Re: [R] R-Logo in \LaTeX
On Thu, Mar 06, 2008 at 06:54:41PM -0500, Charilaos Skiadas wrote: On Mar 6, 2008, at 1:49 PM, Mag. Ferri Leberl wrote: Dear everybody! Is there a command in \LaTeX to display the R-Logo or has anybody made it up? Thank you in advance. Isn't it just an image? Hence you would include it like one usually includes images. Or do you mean something else? Yes, it is an image, but it would be nice to have it in a vector-based format like EPS or SVG. Anyway, here are some versions (=different sizes): http://developer.r-project.org/Logo/ Gabor Yours, sincerely Mag. Ferri Leberl Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]UNIL DGM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] build options for R
If you are calling libR.so, it is an option set when you initialize R. See Rf_initEmbeddedR in 'Writing R Extensions' and the examples in the tests/Embedded directory in the sources. BTW, discussion of embedded R is definitely off topic for R-help: use the R-devel list. On Thu, 6 Mar 2008, Scott Zentz wrote: Hello Everyone, Recently I was given a Java servlet based web calculator that will call R (libR.so to be exact) but I am having trouble trying to disable R from requiring --save, --no-save or --vanilla... Is there any way to build R on linux and disable R from asking --save, --no-save or --vanilla?? Thanks! -scz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] parameters for lbfgsb (function for optimization)
Whoa! If you want to optimise in R, you need to write an R function to define your objective. If that function is already written in C, as appears to be the case, you need to write an interfact to make it available to R. This is not exactly something novices would take on lightly. If you wish to optimise directly in C, then you should look for software support to do it that was intended for the purpose. The R source code is not like that. You can use it if you want, but you must expect it to be difficult and very puzzling, since it is not intended to be free-standing, but part of a large, complex system. Best of luck. Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Kyeongmi Cheon Sent: Friday, 7 March 2008 3:06 PM To: r-help@r-project.org Subject: [R] parameters for lbfgsb (function for optimization) Can anyone help me with lbfgsb (function for optimization)? It takes the following parameters: void lbfgsb (int n, int lmm, double *x, double *lower, double *upper, int *nbd, double *Fmin, optimfn fn, optimgr gr, int *fail, void *ex, double factr, double pgtol, int *fncount, int *grcount, int maxit, char *msg, int trace, int nREPORT); What do I put for parameter ex (11th parameter)? I looked at optim.c codes at R sites and it's a structure that has bunch of objects such as SEXP R_fcall, SEXP R_gcall, SEXP R_env, double* ndeps, etc. I cannot figure out what it is about. How about fncount, rcount? R-ext.pdf or R help does not explain it in detail. Any comments would help. Thank you. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Puzzling coefficients for linear fitting to polynom
Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): x - 1:3 y - c(1, 4, 9) performing a linear fit f - lm(y ~ poly(x, 2)) gives weird coefficients: coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.667 5.6568542 0.8164966 However the fitted() result makes sense: fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. -- Firas Swidan, PhD Founder and CEO Olymons: Blessing Machines with Vision (TM) http://www.olymons.com P.O.Box 8125 Nazareth 16480 Israel Cell: +.972.(0)54.733.1788 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzling coefficients for linear fitting to polynom
poly() computes by default orthogonal polynomials; check the online help file for poly() for more info. Probably you want to use the 'raw' argument in this example, i.e., x - 1:3 y - c(1, 4, 9) lm(y ~ poly(x, 2, raw = TRUE)) I hope this helps. Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Firas Swidan, PhD [EMAIL PROTECTED] To: r-help@r-project.org Sent: Friday, March 07, 2008 9:16 AM Subject: [R] Puzzling coefficients for linear fitting to polynom Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): x - 1:3 y - c(1, 4, 9) performing a linear fit f - lm(y ~ poly(x, 2)) gives weird coefficients: coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.667 5.6568542 0.8164966 However the fitted() result makes sense: fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. -- Firas Swidan, PhD Founder and CEO Olymons: Blessing Machines with Vision (TM) http://www.olymons.com P.O.Box 8125 Nazareth 16480 Israel Cell: +.972.(0)54.733.1788 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzling coefficients for linear fitting to polynom
It does help if you read the help information for poly. ?poly x - 1:3 y - c(1, 4, 9) f - lm(y ~ poly(x, 2, raw = TRUE)) ## note raw = TRUE coef(f) (Intercept) poly(x, 2, raw = TRUE)1 poly(x, 2, raw = TRUE)2 0 0 1 You were assuming a power basis for the polynomial, 1, x, x^2. If you want to use that you must declare that using raw = TRUE. The default is to use an orthogonal polynomial basis, and you can expect the coefficients relative to that to be, well, puzzling. Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Firas Swidan, PhD Sent: Friday, 7 March 2008 6:16 PM To: r-help@r-project.org Subject: [R] Puzzling coefficients for linear fitting to polynom Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): x - 1:3 y - c(1, 4, 9) performing a linear fit f - lm(y ~ poly(x, 2)) gives weird coefficients: coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.667 5.6568542 0.8164966 However the fitted() result makes sense: fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. -- Firas Swidan, PhD Founder and CEO Olymons: Blessing Machines with Vision (TM) http://www.olymons.com P.O.Box 8125 Nazareth 16480 Israel Cell: +.972.(0)54.733.1788 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re action Time and Time Series Analysis
Dear, I need to analyse reaction time. The general idea might be described as this: There are 8x8 circles. My program light a circle at one time. The subject then click the lighted circled as fast as possible. After the correct circle is clicked, then the next circle will be lighted for the subject to click on, and so on. My questions are: 1. May I use Time Series Analysis? I read that Time Series may be used only for data with regular measurement interval. Can my experiment be considered as regularly measured? Let's assume there are 30 subjects. Each has 330 measured reaction time. I then plot the response time as time series. 2. I need to know if the plots can be classified to groups with different characteristic. That is, I want to check whether there is model (or models) my plots follow. I've read about ARIMA or somekind of model fitting but have no idea about the general concept. Would someone give me some insight?? How to do that with R-Language? Just for additional request, might I know some paper that use time series for analyzing reaction time? I've searched with Google for nearly 2 hours but most of the paper I'm looking for are not free. I would appreciate if I could get real case study. Thank you for the help. regards, Nathanael Gratias indonesia -- View this message in context: http://www.nabble.com/Reaction-Time-and-Time-Series-Analysis-tp15891094p15891094.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzling coefficients for linear fitting to polynom
On 07-Mar-08 08:16:06, Firas Swidan, PhD wrote: Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): x - 1:3 y - c(1, 4, 9) performing a linear fit f - lm(y ~ poly(x, 2)) gives weird coefficients: coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.667 5.6568542 0.8164966 However the fitted() result makes sense: fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. Have a look at the values returned by poly(x,2). The coefficients you are getting are the results of fitting y = a + b1*poly(x,2)[,1] + b2*poly(x,2)[,2] where poly(x, 2)[,1] # [1] -7.071068e-01 -9.073264e-17 7.071068e-01 poly(x, 2)[,2] # [1] 0.4082483 -0.8164966 0.4082483 which is probably not what you may have thought you were doing! It is certainly not the same as fitting y = a + b1*x + b2*(x^2) though of course the fitted values will be the same. Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 07-Mar-08 Time: 08:40:46 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzling coefficients for linear fitting to polynom
Thanks for the clarifications. It seems the confusion resulted from making one assumption more than necessary regarding the behavior of poly(). Best wishes, Firas. On Fri, 2008-03-07 at 18:33 +1000, [EMAIL PROTECTED] wrote: It does help if you read the help information for poly. ?poly x - 1:3 y - c(1, 4, 9) f - lm(y ~ poly(x, 2, raw = TRUE)) ## note raw = TRUE coef(f) (Intercept) poly(x, 2, raw = TRUE)1 poly(x, 2, raw = TRUE)2 0 0 1 You were assuming a power basis for the polynomial, 1, x, x^2. If you want to use that you must declare that using raw = TRUE. The default is to use an orthogonal polynomial basis, and you can expect the coefficients relative to that to be, well, puzzling. Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Firas Swidan, PhD Sent: Friday, 7 March 2008 6:16 PM To: r-help@r-project.org Subject: [R] Puzzling coefficients for linear fitting to polynom Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): x - 1:3 y - c(1, 4, 9) performing a linear fit f - lm(y ~ poly(x, 2)) gives weird coefficients: coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.667 5.6568542 0.8164966 However the fitted() result makes sense: fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. -- Firas Swidan, PhD Founder and CEO Olymons: Blessing Machines with Vision (TM) http://www.olymons.com P.O.Box 8125 Nazareth 16480 Israel Cell: +.972.(0)54.733.1788 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Puzzling coefficients for linear fitting to polynom
Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): x - 1:3 y - c(1, 4, 9) performing a linear fit f - lm(y ~ poly(x, 2)) gives weird coefficients: coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.667 5.6568542 0.8164966 However the fitted() result makes sense: fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. -- Firas Swidan, PhD Founder and CEO Olymons: Blessing Machines with Vision (TM) http://www.olymons.com P.O.Box 8125 Nazareth 16480 Israel Cell: +.972.(0)54.733.1788 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re action Time and Time Series Analysis
Nathanael, On 7 Mar 2008, at 09:40, nathan3073 wrote: Dear, I need to analyse reaction time. The general idea might be described as this: There are 8x8 circles. My program light a circle at one time. The subject then click the lighted circled as fast as possible. After the correct circle is clicked, then the next circle will be lighted for the subject to click on, and so on. My questions are: 1. May I use Time Series Analysis? I read that Time Series may be used only for data with regular measurement interval. Can my experiment be considered as regularly measured? Is the precise timing of new responses expected to influence the RTs? If so, the answer is probably no. Let's assume there are 30 subjects. Each has 330 measured reaction time. I then plot the response time as time series. 2. I need to know if the plots can be classified to groups with different characteristic. This sounds like you need some kind of cluster or mixture analysis, however That is, I want to check whether there is model (or models) my plots follow. I've read about ARIMA or somekind of model fitting but have no idea about the general concept. Would someone give me some insight?? How to do that with R-Language? ... whether you can do this depends on your specific hypotheses. Possibly a simple linear model answers most of your questions. Best, Ingmar Just for additional request, might I know some paper that use time series for analyzing reaction time? I've searched with Google for nearly 2 hours but most of the paper I'm looking for are not free. I would appreciate if I could get real case study. Thank you for the help. regards, Nathanael Gratias indonesia -- View this message in context: http://www.nabble.com/Reaction-Time- and-Time-Series-Analysis-tp15891094p15891094.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. Ingmar Visser Department of Psychology, University of Amsterdam Roetersstraat 15 1018 WB Amsterdam The Netherlands t: +31-20-5256723 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rpart and bagging - how is it done?
