Re: [R] Time series indexes
Fernando Saldanha schrieb: I tried to assign values to specific elements of a time series and got in trouble. The code below should be almost self-explanatory. I wanted to assign 0 to the first element of x, but instead I assigned zero to the second element of x, which is not what I wanted. Is there a function that will allow me to do this without going into index arithmetic (which would be prone to errors)? FS x- ts(c(1,2,3,4), start = 2) x Time Series: Start = 2 End = 5 Frequency = 1 [1] 1 2 3 4 x[2] - 0 x Time Series: Start = 2 End = 5 Frequency = 1 [1] 1 0 3 4 Hello Fernando, is this what you want: x[time(x)==2] - 0 ThPe __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Re: how to add a column with string values to a data frame while still converting them to factors
avneet singh wrote: i got the first part: agm.data$ProdCategory=as.factor(agm.data$ProdCategory) the second i am still struggling with For the second part, see ?is.na. Uwe Ligges On 4/26/05, avneet singh [EMAIL PROTECTED] wrote: I have 2 questions. In essence i am trying to create product categories based on product description and have it as an additional column of my dataframe. some products dont fit any category and i need a list of them. i am having some trouble in this simple (for most) task. Could you please provide suggestions. Thank you. avneet Question 1) I have a data frame agm.data=read.xls(agm.xls) i add a column to it by this and similar statements agm.data$ProdCategory[agm.data$Product.Description==PGX|agm.data$Product.Description==PGW|agm.data$Product.Description==PS|agm.data$Product.Description==PSX]=Molded Graphite if i do agm.data$ProdCategory i get: [1] Extruded Graphite Molded Graphite Molded Graphite [4] Extruded Graphite Extruded Graphite NA [7] Molded Graphite Extruded Graphite Extruded Graphite [10] Extruded Graphite Extruded Graphite Extruded Graphite [13] Extruded Graphite Porous Iso-Molded Graphite . . . . . [1222] Extruded Graphite Molded Graphite Extruded Graphite [1225] Extruded Graphite Extruded Graphite NA [1228] Iso-Molded Graphite Extruded Graphite NA if i check the class, i get is(agm.data$ProdC) [1] character vector i want this to have.. [1] factor oldClass ..like other columns have Question 2)some values of agm.data$ProdCategory are NA. i want to find corresponding values of agm.data$Product.Description so i give the following command agm.data$Product.Description[agm.data$ProdCategory==NA] i get: [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [17] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA . . . . . . [257] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [273] NA NA NA NA NA 165 Levels: 1.563x24 10x50x60 61x1.25 6x72 890S 8x6 9x7; 10x1.5 AGR ... YBDXX88 -- God created man because he was disappointed over the apes. After that he has given up any further experiments ~Mark Twain __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to modify and compile R sourse codes
Zhongming Yang [EMAIL PROTECTED] writes: You can get the source code from the source package and modify it. You'll have to read the relevant documentation about the tools you need to compile it---see the archives, there was a discussion recently about this, and see the documenation that comes with R. I don't use Windows so I can't give any more advice. Depending on what you need to do you might find that what you need is already in the CVS version of Hmisc. The code was added recently and is mentioned here: http://biostat.mc.vanderbilt.edu/twiki/pub/Main/StatReport A PDF with a few simple examples (created while developing and testing the code) are shown here: http://biostat.mc.vanderbilt.edu/twiki/pub/Main/StatReport/latexFineControl.pdf The new options allow you to use any latex command (declaration) to format row and column names, row and column group labels, and each cell of a table individually. Defining new commands in the preamble of your LaTeX document you can great all sorts of splendid (and awful!) combinations of formats. The current version (3.0-5) has options to format the row and column group labels (note that in the CVS version the option name for the column group labels has changed). In case you are not aware of it I should mention that the Sweave function in the tools package is wonderful for creating automated reports. HTH David Dear All: I am working on writing some R functions to make statistical reports automatically. Dr. Harrell's Hmisc has all the wonderful stuff. But sometimes I need change some formats, so I want to read through it and make some modifications to fit my project. Ideally, I want proceed as following: 1. change some source of Hmisc 2. compile and install the modified Hmisc 3. debug my modified functions. And I am working on windows 2000. Could anyone give some suggestions on the tools and the best way to do this? Thanks __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- David Whiting University of Newcastle upon Tyne, UK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to modify and compile R sourse codes
Zhongming Yang wrote: Dear All: I am working on writing some R functions to make statistical reports automatically. Dr. Harrell's Hmisc has all the wonderful stuff. But sometimes I need change some formats, so I want to read through it and make some modifications to fit my project. Ideally, I want proceed as following: 1. change some source of Hmisc 2. compile and install the modified Hmisc 3. debug my modified functions. And I am working on windows 2000. Could anyone give some suggestions on the tools and the best way to do this? The R Installation and Administration manual of R-2.1.0 gives not only suggestions: it provides all the details on the tools required! The manual Writing R Extensions is relevant as well. Uwe Ligges Thanks __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Re: how to add a column with string values to a data frame while still converting them to factors
Hi On 26 Apr 2005 at 16:18, avneet singh wrote: i got the first part: agm.data$ProdCategory=as.factor(agm.data$ProdCategory) the second i am still struggling with ?is.na is.na(whatever) is what you probably want. Cheers Petr On 4/26/05, avneet singh [EMAIL PROTECTED] wrote: I have 2 questions. In essence i am trying to create product categories based on product description and have it as an additional column of my dataframe. some products dont fit any category and i need a list of them. i am having some trouble in this simple (for most) task. Could you please provide suggestions. Thank you. avneet Question 1) I have a data frame agm.data=read.xls(agm.xls) i add a column to it by this and similar statements agm.data$ProdCategory[agm.data$Product.Description==PGX|agm.data$ Product.Description==PGW|agm.data$Product.Description==PS|agm.d ata$Product.Description==PSX]=Molded Graphite if i do agm.data$ProdCategory i get: [1] Extruded Graphite Molded Graphite Molded Graphite [4] Extruded Graphite Extruded Graphite NA [7] Molded Graphite Extruded Graphite Extruded Graphite [10] Extruded Graphite Extruded Graphite Extruded Graphite [13] Extruded Graphite Porous Iso-Molded Graphite . . . . . [1222] Extruded Graphite Molded Graphite Extruded Graphite [1225] Extruded Graphite Extruded Graphite NA [1228] Iso-Molded Graphite Extruded Graphite NA if i check the class, i get is(agm.data$ProdC) [1] character vector i want this to have.. [1] factor oldClass ..like other columns have Question 2)some values of agm.data$ProdCategory are NA. i want to find corresponding values of agm.data$Product.Description so i give the following command agm.data$Product.Description[agm.data$ProdCategory==NA] i get: [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [17] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA . . . . . . [257] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [273] NA NA NA NA NA 165 Levels: 1.563x24 10x50x60 61x1.25 6x72 890S 8x6 9x7; 10x1.5 AGR ... YBDXX88 -- God created man because he was disappointed over the apes. After that he has given up any further experiments ~Mark Twain -- God created man because he was disappointed over the apes. After that he has given up any further experiments ~Mark Twain __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] making table() work
I belive that the problem is not with the table, but with your predictions which are not 0s and 1s. Ales Ziberna - Original Message - From: Stephen Choularton [EMAIL PROTECTED] To: 'R Help' r-help@stat.math.ethz.ch Sent: Wednesday, April 27, 2005 4:19 AM Subject: [R] making table() work I am trying to do some verification across a large dataset, cuData, that has 23 columns. Column 23 (similarity) is the outcome 0 or 1 and the other columns are the features. I do this: verificationglm.model - glm(formula = similarity ~ ., family=binomial, data=cuData[1:1000,]) and produce the model: summary(verificationglm.model) Call: glm(formula = similarity ~ ., family = binomial, data = cuData[1:1000, ]) Deviance Residuals: Min 1Q Median 3Q Max -2.3885 -0.8943 -0.2918 0.8851 2.7025 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 26.3112869 21.2229690 1.240 0.215066 length-0.6249415 0.1906254 -3.278 0.001044 ** meanPitch -0.0110389 0.0053083 -2.080 0.037565 * minimumPitch 0.0002689 0.0024290 0.111 0.911845 maximumPitch -0.0013454 0.0038149 -0.353 0.724326 meanF1-0.0362153 0.0112499 -3.219 0.001286 ** meanF2 0.0016765 0.0115335 0.145 0.884430 meanF3 0.0073960 0.0076235 0.970 0.331964 meanF4 0.0063015 0.0016820 3.746 0.000179 *** meanF5-0.0022535 0.0024885 -0.906 0.365153 ratioF2ToF1 -1.2322825 7.0036532 -0.176 0.860334 ratioF3ToF1 -4.9643148 4.5973552 -1.080 0.280222 jitter-8.7535283 14.5273818 -0.603 0.546806 shimmer1.6706067 2.6327972 0.635 0.525731 percentUnvoicedFrames -0.4863219 1.1638115 -0.418 0.676042 numberOfVoiceBreaks -0.0335636 0.0634956 -0.529 0.597086 percentOfVoiceBreaks -2.9353239 0.8945600 -3.281 0.001033 ** meanIntensity -0.2931293 0.3355314 -0.874 0.382321 minimumIntensity 0.0689654 0.1531059 0.450 0.652392 maximumIntensity 0.2186570 0.2510906 0.871 0.383848 ratioIntensity-8.1777871 13.1676287 -0.621 0.534565 noSyllsIntensity 0.1714826 0.0695021 2.467 0.013614 * speakingRate -0.3564808 0.1507373 -2.365 0.018034 * startSpeech -1.3537348 6.7337461 -0.201 0.840669 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 1384.0 on 999 degrees of freedom Residual deviance: 1084.7 on 976 degrees of freedom AIC: 1132.7 Number of Fisher Scoring iterations: 5 Now I want to use the model to predict on a different part of the dataset. I try this, and get my prediction: pred - predict(verificationglm.model, cuData[1001:2000,1:23]) pred 1001 1002 1003 1004 1005 1006 1007 -0.495901722 -2.406349629 -0.911082179 -0.965869553 -0.488695693 -1.849622304 -1.637722247 1008 1009 1010 1011 1012 1013 1014 -1.148952722 -0.191538278 -1.511895046 -2.989036645 -2.775775622 0.603852124 -0.838613048 1015 1016 1017 1018 1019 1020 1021 -0.434259674 -2.004230065 -0.234829011 1.666502334 2.039631718 -0.592192326 1.667700087 1022 1023 1024 1025 1026 1027 1028 0.104644531 1.748724399 0.391461247 1.356898357 1.468154760 1.090708994 1.071487227 1029 1030 1031 1032 1033 1034 1035 0.720596788 2.378350706 -0.128248232 0.969373318 0.315142756 1.372108172 -2.399517898 1036 1037 1038 1039 1040 1041 1042 -0.684530171 0.761198819 -1.298372615 1.185368711 -1.148974059 0.358234433 0.671495255 1043 1044 1045 1046 1047 1048 1049 0.683771224 0.663767266 2.009012643 0.196591464 2.063417812 0.823472345 0.696638161 [runs on to 2000] However, I then want to check for classAgreement (an e1701 package function). First I want a table. I do this: t = table(pred,cuData[1001:2000,24]) t pred 0 1 -8.900700989801060 1 -8.0484071844879 0 1 -7.792985487755231 0 -7.183383306090131 0 [runs on] when I expect this 0 1 0 ?? 1 ?? with the ?s being some count. When I look at my slice of cuData it looks like this: cuData[1001:2000,24] [1] 1 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 [38] 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 0 1 1 1 1 0 0 0 1 1 0 [75] 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 0 0 1 0 1 1 0 0 0 1 1 1 0 0 0 1 0 1 0 0 0 0 [112] 0 1 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [149] 1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 [186] 0 1 0 0 0 1 0 0 0 0 1 1 0 1 1 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 1 1 1
Re: [R] survreg with numerical covariates
