 # time x - sin(2*pi*tt/8) # sine save with period 8, i.e. freq = .125 x - x - mean(x) # Calculate the periodogram manually using the fft round(Mod(fft(x))^2/32, 3) [1] 0 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 0 0 0 # Now recalculate using spectrum sp - spectrum(x, taper = 0, detrend = FALSE) # Note the power concentrated at sp$freq[4]=.125 in sp$spec[4]. # Note that the fft version returns the mean in element 1 so element 2 # of the fft version corresponds to element 1 of the spectrum version. # Also in comparing we see that the spectrum version omits the symmetric # completion of the first half. round(sp$spec,2) [1] 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 sp$freq [1] 0.03125 0.06250 0.09375 0.12500 0.15625 0.18750 0.21875 0.25000 0.28125 [10] 0.31250 0.34375 0.37500 0.40625 0.43750 0.46875 0.5 1/sp$freq [1] 32.00 16.00 10.67 8.00 6.40 5.33 4.571429 [8] 4.00 3.56 3.20 2.909091 2.67 2.461538 2.285714 [15] 2.13 2.00 # sum of squares vs. sum of periodogram values. # Note that power defined as sum of squares of demeaned series # equals twice sum of periodogram values returned by spectrum # (since spec.pgram does not return the half that is the same # by symmetry) sum(x*x) [1] 16 sum(sp$spec) [1] 8 Date: Fri, 15 Oct 2004 19:55:40 -0400 From: Giuseppe Pagnoni [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: [R] power in a specific frequency band Dear R users I have a really simple question (hoping for a really simple answer :-): Having estimated the spectral density of a time series x (heart rate data) with: x.pgram - spectrum(x,method=pgram) I would like to compute the power in a specific energy band. Assuming that frequency(x)=4 (Hz), and that I am interested in the band between f1 and f2, is the power in the band simply the following? sum(x.pgram$spec[(x.pgram$freq f1/frequency(x)) (x.pgram$freq = f2/frequency(x))]) If it is so, are the returned units the same units as the original time series, but squared (if x is bpm, then the power is in bpm^2)? I own a copy of Venables and Ripley (MASS 2003), but I was not able to extract this information from the time series chapter thanks for any help (please cc to my e-mail, if possible) giuseppe Giuseppe Pagnoni Dept. Psychiatry and Behavioral Sciences Emory University 1639 Pierce Drive, Suite 4000 Atlanta, GA, 30322 tel: 404.712.8431 fax: 404.727.3233 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] returning parameter values by CoCo
Hello, as anyone used CoCo? I used 'backward' function to find a loglinear model for my data, wich I did. Now I´m trying to get the estimates of the parameters of the model. I tried 'returnExpression' but I can´t understand the output. Besides the same model I get the value TRUE. Can someone explain me? Thanks in advance. Assunção __ Email gratuito com 2 000 MB Espaço para guardar 20 anos de correio http://www.portugalmail.pt/2000mb __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Lazy loading... advices
UweL == Uwe Ligges [EMAIL PROTECTED]
on Sat, 16 Oct 2004 18:31:29 +0200 writes:
UweL Philippe Grosjean wrote:
Hello,
I am looking for more information about lazy loading
introduced in R 2.0.0. Doing ?lazyLoad I got some and
there is a 'see also' section that points to
'makeLazyLoading'... But I cannot reach this page.
My problem is: I recompiled a library
UweL Philippe,
UweL citing Doug Bates: Someone named Martin Maechler will
UweL shortly be sending you email regarding the distinction
UweL between 'library' and 'package'. ;-)
of course, now I have to
Note that library(package) attaches and loads a *package* from
one of possibly several libraries { = the directories listed,
e.g., by .libPaths() !}
that uses a lot of functions from other libraries (of
course I can give details if needed). I load it in my
computer: library(svGUI), and it takes something like 20
seconds to load. In R 1.9.1 it took 3-4 seconds on the
same machine (Windows XP). So, I try now to understand
the mechanism and to find a way to lower the loading time
of this library with lazy loading (its goal is to load
faster, isn't, so I probably do something wrong).