I believe that the procedure you describe at the end (resampling the cases) is the original interpretation of bagging, and that using weighting is equivalent when a procedure uses case weights. If you are getting different results when replicating cases and when using weights then rpart is not using its weights strictly as case weights and it would be preferable to replicate cases. But I am getting identical predictions by the two routes: ind - sample(1:81, replace=TRUE) rpart(Kyphosis ~ Age + Number + Start, data=kyphosis[ind,], xval=0) rpart(Kyphosis ~ Age + Number + Start, data=kyphosis, weights=tabulate(ind, nbins=81), xval=0) My memory is that rpart uses unweighted numbers for its control params (unlike tree) and hence is not strictly using case weights. I believe you can avoid that by setting the control params to their minimum and relying on pruning. BTW, it is inaccurate to call these trees 'non-pruned' -- the default setting of cp is still (potentially) doing quite a lot of pruning. Torsten Hothorn can explain why he chose to do what he did. There's a small (but only small) computational advantage in using case weights, but the tricky issue for me is how precisely tree growth is stopped, and I don't think that rpart at its default settings is mimicing what Breiman was doing (he would have been growing much larger trees). On Thu, 6 Mar 2008, [EMAIL PROTECTED] wrote: Hi there. I was wondering if somebody knows how to perform a bagging procedure on a classification tree without running the classifier with weights. Let me first explain why I need this and then give some details of what I have found out so far. I am thinking about implementing the bagging procedure in Matlab. Matlab has a simple classification tree function (in their Statistics toolbox) but it does not accept weights. A modification of the Matlab procedure to accommodate weights would be very complicated. The rpart function in R accepts weights. This seems to allow for a rather simple implementation of bagging. In fact Everitt and Hothorn in chapter 8 of A Handbook of Statistical Analyses Using R describe such a procedure. The procedure consists in generating several samples with replacement from the original data set. This data set has N rows. The implementation described in the book first fits a non-pruned tree to the original data set. Then it generates several (say, 25) multinomial samples of size N with probabilities 1/N. Then, each sample is used in turn as the weight vector to update the original tree fit. Finally, all the updated trees are combined to produce consensus class predictions. Now, a typical realization of a multinomial sample consists of small integers and several 0's. I thought that the way that weighting worked was this: the observations with weights equal to 0 are omitted and the observations with weights 1 are essentially replicated according to the weight. So I thought that instead of running the rpart procedure with weights, say, starting with (1, 0, 2, 0, 1, ... etc.) I could simply generate a sample data set by retaining row 1, omitting row 2, replicating row 3 twice, omitting row 4, retaining row 5, etc. However, this does not seem to work as I expected. Instead of getting identical trees (from running weighted rpart on the original data set and running rpart on the sample data set described above with no weighting) I get trees that are completely different (different threshold values and different order of variables entering the splits). Moreover, the predictions from these trees can be different so the misclassification rates usually differ. This finally brings me to my question - is there a way to mimic the workings of the weighting in rpart by, for example, modification of the data set or, perhaps, some other means. Thanks in advance for your time, Andy __ Andy Jaworski 518-1-01 Process Laboratory 3M Corporate Research Laboratory - E-mail: [EMAIL PROTECTED] Tel: (651) 733-6092 Fax: (651) 736-3122 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interesting remarks about R back in 1999
Hi, yes it's true Mathematica is too expensive (I think one of the most expensive out there), but yacas has a lng way to go before becoming mathematica. I agree mathematica is not really for data analysis but I think it's to help researchers and inventors invent new ideas, understand concepts etc... However for that there is a cheaper alternative made out of three components: a pen, a paper and a human brain. The Rlink was available for free at some point; now it's not even there anymore! Spencer Graves wrote: Many people love Mathematica, but it's strength is symbolic mathematics, not data analysis. ..(snipped) ...(snipped) Today, if I wanted symbolic mathematics, I might try Yacas (http://yacas.sourceforge.net/homepage.html -- and the Ryacas package). Mathematica is probably superior to Yacas, but I'd have to be convinced that the difference was sufficient to justify the extra expense...(snipped) -- View this message in context: http://www.nabble.com/Interesting-remarks-about-R-back-in-1999-tp15882515p15891088.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with 'memory not mapped'
Hi, I'm no expert programmer at all; I'm running an R script (mariam1_2.R). This scripts calls another script, which contains an R function, which .Call some C code. It runs several times without any problem, but sometimes I get the error: --- *** caught segfault *** address 0x1c404ec8, cause 'memory not mapped' Traceback: 1: .Call(rrfunc, as.double(dx), as.integer(nrows), as.integer(ncols), as.double(deltadt), as.character(binmaps), as.integer(ldd), as.double(downst),as.integer(outl), as.double(maxinte), as.double(nmn)) 2: rrfuncR(dx, nrows, ncols, delta.t,binmaps.pathed,ldd,downst, out.index,interc,n.man) 3: eval.with.vis(expr, envir, enclos) 4: eval.with.vis(ei, envir) 5: source(rrprogv1_2.R) 6: eval.with.vis(expr, envir, enclos) 7: eval.with.vis(ei, envir) 8: source(mariam1_2.R) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: --- I've seen something related in the documentation, but it is not easy to understand it for me. I'm trying to perform a Monte Carlo analysis that calls the function iteratively. My question is why sometimes I can call forty or fifty times the function without any problem, and other times it crashes at the first, second or 6th time. Can you give me any advice? Thanks and best regards, Javier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] using xyplot with groups and panel.linejoin
Dear All, I am using xyplot() with many groups like this: statselect - levels(dat$stat) xyplot(relmse~T|lambda, groups=stat, data=dat, panel = panel.superpose, key=simpleKey(statselect, lines=T)) However, I want lines not scatterplots and if I set panel.groups=panel.linejoin that connects the lines according to the relmse and not according the consecutive values of T. Is there a short solution to this very simple problem, which doesn't require custom made panel function? Thanks in advance, Kati __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with 'memory not mapped'
Dear Javier, From your description, it seems you are not the author of the C function rrfunc, which is where the problem is probably located (the segmentation fault is most likely a pointer related problem where the C code is trying to do something with memory it should not be trying to do). Depending on the exact settings of your simulation, the state of your machine, etc the problem will show up at one point or another. I'd get in touch with the author of the C code and let her/him know that there is a problem in it. Most of these are very quickly located using valgrind. Search on the archives and the manual (Writing R extensions) but a call such as R -d valgrind --tool=memcheck --leak-check=full --log-file=marian.valgrind.log --vanilla mariam1_2.R mariam1_2.with.valgrind.Rout after compiling the C code with the -O1 (not -O2, as per default) flag. HTH, R. On Fri, Mar 7, 2008 at 10:26 AM, [EMAIL PROTECTED] wrote: Hi, I'm no expert programmer at all; I'm running an R script (mariam1_2.R). This scripts calls another script, which contains an R function, which .Call some C code. It runs several times without any problem, but sometimes I get the error: --- *** caught segfault *** address 0x1c404ec8, cause 'memory not mapped' Traceback: 1: .Call(rrfunc, as.double(dx), as.integer(nrows), as.integer(ncols), as.double(deltadt), as.character(binmaps), as.integer(ldd), as.double(downst),as.integer(outl), as.double(maxinte), as.double(nmn)) 2: rrfuncR(dx, nrows, ncols, delta.t,binmaps.pathed,ldd,downst, out.index,interc,n.man) 3: eval.with.vis(expr, envir, enclos) 4: eval.with.vis(ei, envir) 5: source(rrprogv1_2.R) 6: eval.with.vis(expr, envir, enclos) 7: eval.with.vis(ei, envir) 8: source(mariam1_2.R) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: --- I've seen something related in the documentation, but it is not easy to understand it for me. I'm trying to perform a Monte Carlo analysis that calls the function iteratively. My question is why sometimes I can call forty or fifty times the function without any problem, and other times it crashes at the first, second or 6th time. Can you give me any advice? Thanks and best regards, Javier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ramon Diaz-Uriarte Statistical Computing Team Structural Biology and Biocomputing Programme Spanish National Cancer Centre (CNIO) http://ligarto.org/rdiaz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] legend for several graphics