As you can se from the example bellow, survreg works prefeclty fine with numerical values. (I'm runing R2.0.1 on WinXP(SP2) and 32bit AMD with survival version 2.17.). As the posting guide asks, plese provide a small example. Ales Ziberna library(survival) Loading required package: splines data(cancer) survreg(Surv(time, status)~age,data=cancer) Call: survreg(formula = Surv(time, status) ~ age, data = cancer) Coefficients: (Intercept) age 6.88712062 -0.01360829 Scale= 0.7587515 Loglik(model)= -1151.9 Loglik(intercept only)= -1153.9 Chisq= 3.91 on 1 degrees of freedom, p= 0.048 n= 228 - Original Message - From: Richard Mott [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Tuesday, April 26, 2005 11:32 PM Subject: [R] survreg with numerical covariates Does anyone know if the survreg function in the survival package can fit numerical covariates ? When I fit a survival model of the form survreg( Surv(time,censored) ~ x ) then x is always treated as a factor even if it is numeric (and even if I try to force it to be numeric using as.numeric(x). Thus, in the particular example I am analysing, a simple numerical covariate becomes a factor with 190 levels. Is this the expected behaviour ? Am I doing something wrong ? I am running R 2.0.1 on a 64bit Debian Linux system, and version 2.17 of the survival package Thanks Richard Mott -- Richard Mott | Wellcome Trust Centre tel 01865 287588 | for Human Genetics fax 01865 287697 | Roosevelt Drive, Oxford OX3 7BN __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] good editor for R sources ? - Thanks
Thank you very much for all the useful answers. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] libz library missing while installing RMySQL
Hello Trying to install the MySQL package, I get the following error. The help archive contains something on this issue but did not help. I work on linux suse 9.3 Configuration error: Could not locate the library libz required by MySQL. The library libz however is not on any mirrors I checked. Sebastian Leuzinger web http://pages.unibas.ch/botschoen/leuzinger __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] survreg with numerical covariates
Ales - I have identified the problem. It is caused by missing data. In my previous posting I was analysing a dataset which happened to have 190 missing values in the covariate. I compared two models: r0 - survreg(s ~1) r1 - survreg(s~x) anova(r0,r1) I was using anova to investigate the significance of adding covariates to the model, rather than printing out the results of a survreg() directly, and this was the problem. The behaviour of anova() is subtly different when applied to lm() or survreg(). With lm(), if the models compared have different numbers of missing observations, you get a warning, but with survreg(), the difference in observation count shows up as a large change in the df and deviance, giving the appearence that the covariate has been fitted as a factor with 191 levels. So my fault, although possibly anova(survreg()) should be consistent with anova(lm()). -Richard Ales Ziberna wrote: As you can se from the example bellow, survreg works prefeclty fine with numerical values. (I'm runing R2.0.1 on WinXP(SP2) and 32bit AMD with survival version 2.17.). As the posting guide asks, plese provide a small example. Ales Ziberna library(survival) Loading required package: splines data(cancer) survreg(Surv(time, status)~age,data=cancer) Call: survreg(formula = Surv(time, status) ~ age, data = cancer) Coefficients: (Intercept) age 6.88712062 -0.01360829 Scale= 0.7587515 Loglik(model)= -1151.9 Loglik(intercept only)= -1153.9 Chisq= 3.91 on 1 degrees of freedom, p= 0.048 n= 228 - Original Message - From: Richard Mott [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Tuesday, April 26, 2005 11:32 PM Subject: [R] survreg with numerical covariates Does anyone know if the survreg function in the survival package can fit numerical covariates ? When I fit a survival model of the form survreg( Surv(time,censored) ~ x ) then x is always treated as a factor even if it is numeric (and even if I try to force it to be numeric using as.numeric(x). Thus, in the particular example I am analysing, a simple numerical covariate becomes a factor with 190 levels. Is this the expected behaviour ? Am I doing something wrong ? I am running R 2.0.1 on a 64bit Debian Linux system, and version 2.17 of the survival package Thanks Richard Mott -- Richard Mott | Wellcome Trust Centre tel 01865 287588 | for Human Genetics fax 01865 287697 | Roosevelt Drive, Oxford OX3 7BN __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Richard Mott | Wellcome Trust Centre tel 01865 287588 | for Human Genetics fax 01865 287697 | Roosevelt Drive, Oxford OX3 7BN __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] dyn.load(), DLL, Fortran, TLNise software
Dear all, I would like to call TLNise (Two-Level Normal indipendent sampling estimation) software within R. This software estimates a hierarchical model and it can be download from Philip Everson's website at http://www.swarthmore.edu/NatSci/peverso1/TLNise/tlnise.htm;. The TLNise software consists of: 1) a Fortran source code (tlnisemv1.f) and 2) a Splus code (TLNisemv1.src) To use this codes within R: 1) I compiled the source code using g77 (from MinGW). I typed g77 -c tlnisemv1.f to create a Fortran object file (tlnisemv1.o). 2) I edited TLNisemv1.src to point to my copy of tlnise.o. I changed the path in dyn.load(c:\\tlnisemv1.o) 3) I modified the Splus code to be used in R (i.e I changed _ with -) 4) I sourced TLNisemv1.src into R using source(c/TLNisemv1.src) to load the R functions. But when I tried to load the shared library using the dyn.load() function I got an error message: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library c:/programmi/wingw/programs/tlnisemv1.dll: So I decided to build a DLL from the source file tlnisemv1.f. For doing this, I first typed g77 --shared -o tlnisemv1.dll tlnisemv.f using g77 (from MinGW). Then I edited TLNisemv1.src to point to my copy of tlnisemv1.dll. But again when I tried to load the shared library using dyn.load() function I got the same error message. How can I debug this problem? How should I proceed? Have you have used TLNise software in R? Thank you Gabriele Accetta __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] DLL problem when using gcc+gsl
Dear list;
This might sound a bit naive, but then I am new to linking C DLL's to R.
I have built a DLL using GCC and am able to load the DLL in R
(is.loaded(contents)==TRUE). When I include it in my function it returns
NaN for all the variables. My R function is,
dyn.load(c:/data/tempdll.dll)
is.loaded(contents)#returns TRUE
#
contents - function(new.cases){
cases - rep(0, length(new.cases))
case.times - rep(0, length(new.cases))
temp - .C(contents,
as.integer(cases),
as.double(case.times))
}
and the header of my C function is
#ifndef _DLL_H_
#define _DLL_H_
#if BUILDING_DLL
# define DLLIMPORT __declspec (dllexport)
#else /* Not BUILDING_DLL */
# define DLLIMPORT __declspec (dllimport)
#endif /* Not BUILDING_DLL */
DLLIMPORT void contents (int *cases,
double *case_times);
#endif /* _DLL_H_ */
and my C function is.
#include windows.h
#include stdio.h
#include stdlib.h
#include dll.h
#include gsl/gsl_rng.h
#include gsl/gsl_randist.h
void contents(int *cases,
double *case_times)
{
gsl_rng * r;
gsl_rng_env_setup();
r = gsl_rng_alloc (gsl_rng_default);
int i;
for(i = 0; i 1000; i++)
{
cases[i] = gsl_ran_negative_binomial (r, 0.10/2.10, 0.10);
case_times[i] = gsl_ran_weibull (r, 9, 5);
}
gsl_rng_free (r);
}
I am clueless as to where it all goes wrong, as if I make the program an
executable it works.
Thanks again.
Vumani
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Re: [R] dyn.load(), DLL, Fortran, TLNise software
accetta == accetta [EMAIL PROTECTED] on Wed, 27 Apr 2005 11:55:56 +0200 writes: accetta Dear all, I would like to call TLNise (Two-Level accetta Normal indipendent sampling estimation) software accetta within R. accetta This software estimates a hierarchical model and it accetta can be download from Philip Everson's website at accetta http://www.swarthmore.edu/NatSci/peverso1/TLNise/tlnise.htm;. accetta The TLNise software consists of: accetta 1) a Fortran source code (tlnisemv1.f) and 2) a accetta Splus code (TLNisemv1.src) accetta To use this codes within R: accetta 1) I compiled the source code using g77 (from accetta MinGW). I typed g77 -c tlnisemv1.f to create a accetta Fortran object file (tlnisemv1.o). accetta 2) I edited TLNisemv1.src to point to my copy of accetta tlnise.o. I changed the path in accetta dyn.load(c:\\tlnisemv1.o) (which was not ok, as you noticed later) accetta 3) I modified the Splus code to be used in R (i.e I accetta changed _ with -) accetta 4) I sourced TLNisemv1.src into R using accetta source(c/TLNisemv1.src) to load the R functions. accetta But when I tried to load the shared library using accetta the dyn.load() function I got an error message: accetta Error in dyn.load(x, as.logical(local), accetta as.logical(now)) : unable to load shared library accetta c:/programmi/wingw/programs/tlnisemv1.dll: accetta So I decided to build a DLL from the source file accetta tlnisemv1.f. accetta For doing this, I first typed accetta g77 --shared -o tlnisemv1.dll tlnisemv.f using g77 (from MinGW). accetta Then I edited TLNisemv1.src to point to my copy of accetta tlnisemv1.dll. Here, you should have used something like Rcmd SHLIB tlnisemv.f which probably calls g77 itself, but does so in way compatible with your version of R --- which is very important. accetta But again when I tried to load the shared library accetta using dyn.load() function I got the same error accetta message. accetta How can I debug this problem? How should I accetta proceed? Have you have used TLNise software in R? accetta Thank you accetta Gabriele Accetta __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] good editor for R sources ? CE is supported
Just for the record and to avoid confusion: R IS supported in Crimson Editor. Look under syntax files 568 and 314. I've used it for a couple of years to no ill effect. It is fine for light use to moderate use. However, given the updates in R, maybe the syntax files are a little long in tooth in a spot or two. --- vincent [EMAIL PROTECTED] wrote: Dear all, (Sorry if the question has already been answered.) Could someone please suggest a good text editor for writing R sources ? (I know emacs exists ... but I find it a bit heavy). I use crimson (http://www.crimsoneditor.com) which is small and simple, but the R syntax seems not to be supported. Thanks for any advice __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Take a levels
Dear all, How can I take a levels from a dataset. For example, data(iris) x.iris - iris[,1:4] y.iris - iris[,5] y.iris [1] setosa setosa setosa setosa setosa setosa [7] setosa setosa setosa setosa setosa setosa [139] virginica virginica virginica virginica virginica virginica [145] virginica virginica virginica virginica virginica virginica Levels: setosa versicolor virginica I want like, y.iris level are, [] setosa versicolor virginica Best regards, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Take a levels
On Apr 27, 2005, at 8:21 AM, Muhammad Subianto wrote: Dear all, How can I take a levels from a dataset. For example, data(iris) x.iris - iris[,1:4] y.iris - iris[,5] y.iris [1] setosa setosa setosa setosa setosa setosa [7] setosa setosa setosa setosa setosa setosa [139] virginica virginica virginica virginica virginica virginica [145] virginica virginica virginica virginica virginica virginica Levels: setosa versicolor virginica I want like, y.iris level are, [] setosa versicolor virginica Muhammad, See ?unique and ?factor. Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Take a levels
probably you want: levels(y.iris) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Muhammad Subianto [EMAIL PROTECTED] To: R-help@stat.math.ethz.ch Sent: Wednesday, April 27, 2005 2:21 PM Subject: [R] Take a levels Dear all, How can I take a levels from a dataset. For example, data(iris) x.iris - iris[,1:4] y.iris - iris[,5] y.iris [1] setosa setosa setosa setosa setosa setosa [7] setosa setosa setosa setosa setosa setosa [139] virginica virginica virginica virginica virginica virginica [145] virginica virginica virginica virginica virginica virginica Levels: setosa versicolor virginica I want like, y.iris level are, [] setosa versicolor virginica Best regards, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Take a levels
Thanks you very much. levels(y.iris) Best regards, Muhammad Subianto [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Take a levels