UweL It might not always be faster. See Brian Ripley's
UweL article in the most recent R Newsletter. Probably you
UweL are using a lot of functions in the startup directly
UweL after the call to library() (in .First.lib() or
UweL .onLoad() or Hooks or whatever). I think you want to
UweL specify LazyLoad: no in the package's DESCRIPTION
UweL file (cp. Writing R Extensions).
I'm pretty sure the increased startup time comes from the very
long 'Depends:' field in Philippe's package.
I'd bet that -- after reading Brian's article -- Philippe will
find that he can make the 'Depends' much smaller, e.g., by
moving entries to 'Suggests'.
Martin
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[R] Plotcorr: colour the ellipses to emphasize the differences
Hello R users! I began with R and I must say that it is really nice. I have data with a lot of variables and have a problem to extract the pattern from correlation matrix. So I tried with plotcorr and it went fine. While I was reading the help page of this function, I found that ellipse display can be even better with use of different colors (the code is bellow). However I have a problem to understand the process of generating the colors. The function cm.colors() with argument 11 produces scale of colors with 11 points. What is the meaning of [5*xc + 6])? If I ommit this part from the code, I see that ellipses bellow diagonal do not have the same color as above the diagonal. In given example numbers 5 and 6 are given (I think so) since there are 11 variables in dataset mtcars. How can one use this setup for other datasets? For example I have a dataset with 15 variables. I would also like to know if it is possible to use some other scale of colors instead of cm.colors, rainbow, heat.colors, terrain.colors, topo.colors. I would like to have positive correlations in blue and nagative ones in red spectrum of colors. Is it possible? Thank you! # Colour the ellipses to emphasize the differences corr.mtcars - cor(mtcars) ord - order(corr.mtcars[1,]) xc - corr.mtcars[ord, ord] plotcorr( xc, col=cm.colors(11)[5*xc + 6]) With regards, Lep pozdrav Gregor GORJANC __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] how to draw a multivariate function
Hi, Rusers: Thanks for answering my last questions. I am frustrated in plotting a trinomial pmf function f(x,y | n, pa, pb) = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) obviously it is a bivariate function of x and y. But I have put a lot of time on this. ** x - seq(0, n, len = n/2+1) # for now I set it to n/2 to control x+y = n y - seq(0, n, len = n/2+1) f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) wireframe(f ~ x * y, shade = TRUE) ** well, but it plots nothing out. I wonder if you could help me? Seems R is hard to learn without your help. Many thanks, Sun [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to draw a multivariate function
Have you studied the documentation for wireframe including the examples? See also the contour plot examples in Venables and Ripley (2002) Modern Applied Statistics with S, 4th ed. (Springer). It might help if you write your function as follows: f - function(x,y ,n, pa, pb) (factorial(n)/ (factorial(x) * factorial(y) * factorial(n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y))) I hope this fills in enough gaps that you will be able to complete the remaining steps. Good luck. spencer graves Sun wrote: Hi, Rusers: Thanks for answering my last questions. I am frustrated in plotting a trinomial pmf function f(x,y | n, pa, pb) = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) obviously it is a bivariate function of x and y. But I have put a lot of time on this. ** x - seq(0, n, len = n/2+1) # for now I set it to n/2 to control x+y = n y - seq(0, n, len = n/2+1) f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) wireframe(f ~ x * y, shade = TRUE) ** well, but it plots nothing out. I wonder if you could help me? Seems R is hard to learn without your help. Many thanks, Sun [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to draw a multivariate function