Thanks a lot, John, Gavin; Hadley and Greg, for your helpful comments and suggestions. I finally achieved what I wanted using the suggested method from Gavin with corrections from Greg. Out of curiosity (and interest to learn): Hadley, how would you simplify that code using lattice or ggplot and how would you automatically draw the legend? Best, Georg Greg Snow [EMAIL PROTECTED] writes: My modification of your example is: library(TeachingDemos) op - par(mfrow = c(3,3), ## split region oma = c(5,0,4,0) + 0.1, ## create outer margin mar = c(5,4,2,2) + 0.1) ## shrink some margins plot(1:10, main = a, pch = 1:2, col= 1:2) plot(1:10, main = b, pch = 1:2, col= 1:2) tmp1 - cnvrt.coords( 0.5, 0, input='plt' )$tdev # save location for mtext plot(1:10, main = c, pch = 1:2, col= 1:2) plot(1:10, main = d, pch = 1:2, col= 1:2) plot(1:10, main = e, pch = 1:2, col= 1:2) plot(1:10, main = f, pch = 1:2, col= 1:2) plot(1:10, main = g, pch = 1:2, col= 1:2) plot(1:10, main = h, pch = 1:2, col= 1:2) plot(1:10, main = i, pch = 1:2, col= 1:2) ## title mtext(My Plots, side = 3, outer = TRUE, font = 2, line = 1, cex = 1.2, at=tmp1$x) ## draw legend par(xpd=NA) tmp2 - cnvrt.coords( tmp1$x, 0.05, input='tdev' )$usr # get location for legend legend(tmp2$x, tmp2$y, legend = c(Type 1, Type 2), pch = 1:2, col = 1:2, ncol = 2, xjust=0.5, yjust=0.5) par(op) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error: file association for 'doc\html\index.html' not available or invalid
Today HTML help stopped working. The menu command Help Html help usually brings up my default web browser (Opera 9.26), but now R gives the error Error: file association for 'doc\html\index.html' not available or invalid If I try the same menu command a second time, R crashes with the message The instruction at 0x7c9106c3 referenced memory at 0x7c9f4302. The memory could not be written. The same problem occurs when calling help from the command line, e.g. with ?license I'm running R 2.6.2 on Windows XP SP2. Googling for the error turns up nothing, and I have no idea where to look for further diagnosis. I have uninstalled, deleted the R package directories, and checked the registry for metions of \R\, but the problem remains. I can't think of any unusual changes I made before this problem arose. Any help would be appreciated. Best regards, Jon Olav Vik __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Packages micEcon, sampleSelection, and maxLik
Dear R Users: We have splitted up the micEcon package into three packages: a) Package maxLik provides tools for maximum likelihood estimations (see http://www.maxLik.org). b) Package sampleSelection provides tools for estimating Heckman-type sample selection/generalized tobit models (see http://www.sampleSelection.org). c) Package micEcon contains the remainder, i.e. mainly tools for microeconomic demand and firm models, e.g. estimating the Almost Ideal Demand System or the Symmetric Normalized Quadratic / Symmetric Generalized McFadden profit function (see http://www.micEcon.org). All these packages are available for download from CRAN and from their websites (see above). Any comments and suggestions on these packages are welcome! Ott Arne -- 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-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using xyplot with groups and panel.linejoin
k.m.csillery at sms.ed.ac.uk writes: I am using xyplot() with many groups like this: statselect - levels(dat$stat) xyplot(relmse~T|lambda, groups=stat, data=dat, panel = panel.superpose, key=simpleKey(statselect, lines=T)) Add lty=l. And remove the panel=panel.superpose, it's the default when groups are given. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Passing function to tapply as a string
Hi,
Was wondering if it is possible to pass function name as a parameter, smth
along this line
param.to.pass-c(1,'max','h')
dd-function(dfd, param=param.to.pass,...){
ttime.int - format(ttime,fmt)
data.frame(
param[3] = tapply(dfd[,param[1]],ttime.int,param[3]),
...)
}
I know there is a as.formula expression but not quite sure if there is some
way to accomplish what i need.
Thanks
--
View this message in context:
http://www.nabble.com/Passing-function-to-tapply-as-a-string-tp15891151p15891151.html
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Reading microsoft .xls format and openoffice OpenDocument files
1. I have used gdata::read.xls() with much happiness. But every now and then it breaks. I have not, as yet, been able to construct a mental model about the class of .xls files for which it works. Does someone have a simple rule for predicting the circumstances under which it will work? 2. Just like there is a read.xls(), it'd be great if we have a read.ods() which directly reads files from openoffice. This should be easier than grokking Microsoft formats given that openoffice is gpl. I hunted a bit and couldn't find any. Does someone know how we might approach this? Am I correct in thinking that our goal is reading OpenDocument files (http://en.wikipedia.org/wiki/OpenDocument) ? -- Ajay Shah http://www.mayin.org/ajayshah [EMAIL PROTECTED] http://ajayshahblog.blogspot.com *(:-? - wizard who doesn't know the answer. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: file association for 'doc\html\index.html' not available or invalid
On Fri, 7 Mar 2008, Jon Olav Vik wrote: Today HTML help stopped working. The menu command Help Html help usually brings up my default web browser (Opera 9.26), but now R gives the error Error: file association for 'doc\html\index.html' not available or invalid If I try the same menu command a second time, R crashes with the message The instruction at 0x7c9106c3 referenced memory at 0x7c9f4302. The memory could not be written. The same problem occurs when calling help from the command line, e.g. with ?license I'm running R 2.6.2 on Windows XP SP2. Googling for the error turns up nothing, and I have no idea where to look for further diagnosis. I have uninstalled, deleted the R package directories, and checked the registry for metions of \R\, but the problem remains. I can't think of any unusual changes I made before this problem arose. Try re-setting the file association for .html files -- it looks as if it has been corrupted (not by R). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dictionary lookup
Duncan Murdoch wrote:
On 06/03/2008 6:45 PM, Thomas Manke wrote:
Hi,
I have a character-valued vector (old_names) and want to translate
its entries whenever possible, using a dictionary (dict=data.frame).
The translation direction is dict$V3 -- dict$V2, but
some values may be undefined (NA). I suppose this is a very basic
task, but I tried in vain to make it more efficient than below.
In particular I would like to avoid the explicit (and slow) loop
Any help is very much welcome.
Thank you, TM
new_names = old_names
m = match(old_names, dict$V3)
N = length(old_names)
for (i in 1:N) {
if (is.na(m[i])) { next ; }
nn = as.vector(dict$V2)[m[i]];
if (nn == ) { next; }
new_names[i] = nn
}
You can vectorize this and it should be fast. Here's a
straightforward replacement for the loop. It keeps the first 2 lines,
and replaces the rest with two more:
new_names - old_names
m - match(old_names, dict$V3)
change - !is.na(m)
new_names[change] - dict$V2[m[change]]
Duncan Murdoch
Thank you all for your responses, it certainly pointed me to the right
direction.
For my purposes, I only had to slightly modify Duncan's suggestion
new_names - rownames(E)
m - match(rownames(E), dict$V3)
change - ( !is.na(m) dict$V2[m] != )
new_names[change] - as.vector(dict$V2[m[change]] )
Best wishes, TM
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Passing function to tapply as a string
Yes,
tapply(rnorm(100), gl(5,20), max)
On 07/03/2008, Yuri Volchik [EMAIL PROTECTED] wrote:
Hi,
Was wondering if it is possible to pass function name as a parameter, smth
along this line
param.to.pass-c(1,'max','h')
dd-function(dfd, param=param.to.pass,...){
ttime.int - format(ttime,fmt)
data.frame(
param[3] = tapply(dfd[,param[1]],ttime.int,param[3]),
...)
}
I know there is a as.formula expression but not quite sure if there is some
way to accomplish what i need.