Well... did you look at the help page on factors??? Did you even make a search? The answer is within your question: levels(iris[,5]) [1] setosa versicolor virginica Eric -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Muhammad Subianto Sent: mercredi 27 avril 2005 14:22 To: R-help@stat.math.ethz.ch Subject: [R] Take a levels Dear all, How can I take a levels from a dataset. For example, data(iris) x.iris - iris[,1:4] y.iris - iris[,5] y.iris [1] setosa setosa setosa setosa setosa setosa [7] setosa setosa setosa setosa setosa setosa [139] virginica virginica virginica virginica virginica virginica [145] virginica virginica virginica virginica virginica virginica Levels: setosa versicolor virginica I want like, y.iris level are, [] setosa versicolor virginica Best regards, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Time series indexes
On 4/26/05, Fernando Saldanha [EMAIL PROTECTED] wrote: I tried to assign values to specific elements of a time series and got in trouble. The code below should be almost self-explanatory. I wanted to assign 0 to the first element of x, but instead I assigned zero to the second element of x, which is not what I wanted. Is there a function that will allow me to do this without going into index arithmetic (which would be prone to errors)? FS x- ts(c(1,2,3,4), start = 2) x Time Series: Start = 2 End = 5 Frequency = 1 [1] 1 2 3 4 x[2] - 0 x Time Series: Start = 2 End = 5 Frequency = 1 [1] 1 0 3 4 Any of the following will do it: x[1] - 0 # or window(x, start = 2, end = 2) - 0 # or since time 2 is at the beginning: window(x, end = 2) - 0 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using e1071 svm: NA/NaN/Inf in foreign function call
On Wed, 27 Apr 2005 10:25:55 +0100 João Mendes Moreira wrote: My mistake. I am sending the ImageBeforeError.RData file. No, no, no! Please the read the posting guide and please read the answers that were posted for you. As you obviously did not do that, let me read it to you again: Z Wouldn't it be possible to simply use a data set that is already available in R, *please*? /Z The solution is definitely not to send a huge data file (6.5M) to those who offered advice and to the subsribers of R-help (where it does not get through anyway, I think). If it is really data-dependent, then you might post the data on the web, but even then it is not very helpful to post a file in which there are dozens of objects when all you need is a data frame. To reproduce the error you must load the file and then to do: library(e1071) model - do.call(learner,learner.pars) I am using nu = 0.7. At this moment I do not get an error but the svm function blocks. It was al that night running without results. So there is no error as you claim above (and as you claimed in your previous mail). This is just to report the fact, that your computations are still running. Let me provide a simple reproducible example which does not involve spamming R-helpers with .RData files. You seem to want to report that the svm set.seed(1071) y - rnorm(100) x1 - rnorm(100) x2 - rnorm(100) svm(y ~ x1 + x2) can be fitted very quickly whereas svm(y ~ x1 + x2, cost = 4096, kernel = polynomial, degree = 4) takes much longer. The decisive parameter here is the cost parameter which is unusually large. I'm not sure why the algorithm gets so slow, but you might also want to check whether a cost parameter of the magnitude is appropriate. The other parameter which is important is the degree of the polynomial kernel, in which the complexity is also increasing. So the message is: Be careful in the selection of the hyperparameters of the SVM. Maybe someone else on the list can provide more insight on guidelines for choosing the hyperparameters of a polynomial kernel SVM. Z Using other kernels it used to finish in a few seconds. I have done already thousands of tests with other kernels. Only with the polynomial one I am not able to get results. Thanks for any help. Joao - Original Message - From: Achim Zeileis [EMAIL PROTECTED] To: João Mendes Moreira [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Tuesday, April 26, 2005 4:09 PM Subject: Re: [R] Error using e1071 svm: NA/NaN/Inf in foreign function call On Tue, 26 Apr 2005 15:46:20 +0100 João Mendes Moreira wrote: Hello, As far I saw in archive mailing list, I am not the first person with this problem. Anyway I was not able to pass this error once the information I got from the archive it is not very conclusive for this case. I have used linear, radial and sigmoid kernels for the same data in the same conditions and everything is ok. This problem just happens with the polynomial kernel. I send the debuging result from a reproducible example. The error message is at the end. I receive a different error message: Error in eval(expr, envir, enclos) : Object Fim not found So much for the reproducibility... Wouldn't it be possible to simply use a data set that is already available in R, *please*? Anyways, it seems that your specification of `nu' causes the problem: 0 might be a little bit too small. Z __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Split dataset as factor and numeric
Dear all, If I have dataset like, A125 B 2 AA0 C 0 A11 B 3 A10 C 3 B 0 A145 A17 C 0 B 0 A10 B 6 B 3 Is there any function to split like, A125 A11 A10 A145 A17 A10 or AA0 C 0 A10 B 0 B 0 C 0 B 0 A10 Thanks, Jan Sabee __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R for Windows MSI Installation File
Hi, Just wondering if anyone has created (or knows where one is available from) an MSI installation file for R v2.0.1? Many thanks, Richard. Richard Abraham Senior Information Systems Officer Department of Social Medicine, University of Bristol Tel: 44(0)117 928 7254 Fax: 44(0)117 928 7265 Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Split dataset as factor and numeric
Jan Sabee wrote on 4/27/2005 5:58 AM: Dear all, If I have dataset like, A125 B 2 AA0 C 0 A11 B 3 A10 C 3 B 0 A145 A17 C 0 B 0 A10 B 6 B 3 Is there any function to split like, A125 A11 A10 A145 A17 A10 or AA0 C 0 A10 B 0 B 0 C 0 B 0 A10 Thanks, Jan Sabee Jan, ?subset subset(x, V1 == A1) subset(x, V2 == 0) HTH, --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] date format
Hi, I'm trying to convert a vector containing dates in character format (dd/mm/yy): mdy.date (from date package) seems to be able to do that, but it returns to me a vector containing julian dates... but negative! for example: 16/12/03 is converted into -20470 it is because R recognizes year ../03 as 1903, instead of 2003, but how can I do to solve this problem? (of course, I could add 36525 to each data, but it's not very elegant) Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] lattice plot problem!
On Tuesday 26 April 2005 21:52, Ivy_Li wrote:
Hello everybody,
Could I consult you two questions?
Recently I write some code about lattice plot.
1) bwplot function
I know the lattice default background color is grey and the box
color is green, but I don't like the color. So I change the
background color to white use the
trellis.device(bg=white)
That doesn't (and is not intended to) change the lattice background
color to white. If it does, you are using an old version of R, please
consider upgrading. In recent versions, the help page for
trellis.device discusses this issue in some detail. Most of the
commentary applies to earlier versions as well; if you are unwilling to
upgrade R, you can still read it at
http://cran.r-project.org/doc/packages/lattice.pdf
then I modify the
panel=function(...)
+{
+ panel.bwplot(col=black...)
+ }
But I find the the box color is still green, just change the color
of point which is in the box, I guss it is the mean or median value
of the data. So How to change the color of box, and how to change the
shape of the point in the box. I want it like other boxplot, like a
horizontal line.
2) In every lattices, I want to add a mean line of the group. I try
the
panel.abline(h)
But I find in very lattice add the same line. Then I try the
panel.linejoin etc. But I can not get my wish plot.
Tom Mulholland has already responded to this part.
Deepayan
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Re: [R] date format
you could use as.Date(), i.e., ss - 16/12/03 as.Date(ss, %d/%m/%y) julian(as.Date(ss, %d/%m/%y)) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: alessandro carletti [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Wednesday, April 27, 2005 3:08 PM Subject: [R] date format Hi, I'm trying to convert a vector containing dates in character format (dd/mm/yy): mdy.date (from date package) seems to be able to do that, but it returns to me a vector containing julian dates... but negative! for example: 16/12/03 is converted into -20470 it is because R recognizes year ../03 as 1903, instead of 2003, but how can I do to solve this problem? (of course, I could add 36525 to each data, but it's not very elegant) Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] date format
You haven't even begun to tell us how you are doing this. R does not itself convert dates to numbers, to wit: as.Date(16/12/03, %d/%m/%y) [1] 2003-12-16 Here's one way: x - as.Date(16/12/03, %d/%m/%y) xx - as.POSIXlt(x) xx$year - xx$year-100 as.Date(xx) On Wed, 27 Apr 2005, alessandro carletti wrote: Hi, I'm trying to convert a vector containing dates in character format (dd/mm/yy): mdy.date (from date package) seems to be able to do that, but it returns to me a vector containing julian dates... but negative! for example: 16/12/03 is converted into -20470 it is because R recognizes year ../03 as 1903, instead of 2003, but how can I do to solve this problem? (of course, I could add 36525 to each data, but it's not very elegant) Not to say wrong. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R for Windows MSI Installation File
Richard Abraham wrote: Hi, Just wondering if anyone has created (or knows where one is available from) an MSI installation file for R v2.0.1? Many thanks, Richard. The setup program build by Inno Setup, as you find it in the binaries section of CRAN is much better than what you would get with the MSI installer, and it has all required features. Best, Philippe Grosjean Richard Abraham Senior Information Systems Officer Department of Social Medicine, University of Bristol Tel: 44(0)117 928 7254 Fax: 44(0)117 928 7265 Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R for Windows MSI Installation File
On Wed, 27 Apr 2005, Richard Abraham wrote: Brian, I can use an MSI installer package to install R on any number of Windows PCs without leaving my desk by using Windows' Group Policy. Is this possible using R's own scriptable installer? The FAQ doesn't say. Well, did you look at the reference given there? Professor Ripley. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to add some of data in the first place dataset
Dear R-help, First I apologize if my question is quite simple. I need add some of data in the first place my dataset, how can I do that. I have tried with rbind, but I did not succes. 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose For example, data(iris) iris[1:10,] Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa 7 4.6 3.4 1.4 0.3 setosa 8 5.0 3.4 1.5 0.2 setosa 9 4.4 2.9 1.4 0.2 setosa 10 4.9 3.1 1.5 0.1 setosa The result something like this, 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose 5.1 3.5 1.4 0.2 setosa 4.9 3.0 1.4 0.2 setosa 4.7 3.2 1.3 0.2 setosa 4.6 3.1 1.5 0.2 setosa 5.0 3.6 1.4 0.2 setosa 5.4 3.9 1.7 0.4 setosa 4.6 3.4 1.4 0.3 setosa 5.0 3.4 1.5 0.2 setosa 4.4 2.9 1.4 0.2 setosa 4.9 3.1 1.5 0.1 setosa Sincerely, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R for Windows MSI Installation File
Brian, I can use an MSI installer package to install R on any number of Windows PCs without leaving my desk by using Windows' Group Policy. Is this possible using R's own scriptable installer? The FAQ doesn't say. Thanks, Rich. --On 27 April 2005 14:17 +0100 Prof Brian Ripley [EMAIL PROTECTED] wrote: Note that I know of: why would one want a lower-quality installer than that provided? The installer is scriptable: see the rw-FAQ. On Wed, 27 Apr 2005, Richard Abraham wrote: Just wondering if anyone has created (or knows where one is available from) an MSI installation file for R v2.0.1? Many thanks, Richard. Of course, 2.1.0 is current. -- 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 Richard Abraham Senior Information Systems Officer Department of Social Medicine, University of Bristol Tel: 44(0)117 928 7254 Fax: 44(0)117 928 7265 Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Time series indexes
Thanks, Gabor. Reading your suggestion (and a previous one as well) I
realized I surely expressed myself quite badly when asking the
question.