Hi, All: Thanks. Here is the code n = 30 lamdaa = 4 lamdab = 1.5 pa = lamdaa/n pb = lamdab/n x - seq(0, n/2, len = n/2+1) y - seq(0, n/2, len = n/2+1) f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) wireframe(f ~ x * y, shade = TRUE) The above cannot show anything. Just le t you know that now I changed to cloud, it can display something :) cloud(f ~ x * y, shade = TRUE) I have questions: 1. what does x*y mean here? I don't think it is a vector dot multiplication. I guess it will creat all rows of x and y for all possible combinations? Why wireframe cannot show here? 2. How to show the value on the cloud plot? I have no idea of how much the data value is from the plot. 3. Where can I get resources of R? The help file seems not very helpful to me. For example, the lm () function, its weighted least square option does not say clearly the weight = standard deviation. It said it is to minimize sum w*error^2, which mislead us to think it takes variance. I have to ask experienced people. And everytime the answer depends on luck. Thanks, - Original Message - From: Andrew Ward [EMAIL PROTECTED] To: Sun [EMAIL PROTECTED] Sent: Saturday, October 16, 2004 10:15 PM Subject: RE: [R] how to draw a multivariate function Dear Sun, Could you please provide an example that can be run by readers of the list? What you've given is missing at least n and pa. Regards, Andrew C. Ward,[EMAIL PROTECTED] Senior Analyst (Quantitative), Tel: +61 7 3864 0439 Queensland Studies Authority, Fax: +61 7 3229 3318 295 Ann Street, Brisbane Qld 4000, Australia -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Sun Sent: Sunday, 17 October 2004 12:59 PM To: R-help mailing list Subject: [R] how to draw a multivariate function Hi, Rusers: Thanks for answering my last questions. I am frustrated in plotting a trinomial pmf function f(x,y | n, pa, pb) = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) obviously it is a bivariate function of x and y. But I have put a lot of time on this. ** x - seq(0, n, len = n/2+1) # for now I set it to n/2 to control x+y = n y - seq(0, n, len = n/2+1) f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) wireframe(f ~ x * y, shade = TRUE) ** well, but it plots nothing out. I wonder if you could help me? Seems R is hard to learn without your help. Many thanks, Sun [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ~~ This email (including any attached files) is for the intended recipient(s) only. If you received this email by mistake, please, as a courtesy, tell the sender, then delete this email. The views and opinions are the originator's and do not necessarily reflect those of the Queensland Studies Authority. All reasonable precautions have been taken to ensure that this email contained no viruses at the time it was sent. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to draw a multivariate function
Thanks, Spencer. Turing it into a function seems really to make difference. Well but I don't know what is the difference between data1 = cbind (x, y, f(x,y)) cloud(data1) and cloud(f(x,y) ~ x *y)? They draw different plots. I don't know which one is correct. I am really confused. Here are codes: n = 30 lamdaa = 4 lamdab = 1.5 pa = lamdaa/n pb = lamdab/n x - seq(0, n/2, len = n/2+1) y - seq(0, n/2, len = n/2+1) f - function(x,y) (factorial(n)/ (factorial(x) * factorial(y) * factorial(n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y))) data1 = cbind (x, y, f(x,y)) cloud(data1) cloud(f(x,y) ~ x *y) Thanks a lot - Original Message - From: Spencer Graves [EMAIL PROTECTED] To: Sun [EMAIL PROTECTED] Cc: R-help mailing list [EMAIL PROTECTED] Sent: Saturday, October 16, 2004 11:06 PM Subject: Re: [R] how to draw a multivariate function Have you studied the documentation for wireframe including the examples? See also the contour plot examples in Venables and Ripley (2002) Modern Applied Statistics with S, 4th ed. (Springer). It might help if you write your function as follows: f - function(x,y ,n, pa, pb) (factorial(n)/ (factorial(x) * factorial(y) * factorial(n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y))) I hope this fills in enough gaps that you will be able to complete the remaining steps. Good luck. spencer graves Sun wrote: Hi, Rusers: Thanks for answering my last questions. I am frustrated in plotting a trinomial pmf function f(x,y | n, pa, pb) = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) obviously it is a bivariate function of x and y. But I have put a lot of time on this. ** x - seq(0, n, len = n/2+1) # for now I set it to n/2 to control x+y = n y - seq(0, n, len = n/2+1) f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))* pa^x * pb^y * ((1-pa-pb)^(n-x-y)) wireframe(f ~ x * y, shade = TRUE) ** well, but it plots nothing out. I wonder if you could help me? Seems R is hard to learn without your help. Many thanks, Sun [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to draw a multivariate function
On Saturday 16 October 2004 23:12, Sun wrote:
Hi, All:
Thanks. Here is the code
n = 30
lamdaa = 4
lamdab = 1.5
pa = lamdaa/n
pb = lamdab/n
x - seq(0, n/2, len = n/2+1)
y - seq(0, n/2, len = n/2+1)
Have you looked at what values of x and y these produce? They include
non-integer values. Are you sure you want that?
f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))*
pa^x * pb^y * ((1-pa-pb)^(n-x-y))
wireframe(f ~ x * y, shade = TRUE)
The above cannot show anything.