Thanks
--
View this message in context:
http://www.nabble.com/Passing-function-to-tapply-as-a-string-tp15891151p15891151.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Help with 'memory not mapped'
Dear Ramon, I'm afraid I'm the author of the C function. Although I'm not a 'real' programmer I need to do some programming in my research work. As you say, I've used MAKEFLAGS=CFLAGS=-O1, and valgrind, with the expression you've said. The content of the log file contains three blocks of the type: - ==6464== Invalid read of size 8 ==6464==at 0x401433E: (within /lib64/ld-2.6.1.so) ==6464==by 0x4009631: (within /lib64/ld-2.6.1.so) ==6464==by 0x5CAA804: (within /lib64/libc-2.6.1.so) ==6464==by 0x59B0143: (within /lib64/libdl-2.6.1.so) ==6464==by 0x400C8E5: (within /lib64/ld-2.6.1.so) ==6464==by 0x59B036C: (within /lib64/libdl-2.6.1.so) ==6464==by 0x59B00F9: dlsym (in /lib64/libdl-2.6.1.so) ==6464==by 0x54F60C: R_local_dlsym (dynload.c:214) ==6464==by 0x4172F8: AddDLL (Rdynload.c:565) ==6464==by 0x4179F3: do_dynload (Rdynload.c:895) ==6464==by 0x4C3A44: do_internal (names.c:1120) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464== Address 0x64A7608 is 24 bytes inside a block of size 27 alloc'd ==6464==at 0x4C21D06: malloc (in /usr/lib64/valgrind/amd64-linux/vgpreload_memcheck.so) ==6464==by 0x4172CC: AddDLL (Rdynload.c:557) ==6464==by 0x4179F3: do_dynload (Rdynload.c:895) ==6464==by 0x4C3A44: do_internal (names.c:1120) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464==by 0x495C87: Rf_eval (eval.c:507) ==6464==by 0x4B4C04: Rf_ReplIteration (main.c:263) ==6464==by 0x4B4EB7: R_ReplConsole (main.c:312) ==6464==by 0x4B518F: run_Rmainloop (main.c:975) ==6464==by 0x414AF7: main (Rmain.c:35) -- and after a lot of block of the type 'uninitialised value', it finishes like this: - ==6464== Conditional jump or move depends on uninitialised value(s) ==6464==at 0x5BF324C: (within /lib64/libc-2.6.1.so) ==6464==by 0x5BFB8FA: __printf_fp (in /lib64/libc-2.6.1.so) ==6464==by 0x5BF5587: vfprintf (in /lib64/libc-2.6.1.so) ==6464==by 0x5BFE509: printf (in /lib64/libc-2.6.1.so) ==6464==by 0x88ACE24: rainfallrunoffmodel (rain_runoff_modelv1_2.c:517) ==6464==by 0x4750D1: do_dotcall (dotcode.c:1136) ==6464==by 0x495F36: Rf_eval (eval.c:489) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464==by 0x495C87: Rf_eval (eval.c:507) ==6464==by 0x499B6E: do_set (eval.c:1407) ==6464== ==6464== Invalid write of size 8 ==6464==at 0x88AD228: rainfallrunoffmodel (rain_runoff_modelv1_2.c:595) ==6464==by 0x4750D1: do_dotcall (dotcode.c:1136) ==6464==by 0x495F36: Rf_eval (eval.c:489) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464==by 0x495C87: Rf_eval (eval.c:507) ==6464==by 0x499B6E: do_set (eval.c:1407) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x49938D: do_for (eval.c:1058) ==6464== Address 0x26C7E9E0 is not stack'd, malloc'd or (recently) free'd ==6464== ==6464== Syscall param write(buf) points to uninitialised byte(s) ==6464==at 0x5C6AC40: write (in /lib64/libc-2.6.1.so) ==6464== Address 0x402A3DB is not stack'd, malloc'd or (recently) free'd ==6464== ==6464== ERROR SUMMARY: 35923 errors from 210 contexts (suppressed: 244 from 3) ==6464== malloc/free: in use at exit: 284,245,283 bytes in 14,390 blocks. ==6464== malloc/free: 89,306 allocs, 74,916 frees, 1,051,054,834 bytes allocated. ==6464== For counts of detected errors, rerun with: -v ==6464== searching for pointers to 14,390 not-freed blocks. ==6464== checked 141,122,600 bytes. ==6464== ==6464== ==6464== 64 bytes in 16 blocks are definitely lost in loss record 12 of 54 ==6464==at 0x4C21D06: malloc (in /usr/lib64/valgrind/amd64-linux/vgpreload_memcheck.so) ==6464==by 0x4C21D80: realloc (in /usr/lib64/valgrind/amd64-linux/vgpreload_memcheck.so) ==6464==by 0x51BE04: parse_expression (regex.c:5205) ==6464==by 0x51BFF3: parse_branch (regex.c:4715) ==6464==by 0x51C07F: parse_reg_exp (regex.c:4667) ==6464==by 0x51C9A6: Rf_regcomp (regex.c:4636) ==6464==by 0x43BD06: do_gsub (character.c:1222) ==6464==by 0x4C3A44: do_internal (names.c:1120) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464== ==6464== ==6464== 87 bytes in 2 blocks are definitely lost in loss record 13 of 54 ==6464==at 0x4C21D06: malloc (in /usr/lib64/valgrind/amd64-linux/vgpreload_memcheck.so) ==6464==by 0x8AD7DBE: G_strdup (strings.c:195) ==6464==by 0x8AD5A78: G_site_get_head (sites.c:818) ==6464==by 0x8AD6808: sitesget (sitesget.c:66) ==6464==by 0x47757D: do_dotcall (dotcode.c:868) ==6464==by
[R] How to navigate in layout() created graph?
Hi I created a complex layout with using layout() and it looks exactly as I need it. But I don't want to print in the order in which the subfigure are numbered, but in a different order. How can I navigate in the layout so that I can specify the subfigure in which to plot? At the moment I am using a function which is converting the number to mfg parameter for par, but it does not seem to work as expected. Thanks, Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to plot raster obtained from readRAST6 in grey scale?
Hi I have a raster which I would like to plot in a greyscale instead of colour. Is this possible, and how? Thanks Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error
Hello! I need some help, because I don't know how this error means: Error: variables ‘Output1’, ‘Output2’, ‘Output3’, ‘Output4’, ‘Output5’ were specified with different types from the fit Execution halted Can you help me? Thank You __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with 'memory not mapped'
Dear Javier, On Fri, Mar 7, 2008 at 1:09 PM, [EMAIL PROTECTED] wrote: Dear Ramon, I'm afraid I'm the author of the C function. Although I'm not a 'real' Oh, oh... that's too bad. There is not anyone else to blame, then :-). programmer I need to do some programming in my research work. As you say, I've used MAKEFLAGS=CFLAGS=-O1, and valgrind, with the expression you've said. When looking at the valgrind output, search for lines where your library is mentioned (rain_runoff_modelv ?). Ignore everything else about libraries that are not your own (e.g., system or R). Just stick to the issues that relate to your code. From the output you've sent, there is Conditional jump or move depends on uninitialised value(s) on ==6464==by 0x88ACE24: rainfallrunoffmodel (rain_runoff_modelv1_2.c:517) but more worrisome than that (and I think the immediate cause of the segfault) is: ==6464== Invalid write of size 8 ==6464==at 0x88AD228: rainfallrunoffmodel (rain_runoff_modelv1_2.c:595) so it seems your code is trying to write to a place it shouldn't. Depending on when, what, and to where exactly it is trying to write, your code will crash sooner or later. HTH, R. The content of the log file contains three blocks of the type: - ==6464== Invalid read of size 8 ==6464==at 0x401433E: (within /lib64/ld-2.6.1.so) ==6464==by 0x4009631: (within /lib64/ld-2.6.1.so) ==6464==by 0x5CAA804: (within /lib64/libc-2.6.1.so) ==6464==by 0x59B0143: (within /lib64/libdl-2.6.1.so) ==6464==by 0x400C8E5: (within /lib64/ld-2.6.1.so) ==6464==by 0x59B036C: (within /lib64/libdl-2.6.1.so) ==6464==by 0x59B00F9: dlsym (in /lib64/libdl-2.6.1.so) ==6464==by 0x54F60C: R_local_dlsym (dynload.c:214) ==6464==by 0x4172F8: AddDLL (Rdynload.c:565) ==6464==by 0x4179F3: do_dynload (Rdynload.c:895) ==6464==by 0x4C3A44: do_internal (names.c:1120) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464== Address 0x64A7608 is 24 bytes inside a block of size 27 alloc'd ==6464==at 0x4C21D06: malloc (in /usr/lib64/valgrind/amd64-linux/vgpreload_memcheck.so) ==6464==by 0x4172CC: AddDLL (Rdynload.c:557) ==6464==by 0x4179F3: do_dynload (Rdynload.c:895) ==6464==by 0x4C3A44: do_internal (names.c:1120) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464==by 0x495C87: Rf_eval (eval.c:507) ==6464==by 0x4B4C04: Rf_ReplIteration (main.c:263) ==6464==by 0x4B4EB7: R_ReplConsole (main.c:312) ==6464==by 0x4B518F: run_Rmainloop (main.c:975) ==6464==by 0x414AF7: main (Rmain.c:35) -- and after a lot of block of the type 'uninitialised value', it finishes like this: - ==6464== Conditional jump or move depends on uninitialised value(s) ==6464==at 0x5BF324C: (within /lib64/libc-2.6.1.so) ==6464==by 0x5BFB8FA: __printf_fp (in /lib64/libc-2.6.1.so) ==6464==by 0x5BF5587: vfprintf (in /lib64/libc-2.6.1.so) ==6464==by 0x5BFE509: printf (in /lib64/libc-2.6.1.so) ==6464==by 0x88ACE24: rainfallrunoffmodel (rain_runoff_modelv1_2.c:517) ==6464==by 0x4750D1: do_dotcall (dotcode.c:1136) ==6464==by 0x495F36: Rf_eval (eval.c:489) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464==by 0x495C87: Rf_eval (eval.c:507) ==6464==by 0x499B6E: do_set (eval.c:1407) ==6464== ==6464== Invalid write of size 8 ==6464==at 0x88AD228: rainfallrunoffmodel (rain_runoff_modelv1_2.c:595) ==6464==by 0x4750D1: do_dotcall (dotcode.c:1136) ==6464==by 0x495F36: Rf_eval (eval.c:489) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x497CEB: Rf_applyClosure (eval.c:669) ==6464==by 0x495C87: Rf_eval (eval.c:507) ==6464==by 0x499B6E: do_set (eval.c:1407) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x49698F: do_begin (eval.c:1159) ==6464==by 0x495D65: Rf_eval (eval.c:463) ==6464==by 0x49938D: do_for (eval.c:1058) ==6464== Address 0x26C7E9E0 is not stack'd, malloc'd or (recently) free'd ==6464== ==6464== Syscall param write(buf) points to uninitialised byte(s) ==6464==at 0x5C6AC40: write (in /lib64/libc-2.6.1.so) ==6464== Address 0x402A3DB is not stack'd, malloc'd or (recently) free'd ==6464== ==6464== ERROR SUMMARY: 35923 errors from 210 contexts (suppressed: 244 from 3) ==6464== malloc/free: in use at exit: 284,245,283 bytes in 14,390 blocks. ==6464== malloc/free: 89,306 allocs, 74,916 frees, 1,051,054,834 bytes allocated. ==6464== For counts of detected errors, rerun with: -v ==6464== searching for pointers to 14,390 not-freed blocks. ==6464== checked 141,122,600 bytes. ==6464== ==6464== ==6464== 64 bytes in 16 blocks are definitely lost in loss record
Re: [R] How to navigate in layout() created graph?
On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote: Hi I created a complex layout with using layout() and it looks exactly as I need it. But I don't want to print in the order in which the subfigure are numbered, but in a different order. How can I navigate in the layout so that I can specify the subfigure in which to plot? At the moment I am using a function which is converting the number to mfg parameter for par, but it does not seem to work as expected. Thanks, Rainer Look at documentation of ?layout. The mat argument determines the order in which the figures appear. The example code that talks about scatterplot with marginal histograms should show you how that works. In other words, to the best of my knowledge, you specify in the call to layout in which order the new figures you create will be filling the layout. -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate in layout() created graph?