Luckily one person privately suggested the following solution, which
is exactly what I was looking for:
x[time(x)==2] - 0
This works wonderfully. However, I tested it and it is very slow. So I
am back to index arithmetic, which is fast:
x[2 - start(x)[1] + 1] - 0
However, this is much more prone to errors than the first solution above.
I also tried to define get() and set() functions for time series as follows:
ts.get - function(myts, i) { myts[i - start(myts)[1] + 1] }
ts.set - function(myts, i, val) { myts[i - start(myts)[1] + 1] - val; myts }
Both of these work, and ts.get is quite useful. However, ts.set is
again slow due to the need to copy the whole object. So ts.set cannot
be used just as
ts.set(x, i, val)
but rather one must write
x - ts.set(x, i, val)
In the definition of ts.set, the expression after the semicolon
returns a modified copy of the argument myts. It has to be done this
way (please let me know if there is a better and fast way) since in R
there is no way of passing arguments by reference. That is, one cannot
have side effects on the arguments of a function.
Cheers,
FS
On 4/27/05, Gabor Grothendieck [EMAIL PROTECTED] wrote:
On 4/26/05, Fernando Saldanha [EMAIL PROTECTED] wrote:
I tried to assign values to specific elements of a time series and got
in trouble. The code below should be almost self-explanatory. I wanted
to assign 0 to the first element of x, but instead I assigned zero to
the second element of x, which is not what I wanted. Is there a
function that will allow me to do this without going into index
arithmetic (which would be prone to errors)?
FS
x- ts(c(1,2,3,4), start = 2)
x
Time Series:
Start = 2
End = 5
Frequency = 1
[1] 1 2 3 4
x[2] - 0
x
Time Series:
Start = 2
End = 5
Frequency = 1
[1] 1 0 3 4
Any of the following will do it:
x[1] - 0
# or
window(x, start = 2, end = 2) - 0
# or since time 2 is at the beginning:
window(x, end = 2) - 0
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[R] Error list and debugging R code
Hi, I am trying to debug my code and looking for any tool to help me out with it. My main problem is with the error messages I can not figure out where they come from (what function produced them) and what do they mean. Is there such a think as list of error messages produced by R and standard R packages, and what do they mean? Or any tool to extract information which function produced the run-time error (a call stack would be great)? I looked into debug and trace tools but you seem to have to know where the problem is to use them. Jarek =\ Jarek Tuszynski, PhD. o / \ Science Applications International Corporation \__,| (703) 676-4192 \ [EMAIL PROTECTED] `\ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to add some of data in the first place dataset
Try this: x- matrix(1:12, nrow = 4, byrow = T) y - matrix (13:24, nrow=4, by.row=T) To add the rows of y before the rows of x: rbind(y,x) -- Tyler Smith __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to add some of data in the first place dataset
Muhammad Subianto said the following on 2005-04-27 15:48: Dear R-help, First I apologize if my question is quite simple. I need add some of data in the first place my dataset, how can I do that. I have tried with rbind, but I did not succes. Can you send a reproducible example where rbind didn't succeed? 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose For example, data(iris) iris[1:10,] Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa 7 4.6 3.4 1.4 0.3 setosa 8 5.0 3.4 1.5 0.2 setosa 9 4.4 2.9 1.4 0.2 setosa 10 4.9 3.1 1.5 0.1 setosa Assume that you have the new data in a `data.frame', e.g. (new.data - read.table(clipboard, header = FALSE, col.names = c(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width, Species))) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 0.1 3.6 0.4 0.9rose 2 4.1 4.0 1.2 1.2rose 3 4.4 3.2 1.9 0.5rose 4 4.6 1.1 1.1 0.2rose Then, add.data - rbind(new.data, iris) will do the trick. Confirm this by add.data[1:10, ] Sepal.Length Sepal.Width Petal.Length Petal.Width Species 10.1 3.6 0.4 0.9rose 24.1 4.0 1.2 1.2rose 34.4 3.2 1.9 0.5rose 44.6 1.1 1.1 0.2rose 151 5.1 3.5 1.4 0.2 setosa 210 4.9 3.0 1.4 0.2 setosa 310 4.7 3.2 1.3 0.2 setosa 410 4.6 3.1 1.5 0.2 setosa 55.0 3.6 1.4 0.2 setosa 65.4 3.9 1.7 0.4 setosa HTH, Henric The result something like this, 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose 5.1 3.5 1.4 0.2 setosa 4.9 3.0 1.4 0.2 setosa 4.7 3.2 1.3 0.2 setosa 4.6 3.1 1.5 0.2 setosa 5.0 3.6 1.4 0.2 setosa 5.4 3.9 1.7 0.4 setosa 4.6 3.4 1.4 0.3 setosa 5.0 3.4 1.5 0.2 setosa 4.4 2.9 1.4 0.2 setosa 4.9 3.1 1.5 0.1 setosa __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] libz library missing while installing RMySQL
Thanks Jari although these libraries are installed now, I get the following message: Configuration error: could not find the MySQL installation include and/or library directories. Manually specify the location of the MySQL libraries and the header files and re-run R CMD INSTALL. Somebody has had this problem before but that did not help me. Can somebody give me a hint? Sebastian On Wednesday 27 April 2005 10:40, you wrote: On Wed, 2005-04-27 at 09:43 +0200, Sebastian Leuzinger wrote: Hello Trying to install the MySQL package, I get the following error. The help archive contains something on this issue but did not help. I work on linux suse 9.3 Configuration error: Could not locate the library libz required by MySQL. The library libz however is not on any mirrors I checked. Sebastian, You made some R core people very happy: they have tried very hard to explain that there is a difference between a *library* and a *package*, and that this difference really matters. This seems to be the first case on R-News when this distinction really seems to matter: you need a library. I don't know about SuSE (or with some other really weird capitalization), but in my system libz belongs to package zlib that you must install, and in my system you'd probably need zlib-devel as well (this is FC3). So they are system level libraries that come with SuSe instead of R packages. cheers, jari oksanen -- Sebastian Leuzinger Institute of Botany, University of Basel Schnbeinstr. 6 CH-4056 Basel ph0041 (0) 61 2673511 fax 0041 (0) 61 2673504 email [EMAIL PROTECTED] web http://pages.unibas.ch/botschoen/leuzinger __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to add some of data in the first place dataset
If you want to add rows to a data frame using rbind, your additional rows must be in a data frame with the same (column) names. R data( iris ) R a - data.frame( Sepal.Length=c(1:4), Sepal.Width=c(2:5), Petal.Length=c(3:6), Petal.Width=c(4:7), Species=rep(rosa,4)) R b - iris[1:10,] R rbind(a,b) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 1.0 2.0 3.0 4.0rosa 2 2.0 3.0 4.0 5.0rosa 3 3.0 4.0 5.0 6.0rosa 4 4.0 5.0 6.0 7.0rosa 11 5.1 3.5 1.4 0.2 setosa 21 4.9 3.0 1.4 0.2 setosa 31 4.7 3.2 1.3 0.2 setosa 41 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa 7 4.6 3.4 1.4 0.3 setosa 8 5.0 3.4 1.5 0.2 setosa 9 4.4 2.9 1.4 0.2 setosa 10 4.9 3.1 1.5 0.1 setosa Arne On Wednesday 27 April 2005 15:48, Muhammad Subianto wrote: Dear R-help, First I apologize if my question is quite simple. I need add some of data in the first place my dataset, how can I do that. I have tried with rbind, but I did not succes. 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose For example, data(iris) iris[1:10,] Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa 7 4.6 3.4 1.4 0.3 setosa 8 5.0 3.4 1.5 0.2 setosa 9 4.4 2.9 1.4 0.2 setosa 10 4.9 3.1 1.5 0.1 setosa The result something like this, 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose 5.1 3.5 1.4 0.2 setosa 4.9 3.0 1.4 0.2 setosa 4.7 3.2 1.3 0.2 setosa 4.6 3.1 1.5 0.2 setosa 5.0 3.6 1.4 0.2 setosa 5.4 3.9 1.7 0.4 setosa 4.6 3.4 1.4 0.3 setosa 5.0 3.4 1.5 0.2 setosa 4.4 2.9 1.4 0.2 setosa 4.9 3.1 1.5 0.1 setosa Sincerely, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Arne Henningsen Department of Agricultural Economics University of Kiel Olshausenstr. 40 D-24098 Kiel (Germany) Tel: +49-431-880 4445 Fax: +49-431-880 1397 [EMAIL PROTECTED] http://www.uni-kiel.de/agrarpol/ahenningsen/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to break data in quantiles properly?
Hi,
I would like to break a dataset in n.classes quantiles.
Till now, I used the following code:
Classify.Quantile - function (dataset, nclasses = 10)
{
n.probs - seq(0,1,length=nclasses+1)
n.labels = paste(C, 1:nclasses-1, sep=)
n.rows - nrow(dataset)
n.cols - ncol(dataset)
n.motif - dataset
for (j in 2:n.cols)
{
cat(j, );
discr =
n.labels[unclass(cut(dataset[,j],quantile(dataset[,j],n.probs),include.lowest=T))]
n.motif[,j] = discr
}
res - list(motif=n.motif, labels=n.labels, n.classes=nclasses)
return(res)
}
but if you try to call this with a dataset with a lot of same value, you got a
Error in cut.default(dataset[, j], quantile(dataset[, j], n.probs),
include.lowest = T) :
cut: breaks are not unique
I perfectly understand why but I would like to know how to avoid this behaviour.
for e.g., use this code to raise the error:
x=matrix(0,1000,1)
x[100]=1
Classify.Quantile(x, 10)
of course this dataset is a bit extreme but it happens to get data
with very small variance.
Thanks for any help you could provide
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[R] Re: How to add some of data in the first place dataset
Thanks all for your help. Kind regards, Muhammad Subianto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error list and debugging R code
Tuszynski, Jaroslaw W. wrote: Hi, I am trying to debug my code and looking for any tool to help me out with it. My main problem is with the error messages I can not figure out where they come from (what function produced them) and what do they mean. Is there such a think as list of error messages produced by R and standard R packages, and what do they mean? Or any tool to extract information which function produced the run-time error (a call stack would be great)? I looked into debug and trace tools but you seem to have to know where the problem is to use them. See any good book about R programming, the manuals, R News, and the following help pages: ?traceback ?debug ?recover ?browser and other relevant pages these pages are pointing to. Uwe Ligges Jarek =\ Jarek Tuszynski, PhD. o / \ Science Applications International Corporation \__,| (703) 676-4192 \ [EMAIL PROTECTED] `\ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fitting a kind of Proportional Odds Modell using nlme, polr, lrm or ordgee
Hello,
I'm trying to fit a special kind of proportional odds model from:
Whitehead et al. (2001). Meta-analysis of ordinal outcome using
individual patient data. Statistics in medicine 20: 2243-2260. (model 2)
The data are as follows:
library(nlme)
library(geepack)
library(Design)
library(MASS)
options(contrasts=c(contr.SAS,contr.poly))
counts -
c(2,22,54,29,3,4,23,45,22,2,1,22,35,11,3,14,119,180,54,6,7,16,17,
10,3,13,20,24,10,1,8,24,73,52,13,21,106,175,62,17,2,13,18,7,1,3,14,19,3,0)
category - c(rep(1:5,10))
treatment - (rep(c(0,0,0,0,0,1,1,1,1,1,),5))
study - gl(5,10)
percoftotal - counts/sum(counts)
cout - data.frame(study,treatment,category,counts,percoftotal)
The data are from five controlled clinical trials (study). The patients
were given placebo (treatment=0) or treatment (treatment=1) and had a
response in one of five categories.
Now I want to fit the model given by Whitehead et al. This assumes
proportional odds between treatments, but stratifies by study. This
means that the cut-off points associated with the distribution of the
underlying latent variable for determining the response category are
allowed to vary from study to study but are the same for both treatment
groups within a study.