Just le t you know that now I changed to cloud, it can display
something :) cloud(f ~ x * y, shade = TRUE)
I have questions:
1.
what does x*y mean here? I don't think it is a vector dot
multiplication. I guess it will creat all rows of x and y for all
possible combinations? Why wireframe cannot show here?
Why guess instead of reading the documentation and looking at the
examples? There's a very relevant example in the help page for
wireframe.
You clearly want to evaluate 'f' at all combinations of x and y, yet you
seem to be evaluating it only along the diagonal (x = y). The correct
way to do this is (as studying the examples should have suggested to
you):
g - expand.grid(x = seq(0, n), y = seq(0, n))
g$z - dtrinom(g$x, g$y)
wireframe(z ~ x * y, data = g, shade = TRUE)
where dtrinom could be defined as
dtrinom - function(x, y)
{
ifelse(x + y n,
NA,
factorial(n)/ (factorial(x) *
factorial(y) * factorial (n-x-y))*
pa^x * pb^y * ((1-pa-pb)^(n-x-y)))
}
although I would suggest working on the log scale for numerical
stability:
dtrinom - function(x, y)
{
ifelse(x + y n,
NA,
exp((lfactorial(n) - lfactorial(x) -
lfactorial(y) - lfactorial(n-x-y)) +
x * log(pa) + y * log(pb) +
(n-x-y) * log(1-pa-pb)))
}
2.
How to show the value on the cloud plot? I have no idea of how much
the data value is from the plot.
Read the documentation for the 'scales' argument.
3. Where can I get resources of R? The help file seems not very
helpful to me. For example, the lm () function, its weighted least
square option does not say clearly the weight = standard deviation.
It said it is to minimize sum w*error^2, which mislead us to think it
takes variance. I have to ask experienced people. And everytime the
answer depends on luck.
It's too bad you feel that way. Statistics, and in particular linear
modeling, is a non-trivial subject, and R documentation is not supposed
to serve as a textbook. If you don't understand what minimizing
'sum(w*e^2)' means, you really do need help from 'experienced people'.
Alternatively, look at the references listed in the help page for lm.
Hope that helps,
Deepayan
Thanks,
- Original Message -
From: Andrew Ward [EMAIL PROTECTED]
To: Sun [EMAIL PROTECTED]
Sent: Saturday, October 16, 2004 10:15 PM
Subject: RE: [R] how to draw a multivariate function
Dear Sun,
Could you please provide an example that can be run
by readers of the list? What you've given is
missing at least n and pa.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to draw a multivariate function
Hi, Thanks. But minimizing
'sum(w*e^2)'
means w is the variance instead of the standard deviation. However, the
truth is that R takes standard deviation. R will square it!
R-help document is not that to be proud of. It is not very clear or helpful
sometimes.
- Original Message -
From: Deepayan Sarkar [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: Sun [EMAIL PROTECTED]; Andrew Ward [EMAIL PROTECTED]
Sent: Sunday, October 17, 2004 12:04 AM
Subject: Re: [R] how to draw a multivariate function
On Saturday 16 October 2004 23:12, Sun wrote:
Hi, All:
Thanks. Here is the code
n = 30
lamdaa = 4
lamdab = 1.5
pa = lamdaa/n
pb = lamdab/n
x - seq(0, n/2, len = n/2+1)
y - seq(0, n/2, len = n/2+1)
Have you looked at what values of x and y these produce? They include
non-integer values. Are you sure you want that?
f = factorial(n)/ (factorial(x) * factorial(y) * factorial (n-x-y))*
pa^x * pb^y * ((1-pa-pb)^(n-x-y))
wireframe(f ~ x * y, shade = TRUE)
The above cannot show anything.