On 07/03/2008, Charilaos Skiadas [EMAIL PROTECTED] wrote: On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote: Hi I created a complex layout with using layout() and it looks exactly as I need it. But I don't want to print in the order in which the subfigure are numbered, but in a different order. How can I navigate in the layout so that I can specify the subfigure in which to plot? At the moment I am using a function which is converting the number to mfg parameter for par, but it does not seem to work as expected. Thanks, Rainer Look at documentation of ?layout. The mat argument determines the order in which the figures appear. The example code that talks about scatterplot with marginal histograms should show you how that works. In other words, to the best of my knowledge, you specify in the call to layout in which order the new figures you create will be filling the layout. Yes - I know that I can define the order when I call layout(mat=..., ...), but is there any way that I can skip one when plotting and return to plot that one later? (e.g. plot subfigure 1, then 3 and finally 2) -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa Haris Skiadas Department of Mathematics and Computer Science Hanover College -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] training svm
On Mar 7, 2008, at 2:17 AM, Oldrich Kruza wrote:
Hello Soumyadeep,
if you store the data in a tabular file, then I suggest using standard
text-editing tools like cut (say your file is called data.csv, fields
are separated with commas and you want to get rid of the third and
sixth column):
$ cut --complement --delimiter=, --fields=3,6 data.csv
data_cut.csv
If you're not in an Unix environment but have perl, then you may use a
script like:
open SRC, data.csv or die(couldn't open source);
open DST, data_cut.csv or die(couldn't open destination);
while (SRC) {
chomp;
@fields = split /,/;#substitute the comma for the
delimiter you use
splice @fields, 2, 1;#get rid of third column (they're
zero-based, thus 2 instead of 3)
splice @fields, 5, 1;#get rid of sixth column
print DST join(,, @fields), \n;
}
If you need to do the selection within R, then you can do it by
indexing the data structure. Suppose you have the data in a data.frame
called data. Then:
data - data[,-6]
data - data[,-3]
might do the trick (but since I'm not much of an R hacker, this is
without guarantee). I think it might be better however to do the
preprocessing before the data get into R because then you avoid
loading the columns to discard into memory.
I am guessing that the data is already in R, so it should be easier
to do it in R, especially if he doesn't know which columns are the
ones with all identical values. For instance, suppose the data set is
called x. Then the following would return TRUE for the columns that
have all values the same:
allsame - sapply(x,function(y) length(table(y))==1)
and then the following will take them out
newdata - x[,!allsame]
Hope this helps
~ Oldrich
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
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Re: [R] Error
I need some help, because I don't know how this error means: Error:
variables ?Output1?, ?Output2?, ?Output3?, ?Output4?, ?Output5? were
specified with different types from the fit
Execution halted
Can you help me?
No!
Please read the posting guide at
http://www.R-project.org/posting-guide.html
Please give a reproducible example that tells us exactly what you were
doing when you got the error message (and give a subject that is more
informative than 'Error').
Regards,
Richie.
Mathematical Sciences Unit
HSL
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate in layout() created graph?
On Mar 7, 2008, at 8:02 AM, Rainer M Krug wrote: On 07/03/2008, Charilaos Skiadas [EMAIL PROTECTED] wrote: On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote: Hi I created a complex layout with using layout() and it looks exactly as I need it. But I don't want to print in the order in which the subfigure are numbered, but in a different order. How can I navigate in the layout so that I can specify the subfigure in which to plot? At the moment I am using a function which is converting the number to mfg parameter for par, but it does not seem to work as expected. Thanks, Rainer Look at documentation of ?layout. The mat argument determines the order in which the figures appear. The example code that talks about scatterplot with marginal histograms should show you how that works. In other words, to the best of my knowledge, you specify in the call to layout in which order the new figures you create will be filling the layout. Yes - I know that I can define the order when I call layout(mat=..., ...), but is there any way that I can skip one when plotting and return to plot that one later? (e.g. plot subfigure 1, then 3 and finally 2) Only if you number them in the right order. You might have better luck using split.screen, and the associated screen command, for what you are trying to do. I'm not sure I understand it, why don't you want to just number the subfigures in the order in which you will draw them? -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa Haris Skiadas Department of Mathematics and Computer Science Hanover College -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rpart and bagging - how is it done?
On Fri, 7 Mar 2008, Prof Brian Ripley wrote: I believe that the procedure you describe at the end (resampling the cases) is the original interpretation of bagging, and that using weighting is equivalent when a procedure uses case weights. If you are getting different results when replicating cases and when using weights then rpart is not using its weights strictly as case weights and it would be preferable to replicate cases. But I am getting identical predictions by the two routes: ind - sample(1:81, replace=TRUE) rpart(Kyphosis ~ Age + Number + Start, data=kyphosis[ind,], xval=0) rpart(Kyphosis ~ Age + Number + Start, data=kyphosis, weights=tabulate(ind, nbins=81), xval=0) My memory is that rpart uses unweighted numbers for its control params (unlike tree) and hence is not strictly using case weights. I believe you can avoid that by setting the control params to their minimum and relying on pruning. BTW, it is inaccurate to call these trees 'non-pruned' -- the default setting of cp is still (potentially) doing quite a lot of pruning. Torsten Hothorn can explain why he chose to do what he did. There's a small (but only small) computational advantage in using case weights, but the tricky issue for me is how precisely tree growth is stopped, and I don't think that rpart at its default settings is mimicing what Breiman was doing (he would have been growing much larger trees). its mainly used to avoid repeated formula parsing and other data preprocessing steps everytime a tree is grown (which in my experience can be quite a substancial advantage both with respect to speed and memory consumption). As Brian said, rpart doesn't really interpret weights as case weights and thus the example code from the book is not totally correct. However, for example, party::ctree accepts case weights. Best wishes, Torsten On Thu, 6 Mar 2008, [EMAIL PROTECTED] wrote: Hi there. I was wondering if somebody knows how to perform a bagging procedure on a classification tree without running the classifier with weights. Let me first explain why I need this and then give some details of what I have found out so far. I am thinking about implementing the bagging procedure in Matlab. Matlab has a simple classification tree function (in their Statistics toolbox) but it does not accept weights. A modification of the Matlab procedure to accommodate weights would be very complicated. The rpart function in R accepts weights. This seems to allow for a rather simple implementation of bagging. In fact Everitt and Hothorn in chapter 8 of A Handbook of Statistical Analyses Using R describe such a procedure. The procedure consists in generating several samples with replacement from the original data set. This data set has N rows. The implementation described in the book first fits a non-pruned tree to the original data set. Then it generates several (say, 25) multinomial samples of size N with probabilities 1/N. Then, each sample is used in turn as the weight vector to update the original tree fit. Finally, all the updated trees are combined to produce consensus class predictions. Now, a typical realization of a multinomial sample consists of small integers and several 0's. I thought that the way that weighting worked was this: the observations with weights equal to 0 are omitted and the observations with weights 1 are essentially replicated according to the weight. So I thought that instead of running the rpart procedure with weights, say, starting with (1, 0, 2, 0, 1, ... etc.) I could simply generate a sample data set by retaining row 1, omitting row 2, replicating row 3 twice, omitting row 4, retaining row 5, etc. However, this does not seem to work as I expected. Instead of getting identical trees (from running weighted rpart on the original data set and running rpart on the sample data set described above with no weighting) I get trees that are completely different (different threshold values and different order of variables entering the splits). Moreover, the predictions from these trees can be different so the misclassification rates usually differ. This finally brings me to my question - is there a way to mimic the workings of the weighting in rpart by, for example, modification of the data set or, perhaps, some other means. Thanks in advance for your time, Andy __ Andy Jaworski 518-1-01 Process Laboratory 3M Corporate Research Laboratory - E-mail: [EMAIL PROTECTED] Tel: (651) 733-6092 Fax: (651) 736-3122 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of
Re: [R] Passing function to tapply as a string
Was wondering if it is possible to pass function name as a parameter
Yes. This isn't exactly what you wanted, but it demonstrates the
principle.
x = rnorm(5)
[1] -0.6510448 0.4591730 1.3225205 1.2314391 -0.0888139
myfun - function(fname, x) eval(parse(text=paste(fname,(x),sep=)))
myfun('max',x)
[1] 1.322521
myfun('min',x)
[1] -0.6510448
Regards,
Richie.
Mathematical Sciences Unit
HSL
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate in layout() created graph?
On 07/03/2008, Charilaos Skiadas [EMAIL PROTECTED] wrote: On Mar 7, 2008, at 8:02 AM, Rainer M Krug wrote: On 07/03/2008, Charilaos Skiadas [EMAIL PROTECTED] wrote: On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote: Hi I created a complex layout with using layout() and it looks exactly as I need it. But I don't want to print in the order in which the subfigure are numbered, but in a different order. How can I navigate in the layout so that I can specify the subfigure in which to plot? At the moment I am using a function which is converting the number to mfg parameter for par, but it does not seem to work as expected. Thanks, Rainer Look at documentation of ?layout. The mat argument determines the order in which the figures appear. The example code that talks about scatterplot with marginal histograms should show you how that works. In other words, to the best of my knowledge, you specify in the call to layout in which order the new figures you create will be filling the layout. Yes - I know that I can define the order when I call layout(mat=..., ...), but is there any way that I can skip one when plotting and return to plot that one later? (e.g. plot subfigure 1, then 3 and finally 2) Only if you number them in the right order. You might have better luck using split.screen, and the associated screen command, for what you are trying to do. I'm not sure I understand it, why don't you want to just number the subfigures in the order in which you will draw them? Because i thought it would be easier the other way round? Thanks anyway Rainer -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa Haris Skiadas Department of Mathematics and Computer Science Hanover College -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa Haris Skiadas Department of Mathematics and Computer Science Hanover College -- Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT) Plant Conservation Unit Department of Botany University of Cape Town Rondebosch 7701 South Africa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] LaTeX in R
Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? Best, Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate in layout() created graph?