It is given by
log(Q_{ijk}/(1-Q_{ijk}) = \alpha_{ik} + \beta*x_{ij} (k=1,...,m-1)
This model can be considered as arising from a latent continuos variable
. Assume that the response of the j-th subject in study i is truly equal
to G_{ij} although this latent reponse will never be observerd.
G_{ij} has a logistic distribution with:
Q_{ijk} = P(G_{ij} = \alpha_{ik}) =
1/(1+exp(-(\alpha_{ik}+\beta*x_{ij}) (k=1,...,m-1)
in which Q_{ijk} is the probabilty of having a response for the j-th
subject in category k or better that means p_{ij1}++p_{ijk}= Q_{ijk}
and Q_{ijm}=1, Q_{ij1}=1.
i = 1,...,r (r=5) #number of studies
j = 1,,n_{i} #subject j from study i
k = 1,...,m (m=5)#Category
My problem is how to fit this model in R although I have a SAS code from
Whitehead which is availabe at
http://www.rdg.ac.uk/mps/mps_home/misc/publications.htm . There it is
fit by using Proc NLMIXED. The result for beta is there:
The problem is that alpha_{ik} represents the k_th intercept for the
i_th study. A normal Proportional-Odds model
has only alpha_k which represents the k-th intercept without any
associaton to the study.
I tried to fit it by using the functions of Pinheiro/ Bates nlme, nls,
gnls;
polr by Venables/Ripley ;
lrm by Harrell
ordgee by Yan;
but I came to no conclusion.
I would be very happy if anyone could help me.
Thank you VERY VERY much in advance.
Kindly regards,
Henning Henke
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[R] Data() and CSV files
Hello, For reasons I don't understand, data() imports CSV (Comma-Separated Values) as if they were delimited by semicolons instead of commas. (Are semicolon-separated Comma-Separated-Value files common somewhere?) Given that this is the case, if I choose to put comma-delimited CSV files in my data directory, what is the preferred method of loading these into memory? data(filename) would be nice, but not applicable given the above conversion issue. So, this was the best I came up with: read.csv(file.path(.find.package(pkg), data, paste(filename, csv, sep = .))) However, given that others undoubtedly like to include (non-semicolon) .csv files in their packages and load them easily, I would like to know if there is a more elegant way to load these files. Perhaps an annotation in the data/00Index or /data/datalist file that I am unaware of? Thanks, Robert __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Error list and debugging R code
I have found the tools listed by Uwe to be sufficient for my needs, but you may also be interested in Mark Bravington's debug package. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Uwe Ligges Sent: Wednesday, April 27, 2005 8:00 AM To: Tuszynski, Jaroslaw W. Cc: ([EMAIL PROTECTED]) Subject: Re: [R] Error list and debugging R code Tuszynski, Jaroslaw W. wrote: Hi, I am trying to debug my code and looking for any tool to help me out with it. My main problem is with the error messages I can not figure out where they come from (what function produced them) and what do they mean. Is there such a think as list of error messages produced by R and standard R packages, and what do they mean? Or any tool to extract information which function produced the run-time error (a call stack would be great)? I looked into debug and trace tools but you seem to have to know where the problem is to use them. See any good book about R programming, the manuals, R News, and the following help pages: ?traceback ?debug ?recover ?browser and other relevant pages these pages are pointing to. Uwe Ligges Jarek =\ Jarek Tuszynski, PhD. o / \ Science Applications International Corporation \__,| (703) 676-4192 \ [EMAIL PROTECTED] `\ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R/Splus--Perl Interface ssh
Hi: I'm using RSPerl_0.6-3 calling R from Perl under UNIX system. My perl programs with the RSPerl work well in my computer. If submitting these programs to a UNIX system by 'ssh' or to GNQS system by 'qsub', these programs do not work even though both systems can run R. Details are following. Any suggestion will be highly appreciated. Xiao My perl program 'tt.pl' #!/usr/bin/perl -w use R; use RReferences; R::initR(--silent); R::setDebug(0); R::library(RSPerl); @t = R::call(min, (1,2)); I have a script 'doRSPerltt' for the perl program #!/bin/csh setenv LD_LIBRARY_PATH /usr/lib/R/lib setenv R_HOME /usr/lib/R perl -I/usr/lib/R/site-library/RSPerl/examples/../share/blib/arch -I/usr/lib/R/site-library/RSPerl/examples/../share/blib/lib -I/usr/lib/R/site-library/RSPerl/scripts t t.pl I used ssh to submit my job to machine 'queen' ssh queen 'cd /nfs/fs/clarke/xiaoliu ./doRSPerltt' An error message returns Fatal error: you must specify `--save', `--no-save' or `--vanilla' If using 'qsub' to submit my job to a GNQS system qsub doRSPerltt An error message returns Warning: no access to tty (Bad file descriptor). Thus no job control in this shell. Fatal error: you must specify `--save', `--no-save' or `--vanilla' XIAO LIU wrote: Hi: I'm using RSPerl_0.6-3 calling R from Perl under UNIX system. My perl programs with the RSPerl work well in my computer. However, if submitting to GNQS system, these programs do not work. It requires defining '--vanilla'. Can anyone tell me what I should do? Are you certain this has something to do with RSPerl. Specifically, have you tried running an R program on a GNQS system and ensuring that works without --vanilla (as part of the command line arguments)? If it is only for RSPerl, perhaps you would do well to post the error message that is generated. Thank in advance Xiao __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Duncan Temple Lang[EMAIL PROTECTED] Department of Statistics work: (530) 752-4782 371 Kerr Hall fax: (530) 752-7099 One Shields Ave. University of California at Davis Davis, CA 95616, USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to break data in quantiles properly?
Eric Rodriguez wrote:
Hi,
I would like to break a dataset in n.classes quantiles.
Till now, I used the following code:
Classify.Quantile - function (dataset, nclasses = 10)
{
n.probs - seq(0,1,length=nclasses+1)
n.labels = paste(C, 1:nclasses-1, sep=)
n.rows - nrow(dataset)
n.cols - ncol(dataset)
n.motif - dataset
for (j in 2:n.cols)
{
cat(j, );
discr = n.labels[unclass(cut(dataset[,j],quantile(dataset[,j],n.probs),include.lowest=T))]
n.motif[,j] = discr
}
res - list(motif=n.motif, labels=n.labels, n.classes=nclasses)
return(res)
}
but if you try to call this with a dataset with a lot of same value, you got a
Error in cut.default(dataset[, j], quantile(dataset[, j], n.probs),
include.lowest = T) :
cut: breaks are not unique
I perfectly understand why but I would like to know how to avoid this behaviour.
for e.g., use this code to raise the error:
x=matrix(0,1000,1)
x[100]=1
Classify.Quantile(x, 10)
of course this dataset is a bit extreme but it happens to get data
with very small variance.
Thanks for any help you could provide
The cut2 function in the Hmisc package may help. -FH
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
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Re: [R] Data() and CSV files
McGehee, Robert [EMAIL PROTECTED] writes: Hello, For reasons I don't understand, data() imports CSV (Comma-Separated Values) as if they were delimited by semicolons instead of commas. (Are semicolon-separated Comma-Separated-Value files common somewhere?) Given that this is the case, if I choose to put comma-delimited CSV files in my data directory, what is the preferred method of loading these into memory? Semicolon-separated CSV's are common in various parts of Europe, but usually coincident with comma as decimal separator (cf. read.csv/read.csv2). I can't remember whether there was a reason for the choice in data(). data(filename) would be nice, but not applicable given the above conversion issue. So, this was the best I came up with: read.csv(file.path(.find.package(pkg), data, paste(filename, csv, sep = .))) However, given that others undoubtedly like to include (non-semicolon) .csv files in their packages and load them easily, I would like to know if there is a more elegant way to load these files. Perhaps an annotation in the data/00Index or /data/datalist file that I am unaware of? One trick is that .R files are processed before other files, so a mydata.R file containing mydata - read.csv(mydata.csv) should do the trick. It messes with lazy loading of data sets though. An alternative is to preprocess the data set to (say) .Rda format. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Recursive calculation of a series of values
Dear R-users, I'm felling kind of blocked on a quite simple problem and I wonder if someone could give me a help with it. My problem: x[0] = 100 x[1] = (1+v[1])*x[0] x[2] = (1+v[2])*x[1] ... i.e. x[i] = (1+v[i])*x[i-1] and x[0]=k Given a set of v values I wanted to obtain the corresponding x values in an efficient way (i.e. without a for loop). For instance, if x[0] = 100 and v = c(0.2,-0.1,0.05) then I would get x = c(120,108,113.4) I'm almost sure the function filter() from package tseries is the key for getting these values but I'm really blocked. Any help is much appreciated. Lus Torgo -- Luis Torgo FEP/LIACC, University of Porto Phone : (+351) 22 339 20 93 Machine Learning Group Fax : (+351) 22 339 20 99 R. de Ceuta, 118, 6o email : [EMAIL PROTECTED] 4050-190 PORTO - PORTUGALWWW : http://www.liacc.up.pt/~ltorgo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Question about R initial message
Dear R: First of all, nothing is wrong! I just had a question about the following Natural language support but running in an English locale What does than mean, please? Thanks in Advance R 2.1.0 Windows Laura Holt mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Recursive calculation of a series of values
Algebra: cumprod(1+v)*x[0] -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Luis Torgo Sent: Wednesday, April 27, 2005 7:42 AM To: r-help@stat.math.ethz.ch Subject: [R] Recursive calculation of a series of values Dear R-users, I'm felling kind of blocked on a quite simple problem and I wonder if someone could give me a help with it. My problem: x[0] = 100 x[1] = (1+v[1])*x[0] x[2] = (1+v[2])*x[1] ... i.e. x[i] = (1+v[i])*x[i-1] and x[0]=k Given a set of v values I wanted to obtain the corresponding x values in an efficient way (i.e. without a for loop). For instance, if x[0] = 100 and v = c(0.2,-0.1,0.05) then I would get x = c(120,108,113.4) I'm almost sure the function filter() from package tseries is the key for getting these values but I'm really blocked. Any help is much appreciated. Luís Torgo -- Luis Torgo FEP/LIACC, University of Porto Phone : (+351) 22 339 20 93 Machine Learning Group Fax : (+351) 22 339 20 99 R. de Ceuta, 118, 6o email : [EMAIL PROTECTED] 4050-190 PORTO - PORTUGALWWW : http://www.liacc.up.pt/~ltorgo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] its package: inexplicable date-shifting ?!
Can someone please explain to me why
the dates get shifted by one day
when I create an its ( irregular time-series )
object from a matrix for which I've
assigned row names.
E.g. in the example run below,
why does the its object have dates
one-shifted from my original dates?
install.packages('its')
install.packages('Hmisc')
require(its)
m - matrix(1:2, nrow=2)
m
[,1]
[1,]1
[2,]2
its.format('%Y%m%d')
[1] %Y%m%d
rownames(m) - c('20040813', '20040814')
m
[,1]
200408131
200408142
its(structure(m))
1
20040812 1
20040813 2
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Berton Gunter
Sent: Wednesday, April 27, 2005 1:28 PM
To: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: RE: [R] Recursive calculation of a series of values
Algebra:
cumprod(1+v)*x[0]
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Luis Torgo
Sent: Wednesday, April 27, 2005 7:42 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Recursive calculation of a series of values
Dear R-users,
I'm felling kind of blocked on a quite simple problem and I wonder if
someone could give me a help with it.
My problem:
x[0] = 100
x[1] = (1+v[1])*x[0]
x[2] = (1+v[2])*x[1]
...
i.e.
x[i] = (1+v[i])*x[i-1]
and x[0]=k
Given a set of v values I wanted to obtain the corresponding
x values in
an efficient way (i.e. without a for loop).
For instance, if x[0] = 100 and v = c(0.2,-0.1,0.05) then I would get
x = c(120,108,113.4)
I'm almost sure the function filter() from package tseries is the key
for getting these values but I'm really blocked.