Just le t you know that now I changed to cloud, it can display
something :) cloud(f ~ x * y, shade = TRUE)
I have questions:
1.
what does x*y mean here? I don't think it is a vector dot
multiplication. I guess it will creat all rows of x and y for all
possible combinations? Why wireframe cannot show here?
Why guess instead of reading the documentation and looking at the
examples? There's a very relevant example in the help page for
wireframe.
You clearly want to evaluate 'f' at all combinations of x and y, yet you
seem to be evaluating it only along the diagonal (x = y). The correct
way to do this is (as studying the examples should have suggested to
you):
g - expand.grid(x = seq(0, n), y = seq(0, n))
g$z - dtrinom(g$x, g$y)
wireframe(z ~ x * y, data = g, shade = TRUE)
where dtrinom could be defined as
dtrinom - function(x, y)
{
ifelse(x + y n,
NA,
factorial(n)/ (factorial(x) *
factorial(y) * factorial (n-x-y))*
pa^x * pb^y * ((1-pa-pb)^(n-x-y)))
}
although I would suggest working on the log scale for numerical
stability:
dtrinom - function(x, y)
{
ifelse(x + y n,
NA,
exp((lfactorial(n) - lfactorial(x) -
lfactorial(y) - lfactorial(n-x-y)) +
x * log(pa) + y * log(pb) +
(n-x-y) * log(1-pa-pb)))
}
2.
How to show the value on the cloud plot? I have no idea of how much
the data value is from the plot.
Read the documentation for the 'scales' argument.
3. Where can I get resources of R? The help file seems not very
helpful to me. For example, the lm () function, its weighted least
square option does not say clearly the weight = standard deviation.
It said it is to minimize sum w*error^2, which mislead us to think it
takes variance. I have to ask experienced people. And everytime the
answer depends on luck.
It's too bad you feel that way. Statistics, and in particular linear
modeling, is a non-trivial subject, and R documentation is not supposed
to serve as a textbook. If you don't understand what minimizing
'sum(w*e^2)' means, you really do need help from 'experienced people'.
Alternatively, look at the references listed in the help page for lm.
Hope that helps,
Deepayan
Thanks,
- Original Message -
From: Andrew Ward [EMAIL PROTECTED]
To: Sun [EMAIL PROTECTED]
Sent: Saturday, October 16, 2004 10:15 PM
Subject: RE: [R] how to draw a multivariate function
Dear Sun,
Could you please provide an example that can be run
by readers of the list? What you've given is
missing at least n and pa.
__
[EMAIL PROTECTED] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
[EMAIL PROTECTED] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] problem with axis labels
Paul == Paul Stansell [EMAIL PROTECTED]
on Sat, 16 Oct 2004 19:19:42 +0100 (BST) writes:
Paul Dear R users,
Paul Below are some R commands which produce a y-axis label
Paul that is not wholly in the viewing area of the eps file
Paul (or the x11 window).
yes, that has almost nothing to do with the specific device
label-expression(italic(A)==union(italic(H)[italic(i)],i==1,4))
postscript(labelBug.eps,onefile=F,height=5,width=5,pointsize=12)
plot(1,1,xlab=label,ylab=label)
dev.off()
Paul I have tried experimenting with the postscript
Paul bounding box, and using such R commands as
Paul over(phantom(0),...) but I make the whole y-label
Paul visible.
Paul Does anyone have any suggestions for how I may fix this.
yes. You have to use a larger left margin (i.e. margin # 2),
see the 'mar' section in help(par),
and also re-read the part on graphics in the Introduction to R
manual.
The following will look better {independently of the device,
postscript, X11, windows,...}:
label - expression(italic(A) == union(italic(H)[italic(i)],i==1,4))
op - par(mar = .1 + c(5,5,4,1))
plot(1,1, xlab=label, ylab=label)
par(op)# to reset the graphics settings to the defaults
Martin
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