On Mar 7, 2008, at 8:41 AM, Rainer M Krug wrote: I'm not sure I understand it, why don't you want to just number the subfigures in the order in which you will draw them? Because i thought it would be easier the other way round? Thanks anyway Yes, I agree it should not be as hard as it is ;). I am guessing there might be some technical reasons why layout doesn't make it easy to do this. split.screen is good for that. Rainer Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: mean and sd with number of values?
Hi Martin Kaffanke [EMAIL PROTECTED] napsal dne 06.03.2008 18:05:03: Am Donnerstag, den 06.03.2008, 17:47 +0100 schrieb Petr PIKAL: Hi [EMAIL PROTECTED] napsal dne 06.03.2008 17:41:06: Hi there, When i do mean(fl[1:20], na.rm=T) e.g. sum(!is.na(fl[1:20])) This seems to give me a sum of all the items. But I'd like to have it per column. mean(fl[1:20]) aids angststoerautismus bauchspduekrebs prostata -0.7474884 -1.0475500 -0.6267267 NA -1.7318179 fraudep manndep hautkrebs herzinfarkt leukaemie -1.2919841 -1.5124280 -0.3202036 -0.8246589 0.7550549 lungenkrebs manischdepresiv magenkrebs magersucht mannmagsucht -1.2451753 -1.9726160 1.4548394 -0.6242250 NA osteoporose panik schizophrenieschlaganfall zwangsstoer -1.7974270 NA -1.8634089 NA -1.5056106 mean(fl[1:20], na.rm=T) aids angststoerautismus bauchspduekrebs prostata -0.7474884 -1.0475500 -0.6267267 1.2168804 -1.7318179 fraudep manndep hautkrebs herzinfarkt leukaemie -1.2919841 -1.5124280 -0.3202036 -0.8246589 0.7550549 lungenkrebs manischdepresiv magenkrebs magersucht mannmagsucht -1.2451753 -1.9726160 1.4548394 -0.6242250 -0.3765245 osteoporose panik schizophrenieschlaganfall zwangsstoer -1.7974270 -1.2328361 -1.8634089 -1.7165059 -1.5056106 sum(!is.na(fl[1:20])) [1] 892 Then why not compute it across columns either by apply or by colSums apply(!is.na(fl), 2, sum) colSums(!is.na(fl)) Regards Petr So I'd like to have it like the mean, to see how many data is calculated by na.rm=T per column? Thanks, Martin [příloha signature.asc odstraněna uživatelem Petr PIKAL/CTCAP] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] LaTeX in R
Mario Maiworm wrote: Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? See ?plotmath Uwe Ligges Best, Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error
Carla Rebelo wrote: Hello! I need some help, because I don't know how this error means: Error: variables ‘Output1’, ‘Output2’, ‘Output3’, ‘Output4’, ‘Output5’ were specified with different types from the fit Please read the posting guide! Please tell us at least what produced the error in a reproducable way. I think you are using predict and the type of Output1, e.g., was a different one when estimating the model than when trying to predict from it. Uwe Ligges Execution halted Can you help me? Thank You __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to plot raster obtained from readRAST6 in grey scale?
Rainer M Krug wrote: Hi I have a raster which I would like to plot in a greyscale instead of colour. Is this possible, and how? Probably, and we may tell you how given you tell us what raster means: ?raster No documentation for 'raster' in specified packages and libraries: you could try 'help.search(raster)' Uwe Ligges Thanks Rainer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R code for selecting places spatially and by time
Andrew McFadden Andrew.McFadden at maf.govt.nz writes: Hi all The code of trying to write relates to selecting properties (given by x and y co-ordinates) spatially (distance X from infected properties identified by date) over a certain time period. i.e. what properties are within 3 km from properties infected on 2008-01-01 over the last 14 days. Is any one able to give me some clues on how to write code to solve this problem. Some sample data is as follows: x-rep(c(2660156,2660203,2658165,2659303,2661531,2660914),c(2,2,2,2,1,1) ) y-rep(c(6476767,6475013,6475487,6479659,6477004,6476388),c(2,2,2,2,1,1) ) date-as.character(rep(as.Date(c(2008-01-02,2008-01-17,2008-01-01) ,format = %Y-%m-%d),c(4,4,2))) cbind(x,y,date) This should certainly be possible. Your description of the problem isn't entirely clear to me, but here's a first approximation: dat - data.frame(x,y,date) ## data.frame is better than cbind, it can hold dates and locations infdate - as.Date(2008-01-01) infprem - subset(dat,date==infdate) otherprem - subset(dat,date=infdate) ## or: elapsed - dat$date-infdate otherprem - subset(dat,elapsed0 elapsed14) ## I'm not sure this is what you wanted in terms ## of date restrictions, but you could adjust appropriately dist - sqrt(outer(infprem$x,otherprem$x,-)^2+ outer(infprem$y,otherprem$y,-)^2) mindist - apply(dist,2,min) minval - 1000 ## (I don't know what the units are??) prem - subset(otherprem,mindistminval) ## or prem - otherprem[mindistminval,] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to do a time-stratified case-crossover analysis for air pollution data?
Dear Experts,
I am trying to do a time-stratified case-crossover analysis on air
pollution data and number of myocardial infarctions. In order to avoid
model selection bias, I started with a simple simulation.
I'm still not sure if my simulation is right. But the results I get from
the ts-case-crossover are much more variable than those from a glm.
Is this:
a. Due to the simple relation of log-rate of mi cases being = alpha
+ beta*pm that the glm results are more precise?
b. Due to me using method=approximate instead of default exact
in the clogit?
c. Due to the fact that the approximate method in clogit use
breslow for handling ties and not efron?
d. Due to that I've misunderstood how to arrange data when doing a
case-crossover, such that there is a way of using exact, that wouldn't
go berserk due to the many ties (see #Berserk script at bottom).
e. The prize to pay for using a case-crossover analysis?
It should perhaps be noted that, because of the absence of individual
data, the exposure to air-pollution (pm10) is assumed to be common to
all individuals on a certain day.
I'd be most grateful for any help and ideas on this matter.
Best regards,
Fredrik Nilsson, PhD
PS. I am aware of the limitations that Whitaker et al. presented in
Environmetrics 2007; 18: 157-171, but tried to use the time-stratified
case-crossover as it could be a simple, but fairly correct way of
doing this type of analysis.
Simulation script:
library(survival)
n-2*365
samp-100
ti-1:n
ar1-rnorm(1)
for (i in 2:n){
ar1[i]-0.2*ar1[i-1]+rnorm(1)
}
#old version of air pollution
#pm10-2 + .5*sin(2*pi*ti/365 + 127) + 0.1*ar1
startdate-1992-07-01
date1-as.Date(startdate)
dates-date1 + (ti-1)
tyda-weekdays(dates)
tymo-months(dates)
tyyear-as.character(dates)
for (i in 1:n)
{
ask-tyyear[i]
tyyear[i]-substr(ask,1,4)
}
moeff-cumsum(c(1,31,28,31,30,31,30,31,31,30,31,30))
a-as.Date(1994-12-31)
monthnames-months(a+moeff)
tymofa-match(tymo,monthnames)
moeff-.5*sin(2*pi*moeff/365 + 127)
#new version; to have strictly stationary levels of air-pollution
#within strata
pm10-2 + .5*moeff[tymofa] + 0.1*ar1
#intercept and proportionality to pm10 coefficients of log-rate
al- -2.5
be- 1.4
qres-numeric(samp)
glmres-qres
gamres-qres
for(q in 1:samp)
{
rate-exp(al + be*pm10)
mi-rpois(length(rate),lambda=rate)
date1-as.Date(startdate)
dates-date1 + (ti-1)
tyda-weekdays(dates)
tymo-months(dates)
tyyear-as.character(dates)
for (i in 1:n)
{
ask-tyyear[i]
tyyear[i]-substr(ask,1,4)
}
#replicate cases
air-data.frame(mi, tyda, tymo, tyyear, pm10, dates)
air$stratn-as.numeric(strata(air$tyda, air$tymo,air$tyyear))
lest-unique(air$stratn)
air$nctrl-0
airbase-air
#find the number of controls whithin each stratum
for (i in 1:length(lest))
{
a-which(air$stratn==lest[i])
for (j in 1:length(a))
{
air$nctrl[a[j]]-sum(air$mi[a[-j]])
}
}
#create cases and controls
cami-rep(1,sum(mi))
ctmi-rep(0,sum(air$nctrl))
capm-rep(pm10, mi)
ctpm-rep(pm10, air$nctrl)
cast-rep(air$stratn, mi)
ctst-rep(air$stratn, air$nctrl)
cady-rep(dates, mi)
ctdy-rep(dates, air$nctrl)
cases-c(cami, ctmi)
stranu-c(cast, ctst)
days-c(cady, ctdy)
pmva-c(capm, ctpm)
air2-data.frame(cases, days, pmva,stranu)
air.cl-clogit(cases~pmva + strata(stranu), method=approximate,
data=air2)
air.glm-glm(mi~pm10, family=poisson, data=air)
#air.gam-gam(mi~s(pm10), family=poisson)
qres[q]-as.numeric(coef(air.cl))
glmres[q]-as.numeric(coef(air.glm)[2])
#gamres[q]-as.numeric(coef(air.gam)[2])
}
par(mfrow=c(2,1))
plot(density(qres))
plot(density(glmres))
fivenum(qres)
fivenum(glmres)
# Berserk script
# this shows that even for very few cases, my intended way of doing TS
C-C is quite time-consuming.