Any help is much appreciated.
Luís Torgo
--
Luis Torgo
FEP/LIACC, University of Porto Phone : (+351) 22 339 20 93
Machine Learning Group Fax : (+351) 22 339 20 99
R. de Ceuta, 118, 6o email : [EMAIL PROTECTED]
4050-190 PORTO - PORTUGALWWW :
http://www.liacc.up.pt/~ltorgo
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
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Re: [R] Order of boxes in boxplot()
On Thu, 7 Apr 2005, michael watson (IAH-C) wrote: Sorry for such an inane question - how do I control the order in which the boxes are plotted using boxplot() when I pass it a formula and a data.frame? It seems that the groups are plotted in alphabetical order... I want to change this Mick, Here's the code I use to order boxes by decreasing median value. SubtDays is variable of interest ConChnl is original grouping factor. tMedians is a temp data frame dConChnl is new grouping factor with desired order boxplot(SubtDays ~ ConChnl, . ### Default ordering of boxes tMedians - aggregate(SubtDays, list(ConChnl), median, na.rm = TRUE) dConChnl - factor(ConChnl, levels = tMedians[order(tMedians$x), 1]) boxplot(SubtDays ~ dConChnl, . ### Ordered by decreasing median HTH, Jim Porzak Director of Analytics Loyalty Matrix, Inc. (415) 296-1141 x210 R.LoyaltyMatrix.com www.LoyaltyMatrix.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Density curve over a histogram
Dear All I would like to draw a picture with the density curve of a normal distribution over a histogram of a set of random numbers extracted from the same normal distribution. Is that possible? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R/Splus--Perl Interface ssh
On Wed, 27 Apr 2005, XIAO LIU wrote: I'm using RSPerl_0.6-3 calling R from Perl under UNIX system. My perl programs with the RSPerl work well in my computer. If submitting these programs to a UNIX system by 'ssh' or to GNQS system by 'qsub', these programs do not work even though both systems can run R. Details are following. Any suggestion will be highly appreciated. My perl program 'tt.pl' #!/usr/bin/perl -w use R; use RReferences; R::initR(--silent); Hint: that is not correct for batch use. Please do read `An Introduction to R' for how to invoke R appropriately. [...] Please do read the posting guide and choose an less inappropriate list. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Question about R initial message
On Wed, 27 Apr 2005, Laura Holt wrote: I just had a question about the following Natural language support but running in an English locale What does than mean, please? See the `R Installation and Administration' manual, which does explain this. In short it means that your version of R has NLS aka internationalization support but you are not making use of it in this session. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density curve over a histogram
On Wed, 27 Apr 2005 19:06:07 +0100 Paul Smith wrote: Dear All I would like to draw a picture with the density curve of a normal distribution over a histogram of a set of random numbers extracted from the same normal distribution. Is that possible? To quote Simon `Yoda' Blomberg: This is R. There is no if. Only how. (see fortune(Yoda)) Try: R x - rnorm(100) R hist(x, freq = FALSE) R curve(dnorm, col = 2, add = TRUE) Z Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density curve over a histogram
Paul Smith wrote: Dear All I would like to draw a picture with the density curve of a normal distribution over a histogram of a set of random numbers extracted from the same normal distribution. Is that possible? Yes. If you like to know how, see e.g. ?hist and ?curve. Uew Ligges Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Question about R initial message
Laura Holt [EMAIL PROTECTED] writes: Dear R: First of all, nothing is wrong! I just had a question about the following Natural language support but running in an English locale What does than mean, please? $ LANG=pt_BR.UTF8 ~/r-devel/BUILD/bin/R R : Copyright 2005, The R Foundation for Statistical Computing Version 2.2.0 Under development (unstable) (2005-04-24), ISBN 3-900051-07-0 R é um software livre e vem sem GARANTIA ALGUMA. Você pode redistribuí-lo sob certas circunstâncias. Digite 'licence()' ou 'licence()' para detalhes de distribuição. R é um projeto colaborativo com muitos contribuidores. Digite 'contributors()' para obter mais informações e 'citation()' para saber como citar o R ou pacotes do R em publicações. Digite 'demo()' para demonstrações, 'help()' para o sistema on-line de ajuda, ou 'help.start()' para abrir o sistema de ajuda em HTML no seu navegador. Digite 'q()' para sair do R. Got it? -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density curve over a histogram
Le 27 Avril 2005 14:06, Paul Smith a écrit : I would like to draw a picture with the density curve of a normal distribution over a histogram of a set of random numbers extracted from the same normal distribution. Is that possible? Sure. See curve() with add=TRUE. Don't forget to use prob=TRUE when plotting your histogram, though. Vincent __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density curve over a histogram
On 4/27/05, Achim Zeileis [EMAIL PROTECTED] wrote: I would like to draw a picture with the density curve of a normal distribution over a histogram of a set of random numbers extracted from the same normal distribution. Is that possible? To quote Simon `Yoda' Blomberg: This is R. There is no if. Only how. (see fortune(Yoda)) Try: R x - rnorm(100) R hist(x, freq = FALSE) R curve(dnorm, col = 2, add = TRUE) Fantastic! Thanks a lot, Achim. Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Density curve over a histogram
Paul Smith [EMAIL PROTECTED] writes: Dear All I would like to draw a picture with the density curve of a normal distribution over a histogram of a set of random numbers extracted from the same normal distribution. Is that possible? Yes. If you look at the scripts that go with the ISwR package, you'll find a detailed example in ch01.R (end of 1.3/beginning of 1.4). Or you could read the book, of course... -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] its package: inexplicable date-shifting ?!
I don't use 'its' enough to really know but usually when you see dates off
by one and are using POSIXct (which has time zone support) -- POSIXct is
what
'its' uses, it signals that the time zone is different than what you
expected.
Try to specify the time zone explicitly if 'its' supports it. Another
possibility
is to use the 'zoo' package which can handle 'Date' class dates (as well as
'POSIXct' and others). Since 'Date' does not support time zones you can't
get into that sort of trouble in the first place.
See the article in RNews 4/1 on dates classes for background info on dates
in R.
On 4/27/05, Chalasani, Prasad [EMAIL PROTECTED] wrote:
Can someone please explain to me why
the dates get shifted by one day
when I create an its ( irregular time-series )
object from a matrix for which I've
assigned row names.
E.g. in the example run below,
why does the its object have dates
one-shifted from my original dates?
install.packages('its')
install.packages('Hmisc')
require(its)
m - matrix(1:2, nrow=2)
m
[,1]
[1,] 1
[2,] 2
its.format('%Y%m%d')
[1] %Y%m%d
rownames(m) - c('20040813', '20040814')
m
[,1]
20040813 1
20040814 2
its(structure(m))
1
20040812 1
20040813 2
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Berton Gunter
Sent: Wednesday, April 27, 2005 1:28 PM
To: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Subject: RE: [R] Recursive calculation of a series of values
Algebra:
cumprod(1+v)*x[0]
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Luis Torgo
Sent: Wednesday, April 27, 2005 7:42 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Recursive calculation of a series of values
Dear R-users,
I'm felling kind of blocked on a quite simple problem and I wonder if
someone could give me a help with it.
My problem:
x[0] = 100
x[1] = (1+v[1])*x[0]
x[2] = (1+v[2])*x[1]
...
i.e.
x[i] = (1+v[i])*x[i-1]
and x[0]=k
Given a set of v values I wanted to obtain the corresponding
x values in
an efficient way (i.e. without a for loop).
For instance, if x[0] = 100 and v = c(0.2,-0.1,0.05) then I would get
x = c(120,108,113.4)
I'm almost sure the function filter() from package tseries is the key
for getting these values but I'm really blocked.
Any help is much appreciated.
Luís Torgo
--
Luis Torgo
FEP/LIACC, University of Porto Phone : (+351) 22 339 20 93
Machine Learning Group Fax : (+351) 22 339 20 99
R. de Ceuta, 118, 6o email : [EMAIL PROTECTED]
4050-190 PORTO - PORTUGAL WWW :
http://www.liacc.up.pt/~ltorgo
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
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PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
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PLEASE do read the posting guide!
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[R] Closing RGui help windows
Hi. I often wind up with many help windows cluttering my RGui screen when running Windows R 2.0.1. Is there an R instruction to close one or more help windows, or an RGui command to close all help windows? Yours truly, /Ronnen. /P.S. E-mailed CC:s of posted replies appreciated. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] interval censoring case 1 or current status data
Dear all, Is there a function in R dealing with the NPMLE for current status data or interval censored case 1 data? Thanks, Jimmy [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Defining binary indexing operators
Assume we have a function like: foo - function(x, y) how is it possible to define a binary indexing operator, denoted by $, so that x$y functions the same as foo(x, y) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] assign to an element of a vector
I am trying to find a way to assign values to elements of a vector
that will be defined by a user. So I don't have the name of the vector
and cannot hard code the assignment in advance. In the example below I
have to get() the vector using its name. When I try to assign to an
element I get an error:
a - c(1,2,3)
get('a')[1] - 0
Error: Target of assignment expands to non-language object
Any suggestions?
FS
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Re: [R] interval censoring case 1 or current status data
Yes, the Icens package handles all forms of censored data (1-dim) On Apr 27, 2005, at 11:59 AM, Chao Zhu wrote: Dear all, Is there a function in R dealing with the NPMLE for current status data or interval censored case 1 data? Thanks, Jimmy [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html +--- + | Robert Gentleman phone: (206) 667-7700 | | Head, Program in Computational Biology fax: (206) 667-1319 | | Division of Public Health Sciences office: M2-B865 | | Fred Hutchinson Cancer Research Center | | email: [EMAIL PROTECTED] | +--- + __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] assign to an element of a vector
how about
assign( 'a', { z - get('a'); z[1] - 0; z } )
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Fernando Saldanha
Sent: Wednesday, April 27, 2005 3:22 PM
To: Submissions to R help
Subject: [R] assign to an element of a vector
I am trying to find a way to assign values to elements of a vector that will
be defined by a user. So I don't have the name of the vector and cannot hard
code the assignment in advance. In the example below I have to get() the
vector using its name. When I try to assign to an element I get an error:
a - c(1,2,3)
get('a')[1] - 0
Error: Target of assignment expands to non-language object
Any suggestions?
FS
__
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[R] [R-pkgs] GPArotation package
We have just put a package GPArotation on CRAN. The functions in this package perform an number of different orthogonal and oblique rotations for factor analysis, using the gradient projection algorithm described in Coen A. Bernaards and Robert I. Jennrich (2005), Gradient Projection Algorithms and Software for Arbitrary Rotation Criteria in Factor Analysis, , Educational and Psychological Measurement (in press). Additional details are available on the web site http://www.stat.ucla.edu/research/gpa. The following rotations are available: oblimin oblique oblimin family quartimin oblique targetT orthogonaltarget rotation targetQ oblique target rotation pstT orthogonalpartially specified target rotation pstQ oblique partially specified target rotation oblimax oblique entropy orthogonalminimum entropy quartimax orthogonal varimax orthogonal simplimax oblique bentlerTorthogonalBentler's invariant pattern simplicity criterion bentlerQoblique Bentler's invariant pattern simplicity criterion tandemI orthogonalTandem Criterion tandemIIorthogonalTandem Criterion geominT orthogonal geominQ oblique cfT orthogonalCrawford-Ferguson family cfQ oblique Crawford-Ferguson family infomaxTorthogonal infomaxQoblique mccammonorthogonalMcCammon minimum entropy ratio Other rotations can be fairly easily added simply by coding a small function to calculate the rotation criterion objective function, and its gradient. A detailed example of gradient computation is in the paper mentioned above. Paul Gilbert and Coen Bernaards ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Defining binary indexing operators
That sounds like a recipe for headaches. If you want to use x$y because you want a certain kind of x to act like a list with components for certain y, then you probably want to make a class of objects (x) which have x$y implemented as foo(x,y). That way you won't break existing code. Reid Huntsinger -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ali - Sent: Wednesday, April 27, 2005 3:11 PM To: r-help@stat.math.ethz.ch Subject: [R] Defining binary indexing operators Assume we have a function like: foo - function(x, y) how is it possible to define a binary indexing operator, denoted by $, so that x$y functions the same as foo(x, y) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Closing RGui help windows
No, and in 6 years of that interface, no one else has asked. (The information on open windows is not even retained.) You might like to try the single-window pager option in the preferences. On Wed, 27 Apr 2005, Ronnen Levinson wrote: I often wind up with many help windows cluttering my RGui screen when running Windows R 2.0.1. Is there an R instruction to close one or more help windows, or an RGui command to close all help windows? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Defining binary indexing operators
I should have added that if you're not wedded to $ you can do
$ %f% - function(x,y) foo(x,y)
for whatever name f you want, and then %f% is a binary infix operator form
of foo().