#
#case-c(5,0,2,1)
#airq-c(10,2,3,3)
#
#nctrl-0*case
#for (i in 1:length(case))
#{
# nctrl[i]-sum(case[-i])
#}
#
#airca-rep(airq,case)
#airct-rep(airq,nctrl)
#
#cases-rep(1,sum(case))
#ctrls-rep(0,sum(nctrl))
#
#mi-c(cases,ctrls)
#airq-c(airca,airct)
#a-Sys.time()
#air.cl-clogit(mi~airq)
#b-Sys.time()
#b-a
#a-Sys.time()
#air.cl-clogit(mi~airq, method=approximate)
#b-Sys.time()
#b-a
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-Logo in \LaTeX (Mag. Ferri Leberl)
Dear Mag. Ferri Leberl,
I'm using something like:
--- tex.tex ---
\documentclass{article}
\usepackage{graphicx}
\usepackage{fancyvrb}
\newcommand{\Rlogo}{\protect\includegraphics[height=1.8ex,keepaspectratio]{Rlogo.pdf}}
\newcommand{\myinput}[1] {\begin{scriptsize}
\VerbatimInput[frame=single,label=#1]{#1}
\end{scriptsize}}
\title{The R logo, \Rlogo, in \LaTeX}
\author{J.R. Lobry}
\begin{document}
\maketitle
\section{Introduction to \Rlogo:}
This is about the \Rlogo~sofware suite\footnote{
\Rlogo~is available at...}.
% include source in final document
\myinput{tex.tex}
\end{document}
--
with:
unix$ pdflatex tex.tex
this is what I get:
http://pbil.univ-lyon1.fr/members/lobry/tmp/tex.pdf
The PDF version of the R-Logo I'm using is here:
http://pbil.univ-lyon1.fr/members/lobry/tmp/Rlogo.pdf
HTH,
Jean
--
Jean R. Lobry([EMAIL PROTECTED])
Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - LYON I,
43 Bd 11/11/1918, F-69622 VILLEURBANNE CEDEX, FRANCE
allo : +33 472 43 27 56 fax: +33 472 43 13 88
http://pbil.univ-lyon1.fr/members/lobry/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How to do a time-stratified case-crossover analysis for air pollution data? Unformatted text-version, with an additional note
Dear Experts,
I am trying to do a time-stratified case-crossover analysis on air pollution
data and number of myocardial infarctions. In order to avoid model selection
bias, I started with a simple simulation.
I'm still not sure if my simulation is right. But the results I get from the
ts-case-crossover are much more variable than those from a glm.
Is this:
a. Due to the simple relation of log-rate of mi cases being = alpha + beta*pm
that the glm results are more precise?
b. Due to me using method=approximate instead of default exact in the
clogit?
c. Due to the fact that the approximate method in clogit use breslow for
handling ties and not efron?
d. Due to that I've misunderstood how to arrange data when doing a
case-crossover, such that there is a way of using exact, that wouldn't go
berserk due to the many ties (see #Berserk script at bottom).
e. The prize to pay for using a case-crossover analysis?
f. Or, are my simulation results for glm overly precise due to the positive
autocorrelation (changed to -0.9 which made the time-stratifed be more precise,
but not the glm). This note was not in the HTML-version. Sorry for that.
It should perhaps be noted that, because of the absence of individual data, the
exposure to air-pollution (pm10) is assumed to be common to all individuals on
a certain day.
I'd be most grateful for any help and ideas on this matter.
Best regards,
Fredrik Nilsson, PhD
PS. I am aware of the limitations that Whitaker et al. presented in
Environmetrics 2007; 18: 157-171, but tried to use the time-stratified
case-crossover as it could be a simple, but fairly correct way of doing this
type of analysis.
Simulation script:
library(survival)
n-2*365
samp-100
ti-1:n
ar1-rnorm(1)
for (i in 2:n){
ar1[i]-0.2*ar1[i-1]+rnorm(1)
}
#old version of air pollution
#pm10-2 + .5*sin(2*pi*ti/365 + 127) + 0.1*ar1
startdate-1992-07-01
date1-as.Date(startdate)
dates-date1 + (ti-1)
tyda-weekdays(dates)
tymo-months(dates)
tyyear-as.character(dates)
for (i in 1:n)
{
ask-tyyear[i]
tyyear[i]-substr(ask,1,4)
}
moeff-cumsum(c(1,31,28,31,30,31,30,31,31,30,31,30))
a-as.Date(1994-12-31)
monthnames-months(a+moeff)
tymofa-match(tymo,monthnames)
moeff-.5*sin(2*pi*moeff/365 + 127)
#new version; to have strictly stationary levels of air-pollution
#within strata
pm10-2 + .5*moeff[tymofa] + 0.1*ar1
#intercept and proportionality to pm10 coefficients of log-rate
al- -2.5
be- 1.4
qres-numeric(samp)
glmres-qres
gamres-qres
for(q in 1:samp)
{
rate-exp(al + be*pm10)
mi-rpois(length(rate),lambda=rate)
date1-as.Date(startdate)
dates-date1 + (ti-1)
tyda-weekdays(dates)
tymo-months(dates)
tyyear-as.character(dates)
for (i in 1:n)
{
ask-tyyear[i]
tyyear[i]-substr(ask,1,4)
}
#replicate cases
air-data.frame(mi, tyda, tymo, tyyear, pm10, dates)
air$stratn-as.numeric(strata(air$tyda, air$tymo,air$tyyear))
lest-unique(air$stratn)
air$nctrl-0
airbase-air
#find the number of controls whithin each stratum
for (i in 1:length(lest))
{
a-which(air$stratn==lest[i])
for (j in 1:length(a))
{
air$nctrl[a[j]]-sum(air$mi[a[-j]])
}
}
#create cases and controls
cami-rep(1,sum(mi))
ctmi-rep(0,sum(air$nctrl))
capm-rep(pm10, mi)
ctpm-rep(pm10, air$nctrl)
cast-rep(air$stratn, mi)
ctst-rep(air$stratn, air$nctrl)
cady-rep(dates, mi)
ctdy-rep(dates, air$nctrl)
cases-c(cami, ctmi)
stranu-c(cast, ctst)
days-c(cady, ctdy)
pmva-c(capm, ctpm)
air2-data.frame(cases, days, pmva,stranu)
air.cl-clogit(cases~pmva + strata(stranu), method=approximate, data=air2)
air.glm-glm(mi~pm10, family=poisson, data=air)
#air.gam-gam(mi~s(pm10), family=poisson)
qres[q]-as.numeric(coef(air.cl))
glmres[q]-as.numeric(coef(air.glm)[2])
#gamres[q]-as.numeric(coef(air.gam)[2])
}
par(mfrow=c(2,1))
plot(density(qres))
plot(density(glmres))
fivenum(qres)
fivenum(glmres)
# Berserk script
# this shows that even for very few cases, my intended way of doing TS C-C is
quite time-consuming.
#
#case-c(5,0,2,1)
#airq-c(10,2,3,3)
#
#nctrl-0*case
#for (i in 1:length(case))
#{
# nctrl[i]-sum(case[-i])
#}
#
#airca-rep(airq,case)
#airct-rep(airq,nctrl)
#
#cases-rep(1,sum(case))
#ctrls-rep(0,sum(nctrl))
#
#mi-c(cases,ctrls)
#airq-c(airca,airct)
#a-Sys.time()
#air.cl-clogit(mi~airq)
#b-Sys.time()
#b-a
#a-Sys.time()
#air.cl-clogit(mi~airq, method=approximate)
#b-Sys.time()
#b-a
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] LaTeX in R
Thank you, uwe and jeremy. I was actually looking exactly for that! But something still doesn't work: I want to plot a symbol in a legend of a plot, lets say \sigma = 2. 2 should be the value of a variable. So, when I try mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=paste(expression(sigma), = ,mySigma),lty=1) , the sigma is not plotted as a symbol. This version: mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=expression(paste(sigma, = ,mySigma)),lty=1) gives me a 'real' sigma but the mySigma variable is not evaluated. Any ideas? Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 15:27 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: [R] LaTeX in R Mario Maiworm wrote: Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? See ?plotmath Uwe Ligges Best, Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-Logo in \LaTeX (Mag. Ferri Leberl)
Jean,
this is nice, but 1) the logo is a bitmap, it is ugly if you
resize it, 2) you don't need a pdf version for pdflatex, it
handles jpg (and maybe also png as well), so you can
just use the logos at the R developer site.
It would be really nice to have a non-bitmap version, though.
If it exists.
Gabor
On Fri, Mar 07, 2008 at 04:08:21PM +0100, Jean lobry wrote:
Dear Mag. Ferri Leberl,
I'm using something like:
--- tex.tex ---
\documentclass{article}
\usepackage{graphicx}
\usepackage{fancyvrb}
\newcommand{\Rlogo}{\protect\includegraphics[height=1.8ex,keepaspectratio]{Rlogo.pdf}}
\newcommand{\myinput}[1] {\begin{scriptsize}
\VerbatimInput[frame=single,label=#1]{#1}
\end{scriptsize}}
\title{The R logo, \Rlogo, in \LaTeX}
\author{J.R. Lobry}
\begin{document}
\maketitle
\section{Introduction to \Rlogo:}
This is about the \Rlogo~sofware suite\footnote{
\Rlogo~is available at...}.