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Huntsinger, Reid
Sent: Wednesday, April 27, 2005 4:10 PM
To: 'Ali -'; r-help@stat.math.ethz.ch
Subject: RE: [R] Defining binary indexing operators
That sounds like a recipe for headaches. If you want to use x$y because
you want a certain kind of x to act like a list with components for
certain y, then you probably want to make a class of objects (x) which
have x$y implemented as foo(x,y). That way you won't break existing code.
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ali -
Sent: Wednesday, April 27, 2005 3:11 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Defining binary indexing operators
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $, so
that
x$y
functions the same as
foo(x, y)
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Re: [R] Defining binary indexing operators
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $, so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i)); foo(x, i) }
x$y # structure(8, class = myclass)
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RE: [R] Advice for calling a C function
You have the dimensions switched, in
double x [*MATDESC][*OBJ];
so when the dimensions aren't equal you do get odd things.
You might be better off defining functions to index into mat with a pair of
subscripts directly (.C() copies the argument anyway). Come to think of it,
there might be macros/functions for this in Rinternals.h. Then you don't
need to worry about row-major vs column-major order and related issues.
Finally, as this is a C programming question, it should go to R-devel.
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Tyler Smith
Sent: Tuesday, April 26, 2005 11:02 AM
To: R-Help
Subject: [R] Advice for calling a C function
Hi,
I'm having some trouble with a bit of combined C R code. I'm trying to
write a C function to handle the for loops in a function I'm working on
to calculate a similarity matrix. Jari Oksanen has kindly added the
necessary changes to the vegan package so that I can use the vegdist
function, so this isn't absolutely necessary. However, I'm stubborn and
want to know why Jari's code works and mine doesn't! Other than, of
course, the obvious - one of us knows what their doing and the other
doesn't. I would appreciate any help. What I've done is:
pass a matrix x to my C function, as a double:
.C(gowsim, as.double(mat), as.integer(nrow(mat)), as.integer(ncol(mat)))
Then I try and reconstruct the matrix, in the form of a C array:
#include R.h
#include Rmath.h
#include math.h
void gowsim ( double *mat, int *OBJ, int *MATDESC)
{
double x [*MATDESC][*OBJ];
int i, j, nrow, ncol;
nrow = *OBJ;
ncol = *MATDESC;
/* Rebuild Matrix */
for (j=0; j ncol; j++) {
for (i=0; i nrow; i++) {
x[i][j] = *mat;
Rprintf(row %d col %d value %f\n, i, j, x[i][j]);
mat++;
}
}
for (i=0; i nrow; i++) {
Rprintf(%f %f %f %f\n, x[i][0], x[i][1], x[i][2], x[i][3]);
}
}
The Rprintf statements display what's going on at each step. It looks
for all the world as if the assignments are working properly, but when I
try and print the matrix I get very strange results. If mat is 3x3 or
4x4 everything seems ok. But if mat is not symetrical the resulting x
matrix is very strange. In the case of a 5x4 mat only the first column
works out, and for 3x4 mat the second and third positions in the first
column are replaced by the first and second positions of the last
column. I'm guessing that I've messed up something in my use of
pointers, or perhaps the for loop, but I can't for the life of me figure
out what!! Once I sort this out I'll be calculating the differences
between rows in the x array en route to producing a similarity matrix. I
looked at the vegdist code, which is fancier than this, and manages to
avoid rebuilding the matrix entirely, but it's a bit beyond me.
I'm using WindowsXP, R 2.1.0 (2005-04-18), and the MinGW compiler.
Thanks for your continued patience,
Tyler
--
Tyler Smith
PhD Candidate
Department of Plant Science
McGill University
21,111 Lakeshore Road
Ste. Anne de Bellevue, Quebec
H9X 3V9
CANADA
Tel: 514 398-7851 ext. 8726
Fax: 514 398-7897
[EMAIL PROTECTED]
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Re: [R] Defining binary indexing operators
It's not necessary to be that complicated, is it? AFAIK, the '$'
operator is treated specially by the parser so that its RHS is treated
as a string, not a variable name. Hence, a method for $ can just take
the indexing argument directly as given -- no need for any fancy
language tricks (eval(), etc.)
x - structure(3, class = myclass)
y - 5
foo - function(x,y) paste(x, indexed by ', y, ', sep=)
foo(x, y)
[1] 3 indexed by '5'
$.myclass - foo
x$y
[1] 3 indexed by 'y'
The point of the above example is that foo(x,y) behaves differently from
x$y even when both call the same function: foo(x,y) uses the value of
the variable 'y', whereas x$y uses the string y. This is as desired
for an indexing operator $.
-- Tony Plate
Gabor Grothendieck wrote:
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $, so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i)); foo(x, i) }
x$y # structure(8, class = myclass)
[[alternative HTML version deleted]]
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Re: [R] Defining binary indexing operators
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $,
so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i)); foo(x, i)
}
x$y # structure(8, class = myclass)
what about this approach:
foo - function(x, y) x+y
assign($, foo)
would this overwrite $ and make R to forget its definitions in the global
environment?
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[R] tcl/tk problem
I have JUST started using R. I am using R 2.0.1 on a mac (os x) to run another program (called GRASPER, Generalized Regression Analysis and Spatial Prediction for R). When I try to run Grasper, I get an error for tcl/tk: library(grasper) Loading required package: mgcv This is mgcv 1.1-8 Loading required package: MASS Loading required package: tcltk Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /Library/Frameworks/R.framework/Resources/library/tcltk/libs/tcltk.so: dlcompat: dyld: /Applications/R.app/Contents/MacOS/R can't open library: /usr/X11R6/lib/libX11.6.dylib (No such file or directory, errno = 2) Error: .onLoad failed in loadNamespace for 'tcltk' Error: package 'tcltk' could not be loaded However, tcl/tk appears to be there: capabilities() jpeg pngtcltk X11GNOME libz http/ftp sockets TRUE TRUE TRUE TRUEFALSE TRUE TRUE TRUE libxml fifo cledit IEEE754bzip2 PCRE TRUE TRUE TRUE TRUE TRUE TRUE Does anyone know of a solution to this? Thanks, Jennifer Skene __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Getting the name of an object as character
This could be really trivial, but I cannot find the right function to get the name of an object as a character. Assume we have a function like: getName - function(obj) Now if we call the function like: getName(blabla) and 'blabla' is not a defined object, I want getName to return blabla. In other word, if paste(blabla) returns blabla I want to define a paste function which returns the same character by: paste(blabla) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to specify the hierarchical structure of a split plot using lmer ??
I have a problem getting the lmer function of the lme4 package to use the appropriate degrees of freedom for testing. Consider the Semiconductor data from the SASmixed package: library(SASmixed) Semi.lme - lme(resistance ~ ET * position, random=~1|Grp, data=Semiconductor) anova(Semi.lme) numDF denDF F-value p-value (Intercept) 124 3237.261 .0001 ET 3 81.942 0.2015 position 3243.385 0.0345 ET:position 9240.809 0.6125 Here, the ET effect is (correctly) tested on 8 denominator degrees of freedom. In the example in the SASmixed package, the following code is presented: (fm1Semi - lmer(resistance ~ ET * position + (1|Grp), Semiconductor)) anova(fm1Semi) Analysis of Variance Table Df Sum Sq Mean Sq Denom F value Pr(F) ET 3 0.647 0.216 32.000 1.9415 0.14273 position 3 1.129 0.376 32.000 3.3855 0.02991 * ET:position 9 0.809 0.090 32.000 0.8092 0.61127 So here, all effects are tested with 32 denominator degrees of freedom. I have looked at the help page for lmer but have been unable to figure out how to specify the hierarchical structure of a split plot experiment. Also, I have Googled but without any luck. Any help will be appreciated. Søren __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Defining binary indexing operators
Excuse me! I misunderstood the question, and indeed, it is necessary be
that complicated when you try to make x$y behave the same as foo(x,y),
rather than foo(x,y) (doing the former would be inadvisible, as I
think someelse pointed out too.)
Tony Plate wrote:
It's not necessary to be that complicated, is it? AFAIK, the '$'
operator is treated specially by the parser so that its RHS is treated
as a string, not a variable name. Hence, a method for $ can just take
the indexing argument directly as given -- no need for any fancy
language tricks (eval(), etc.)
x - structure(3, class = myclass)
y - 5
foo - function(x,y) paste(x, indexed by ', y, ', sep=)
foo(x, y)
[1] 3 indexed by '5'
$.myclass - foo
x$y
[1] 3 indexed by 'y'
The point of the above example is that foo(x,y) behaves differently from
x$y even when both call the same function: foo(x,y) uses the value of
the variable 'y', whereas x$y uses the string y. This is as desired
for an indexing operator $.
-- Tony Plate
Gabor Grothendieck wrote:
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by
$, so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i));
foo(x, i) }
x$y # structure(8, class = myclass)
[[alternative HTML version deleted]]
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[R] Is this a bug in R?
Dear all, I am trying to fit a nonlinear model with a autocorrelation term, but everytime I type in the command, I got an error message from Winwows and R closes itself. The command line is as follows: mod1-nlme(V~A*exp(-B*A.O)*Vac.t.1.,data,fixed=A+B~1,random=A+B~1|ORDINAL,+ correlation=corCAR1(0.3179,~A.O|ORDINAL,TRUE),start=c(A=1.2,B=0.2)) I have already fitted this model allowing Phi to vary while optimizing, and it was fine, but as soon as I try to keep it fixed (argument TRUE), I simply can't I don't get any error message from R, just a Windows error seying something like R for windows GUI front-end has detected a problem and has to close. And thats it, R is over! I don't know if I am doing anything wrong, or if it has to be with my system (I have Windows XP Pro), but it looks like a bug in R. Do you know anything else about this. Thank you very much, Antonio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Getting the name of an object as character
On Wed, 2005-04-27 at 23:03 +, Ali - wrote: This could be really trivial, but I cannot find the right function to get the name of an object as a character. Assume we have a function like: getName - function(obj) Now if we call the function like: getName(blabla) and 'blabla' is not a defined object, I want getName to return blabla. In other word, if paste(blabla) returns blabla I want to define a paste function which returns the same character by: paste(blabla) Do you mean: exists(plot.default) [1] TRUE deparse(substitute(plot.default)) [1] plot.default exists(MyPlot.Default) [1] FALSE deparse(substitute(MyPlot.Default)) [1] MyPlot.Default x - 1:10 x [1] 1 2 3 4 5 6 7 8 9 10 deparse(substitute(x)) [1] x Does that get what you want? If so, see ?deparse and ?substitute HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Defining binary indexing operators
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $,
so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i)); foo(x,
i)
}
x$y # structure(8, class = myclass)
what about this approach:
foo - function(x, y) x+y
assign($, foo)
would this overwrite $
Yes.
and make R to forget its definitions in the global
environment?
Yes.
Your construct might still be used in a local environment.
f - function(x,y) {
$ - function(x,y) x+y
x$y
}
f(1,2) # 3
# or
z - local({
$ - function(x,y) x+y
x - y - 3
x$y
})
z # 6
works and outside of f and the local, $ works normally.
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RE: [R] assign to an element of a vector
You did not explain the full context of what you are trying to do. Perhaps
this could help:
varName - as.name(bahbah)
varName
bahbah
substitute(a[1] - 0, list(a=varName))
bahbah[1] - 0
So you could perhaps eval() this expression.
Andy
From: Fernando Saldanha
I am trying to find a way to assign values to elements of a vector
that will be defined by a user. So I don't have the name of the vector
and cannot hard code the assignment in advance. In the example below I
have to get() the vector using its name. When I try to assign to an
element I get an error:
a - c(1,2,3)
get('a')[1] - 0
Error: Target of assignment expands to non-language object
Any suggestions?
FS
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Re: [R] How to specify the hierarchical structure of a split plot using lmer ??
Søren Højsgaard wrote: I have a problem getting the lmer function of the lme4 package to use the appropriate degrees of freedom for testing. Consider the Semiconductor data from the SASmixed package: library(SASmixed) Semi.lme - lme(resistance ~ ET * position, random=~1|Grp, data=Semiconductor) anova(Semi.lme) numDF denDF F-value p-value (Intercept) 124 3237.261 .0001 ET 3 81.942 0.2015 position 3243.385 0.0345 ET:position 9240.809 0.6125 Here, the ET effect is (correctly) tested on 8 denominator degrees of freedom. In the example in the SASmixed package, the following code is presented: (fm1Semi - lmer(resistance ~ ET * position + (1|Grp), Semiconductor)) anova(fm1Semi) Analysis of Variance Table Df Sum Sq Mean Sq Denom F value Pr(F) ET 3 0.647 0.216 32.000 1.9415 0.14273 position 3 1.129 0.376 32.000 3.3855 0.02991 * ET:position 9 0.809 0.090 32.000 0.8092 0.61127 So here, all effects are tested with 32 denominator degrees of freedom. I have looked at the help page for lmer but have been unable to figure out how to specify the hierarchical structure of a split plot experiment. Also, I have Googled but without any luck. Any help will be appreciated. Søren Unfortunately the answer is quite simple. The current version of the lme4 package does not attempt to get the degrees of freedom right - it just gives an upper bound. The reason is more than simple laziness on my part. The lmer function can fit models with non-nested grouping factors and it is not easy to define sensible values for the degrees of freedom in such cases. Hence I have put that problem aside while dealing with other issues. (Also - as long time readers of this list may know - the topic of denominator degrees of freedom is one of my least favorite topics.) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Advice for calling a C function
Thank you. I can't believe how much time I spent going over that short
bit of code without noticing that I had switched the dimensions. I was
sure there was some arcane bit of pointer-lore that was eluding me.
Patrick Burns pointed out an alternative approach to me, leaving the
data in the form of a double vector. The index in the double vector
corresponding to the matrix location can be calculated from:
Rmatrix [row,col] is equivalent to: Cvector [(row-1) +
(col-1)*nrows(Rmatrix)]
This is easily inserted inside a for loop, sidestepping the whole issue
of rebuilding a matrix. This has the added bonus that I don't have to
break the matrix back down into a vector to pass it back to R.
Thank you all for your very helpful advice. Since most of this
discussion has been in the R-help list, that's where I sent this post.
In future I'll direct my C questions to the devel list. Hopefully, any
further questions I have won't involve anything so silly as switching
indexes.
Cheers,
Tyler
Tyler Smith
PhD Candidate
Department of Plant Science
McGill University
21,111 Lakeshore Road
Ste. Anne de Bellevue, Quebec
H9X 3V9
CANADA
Tel: 514 398-7851 ext. 8726
Fax: 514 398-7897
[EMAIL PROTECTED]
Huntsinger, Reid wrote:
You have the dimensions switched, in
double x [*MATDESC][*OBJ];
so when the dimensions aren't equal you do get odd things.
You might be better off defining functions to index into mat with a pair of
subscripts directly (.C() copies the argument anyway). Come to think of it,
there might be macros/functions for this in Rinternals.h. Then you don't
need to worry about row-major vs column-major order and related issues.
Finally, as this is a C programming question, it should go to R-devel.
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Tyler Smith
Sent: Tuesday, April 26, 2005 11:02 AM
To: R-Help
Subject: [R] Advice for calling a C function
Hi,
I'm having some trouble with a bit of combined C R code. I'm trying to
write a C function to handle the for loops in a function I'm working on
to calculate a similarity matrix. Jari Oksanen has kindly added the
necessary changes to the vegan package so that I can use the vegdist
function, so this isn't absolutely necessary. However, I'm stubborn and
want to know why Jari's code works and mine doesn't! Other than, of
course, the obvious - one of us knows what their doing and the other
doesn't. I would appreciate any help. What I've done is:
pass a matrix x to my C function, as a double:
.C(gowsim, as.double(mat), as.integer(nrow(mat)), as.integer(ncol(mat)))
Then I try and reconstruct the matrix, in the form of a C array:
#include R.h
#include Rmath.h
#include math.h
void gowsim ( double *mat, int *OBJ, int *MATDESC)
{
double x [*MATDESC][*OBJ];
int i, j, nrow, ncol;
nrow = *OBJ;
ncol = *MATDESC;
/* Rebuild Matrix */
for (j=0; j ncol; j++) {
for (i=0; i nrow; i++) {
x[i][j] = *mat;
Rprintf(row %d col %d value %f\n, i, j, x[i][j]);
mat++;
}
}
for (i=0; i nrow; i++) {
Rprintf(%f %f %f %f\n, x[i][0], x[i][1], x[i][2], x[i][3]);
}
}
The Rprintf statements display what's going on at each step. It looks
for all the world as if the assignments are working properly, but when I
try and print the matrix I get very strange results. If mat is 3x3 or
4x4 everything seems ok. But if mat is not symetrical the resulting x
matrix is very strange. In the case of a 5x4 mat only the first column
works out, and for 3x4 mat the second and third positions in the first
column are replaced by the first and second positions of the last
column. I'm guessing that I've messed up something in my use of
pointers, or perhaps the for loop, but I can't for the life of me figure
out what!! Once I sort this out I'll be calculating the differences
between rows in the x array en route to producing a similarity matrix. I
looked at the vegdist code, which is fancier than this, and manages to
avoid rebuilding the matrix entirely, but it's a bit beyond me.
I'm using WindowsXP, R 2.1.0 (2005-04-18), and the MinGW compiler.
Thanks for your continued patience,
Tyler
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Re: [R] Defining binary indexing operators
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i)); foo(x,
i)
}
x$y # structure(8, class = myclass)
If I got it right, in the above example you provided '$' is defined as a
method for a S3 class. How is it possible to do the same with a S4 class. If
this is not possible, what is the best way to define the '$' operator whose
first arguments is a S4 object and doesn't overwrite the global definition?
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Re: [R] its package: inexplicable date-shifting ?!
On 27 April 2005 at 13:45, Chalasani, Prasad wrote:
| Can someone please explain to me why
| the dates get shifted by one day
| when I create an its ( irregular time-series )
| object from a matrix for which I've
| assigned row names.
I think you initiated the its() object the wrong way -- the date object needs
to be supplied, you were sort-of hiding that in the matrix rownames:
m - matrix(1:2, nrow=2)
its(m, as.POSIXct(strptime(c('20040813', '20040814'), %Y%m%d)))
1
2004-08-13 1
2004-08-14 2
Hth, Dirk
| E.g. in the example run below,
| why does the its object have dates
| one-shifted from my original dates?
|
| install.packages('its')
| install.packages('Hmisc')
| require(its)
| m - matrix(1:2, nrow=2)
| m
| [,1]
| [1,]1
| [2,]2
| its.format('%Y%m%d')
| [1] %Y%m%d
| rownames(m) - c('20040813', '20040814')
| m
| [,1]
| 200408131
| 200408142
| its(structure(m))
| 1
| 20040812 1
| 20040813 2
|
|
| -Original Message-
| From: [EMAIL PROTECTED]
| [mailto:[EMAIL PROTECTED] On Behalf Of Berton Gunter
| Sent: Wednesday, April 27, 2005 1:28 PM
| To: [EMAIL PROTECTED]
| Cc: r-help@stat.math.ethz.ch
| Subject: RE: [R] Recursive calculation of a series of values
|
|
| Algebra:
| cumprod(1+v)*x[0]
|
| -- Bert Gunter
| Genentech Non-Clinical Statistics
| South San Francisco, CA
|
| The business of the statistician is to catalyze the scientific learning
| process. - George E. P. Box
|
|
|
| -Original Message-
| From: [EMAIL PROTECTED]
| [mailto:[EMAIL PROTECTED] On Behalf Of Luis Torgo
| Sent: Wednesday, April 27, 2005 7:42 AM
| To: r-help@stat.math.ethz.ch
| Subject: [R] Recursive calculation of a series of values
|
| Dear R-users,
|
| I'm felling kind of blocked on a quite simple problem and I wonder if
| someone could give me a help with it.
|
| My problem:
|
| x[0] = 100
| x[1] = (1+v[1])*x[0]
| x[2] = (1+v[2])*x[1]
| ...
|
| i.e.
|
| x[i] = (1+v[i])*x[i-1]
| and x[0]=k
|
| Given a set of v values I wanted to obtain the corresponding
| x values in
| an efficient way (i.e. without a for loop).
|
| For instance, if x[0] = 100 and v = c(0.2,-0.1,0.05) then I would get
| x = c(120,108,113.4)
|
| I'm almost sure the function filter() from package tseries is the key
| for getting these values but I'm really blocked.
|
| Any help is much appreciated.
|
| Luís Torgo
|
| --
| Luis Torgo
| FEP/LIACC, University of Porto Phone : (+351) 22 339 20 93
| Machine Learning Group Fax : (+351) 22 339 20 99
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--
Better to have an approximate answer to the right question than a precise
answer to the wrong question. -- John Tukey as quoted by John Chambers
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Re: [R] Defining binary indexing operators
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i));
foo(x,
i)
}
x$y # structure(8, class = myclass)
If I got it right, in the above example you provided '$' is defined as a
method for a S3 class. How is it possible to do the same with a S4 class.
If
this is not possible, what is the best way to define the '$' operator
whose
first arguments is a S4 object and doesn't overwrite the global
definition?
The myclass example is defined as an S3 method. In the above its defined
as a function, not a method.
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Re: [R] Is this a bug in R?
You are in fact using the contributed package 'nlme', not just R. Please read both the section on BUGS in the FAQ and the posting guide, and send a reproducible example to the nlme maintainer. One thing the posting guide asks for is a useful subject line. Something like `A crash when using nlme'. On Thu, 28 Apr 2005, Revilla,AJ (pgt) wrote: Dear all, I am trying to fit a nonlinear model with a autocorrelation term, but everytime I type in the command, I got an error message from Winwows and R closes itself. The command line is as follows: mod1-nlme(V~A*exp(-B*A.O)*Vac.t.1.,data,fixed=A+B~1,random=A+B~1|ORDINAL,+ correlation=corCAR1(0.3179,~A.O|ORDINAL,TRUE),start=c(A=1.2,B=0.2)) I have already fitted this model allowing Phi to vary while optimizing, and it was fine, but as soon as I try to keep it fixed (argument TRUE), I simply can't I don't get any error message from R, just a Windows error seying something like R for windows GUI front-end has detected a problem and has to close. And that´s it, R is over! I don't know if I am doing anything wrong, or if it has to be with my system (I have Windows XP Pro), but it looks like a bug in R. Do you know anything else about this. Thank you very much, Antonio -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