% include source in final document
\myinput{tex.tex}
\end{document}
--
with:
unix$ pdflatex tex.tex
this is what I get:
http://pbil.univ-lyon1.fr/members/lobry/tmp/tex.pdf
The PDF version of the R-Logo I'm using is here:
http://pbil.univ-lyon1.fr/members/lobry/tmp/Rlogo.pdf
HTH,
Jean
--
Jean R. Lobry([EMAIL PROTECTED])
Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - LYON I,
43 Bd 11/11/1918, F-69622 VILLEURBANNE CEDEX, FRANCE
allo : +33 472 43 27 56 fax: +33 472 43 13 88
http://pbil.univ-lyon1.fr/members/lobry/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Csardi Gabor [EMAIL PROTECTED]UNIL DGM
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] boxcox.fit error
Hi, Thakns all for your help I am doing the next in my dataframe tabla, column pend1, because the Lilliefors (Kolmogorov-Smirnov) test give me a pvalue alfa. (data no normal distribution). I need do a transformation with box-cox or other: bc - boxcox.fit(tabla$pend1) R send to me: Error in boxcox.fit(tabla$pend1) : Transformation requires positive data The summary for my data is: summary(tabla$pend1) Min. 1st Qu. MedianMean 3rd Qu.Max. 0.012.819.820.227.257.9 length(tabla$pend1) [1] 4408 Someone can help me?. What is the error? Thanks Alexys H __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] locate the rows in a dataframe with some criteria
Hi, netters, This is probably a rookie question but I couldn't find the answer after hours of searching and trying. Suppose there'a a dataframe M: x y 10 A 13 B 8 A 11 A I want to locate the rows where x =10 and y=A. I know how to do it to vectors by using which, but how to do it with the dataframe? Thank you very much! Zhihua Li _ MSN ÖÐÎÄÍø£¬×îÐÂʱÉÐÉú»î×ÊѶ£¬°×Áì¾Û¼¯ÃÅ»§¡£ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] locate the rows in a dataframe with some criteria
On Mar 7, 2008, at 10:50 AM, zhihuali wrote: Hi, netters, This is probably a rookie question but I couldn't find the answer after hours of searching and trying. Suppose there'a a dataframe M: x y 10 A 13 B 8 A 11 A I want to locate the rows where x =10 and y=A. I know how to do it to vectors by using which, but how to do it with the dataframe? Does ?subset do what you want? Thank you very much! Zhihua Li Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] locate the rows in a dataframe with some criteria
Hi Zhihua,
M - data.frame (x=c(10, 13, 8, 11), y=c('A', 'B', 'A', 'A'))
which (M$x = 10 M$y == 'A')
# [1] 1 4
Hope it helps,
Nael
2008/3/7 N. Lapidus [EMAIL PROTECTED]:
Hi Zhihua,
M - data.frame (x=c(10, 13, 8, 11), y=c('A', 'B', 'A', 'A'))
which (M$x = 10 M$y == 'A')
# [1] 1 4
Hope it helps,
Nael
2008/3/7 zhihuali [EMAIL PROTECTED]:
Hi, netters,
This is probably a rookie question but I couldn't find the answer after
hours of searching and trying.
Suppose there'a a dataframe M:
x y
10 A
13 B
8 A
11 A
I want to locate the rows where x =10 and y=A. I know how to do it to
vectors by using
which, but how to do it with the dataframe?
Thank you very much!
Zhihua Li
_
MSN ÖÐÎÄÍø£¬×îÐÂʱÉÐÉú»î×ÊѶ£¬°×Áì¾Û¼¯ÃÅ»§¡£
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] LaTeX in R
You might want to read Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation in Plots. R News 2 (3), 32-34. with an example at the end that meets your requirements: (please note that I removed those ugly ; mySigma[1] - 2 mySigma[2] - 3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10, sd = mySigma[2]), lty = 2) legend1 - substitute(sigma == myS, list(myS = mySigma[1])) legend2 - substitute(sigma == myS, list(myS = mySigma[2])) legend(x = topright, lty = c(1,2), legend = do.call(expression, list(legend1, legend2))) Uwe Ligges Mario Maiworm wrote: Finally, this should work for an array of sigmas. I just realized that the substitute()-command is not evaluated within a c()-environment :( mySigma[1] - 2; mySigma[2] - 3; plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') ; lines(dnorm(1:10,sd = mySigma[2]),lty = 2); legend(x = topright, lty = c(1,2),legend = c(substitute(sigma == myS, list(myS = mySigma[1])),substitute(sigma == myS, list(myS = mySigma[2] Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 16:30 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: AW: [R] LaTeX in R Mario Maiworm wrote: Thank you, uwe and jeremy. I was actually looking exactly for that! But something still doesn't work: I want to plot a symbol in a legend of a plot, lets say \sigma = 2. 2 should be the value of a variable. So, when I try mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=paste(expression(sigma), = ,mySigma),lty=1) , the sigma is not plotted as a symbol. This version: mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=expression(paste(sigma, = ,mySigma)),lty=1) gives me a 'real' sigma but the mySigma variable is not evaluated. Any ideas? Yes: mySigma - 2 plot(1:10, dnorm(1:10, sd = mySigma), type='l') legend(x = topright, lty = 1, legend = substitute(sigma == myS, list(myS = mySigma))) Uwe Ligges Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 15:27 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: [R] LaTeX in R Mario Maiworm wrote: Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? See ?plotmath Uwe Ligges Best, Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] polygon shapefile from line edge coordinate list
Hello, I am looking for advice on a task I am trying to complete. I have a 4 column dataframe defining the start and end coordinates of line edges (from a CGAL alpha shapes function to define concave hulls from point clusters). I would like to create polygon shapefiles from these line edges, presumably creating lines first and then polygons. e.g. columns are: startX startY endX endY where each row represents start and end coordinates of a line segment. I am new to R spatial packages and I am not sure which one is best suited to this task. There seems to be a lot of options, which is great but hard to know which one to start with. Any suggestions on the best way to proceed with this? One further challenge is that the list includes line segments defining multiple polygons. Thanks for any advice. Murray __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] LaTeX in R
Or my personal favorite if the length of mySigma is variable: mySigma - 2:3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10,sd = mySigma[2]),lty = 2) leg - as.expression(lapply(mySigma, function(x) bquote(sigma == .(x legend(x = topright, lty = c(1,2),legend = leg) Thanks, --sundar Uwe Ligges said the following on 3/7/2008 8:15 AM: You might want to read Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation in Plots. R News 2 (3), 32-34. with an example at the end that meets your requirements: (please note that I removed those ugly ; mySigma[1] - 2 mySigma[2] - 3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10, sd = mySigma[2]), lty = 2) legend1 - substitute(sigma == myS, list(myS = mySigma[1])) legend2 - substitute(sigma == myS, list(myS = mySigma[2])) legend(x = topright, lty = c(1,2), legend = do.call(expression, list(legend1, legend2))) Uwe Ligges Mario Maiworm wrote: Finally, this should work for an array of sigmas. I just realized that the substitute()-command is not evaluated within a c()-environment :( mySigma[1] - 2; mySigma[2] - 3; plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') ; lines(dnorm(1:10,sd = mySigma[2]),lty = 2); legend(x = topright, lty = c(1,2),legend = c(substitute(sigma == myS, list(myS = mySigma[1])),substitute(sigma == myS, list(myS = mySigma[2] Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 16:30 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: AW: [R] LaTeX in R Mario Maiworm wrote: Thank you, uwe and jeremy. I was actually looking exactly for that! But something still doesn't work: I want to plot a symbol in a legend of a plot, lets say \sigma = 2. 2 should be the value of a variable. So, when I try mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=paste(expression(sigma), = ,mySigma),lty=1) , the sigma is not plotted as a symbol. This version: mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=expression(paste(sigma, = ,mySigma)),lty=1) gives me a 'real' sigma but the mySigma variable is not evaluated. Any ideas? Yes: mySigma - 2 plot(1:10, dnorm(1:10, sd = mySigma), type='l') legend(x = topright, lty = 1, legend = substitute(sigma == myS, list(myS = mySigma))) Uwe Ligges Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 15:27 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: [R] LaTeX in R Mario Maiworm wrote: Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? See ?plotmath Uwe Ligges Best, Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with Error!
Hi, Can anyone explain the following error?? Error in FUN(newX[, i], ...) : missing observations in cov/cor In addition: Warning message: In FUN(newX[, i], ...) : NAs introduced by coercion svm_modelSAheart1 - svm(x_training, y_training) is the command i am using.my x/y training are working fine. If anyone needs more information just let me know! Hope to hear from someone soon. Regards. -- View this message in context: http://www.nabble.com/Help-with-Error%21-tp15900037p15900037.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rpart and bagging - how is it done?
I would like to thank Brian Ripley and Torsten Hothorn for their quick and thoughtful responses. I rerun the example given by Professor Ripley by just starting R and sourcing the code below and I got slightly different results. Then I ran it again setting the random seed before the sample command and I got identical results a few times. However, I found the example below that seems to be a reproducible on my system (Win200 Pro, CoreDuo Xeon about a year old). I get the same results in 2.6.2 (patched March 4) and 2.7.0 (version of February 28). Both were compiled from the tarballs in Cygwin and up-to-date Rtools with no errors. I just ran make fullcheck on 2.6.2 and it passes with no problems (just usual stuff - network conectivity fails due to our firewall and slight numercial differences in a few cases. The results from the rpart test are attached included at the bottom of this post. set.seed(123) library(rpart) ind - sample(1:81, replace=TRUE) rpart(Kyphosis ~ Age + Number + Start, data=kyphosis[ind,], xval=0) rpart(Kyphosis ~ Age + Number + Start, data=kyphosis, weights=tabulate(ind, nbins=81), xval=0) Here is what I get: rpart(Kyphosis ~ Age + Number + Start, data=kyphosis[ind,], xval=0) n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 14 absent (0.8271605 0.1728395) * rpart(Kyphosis ~ Age + Number + Start, data=kyphosis, +weights=tabulate(ind, nbins=81), xval=0) n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 14 absent (0.8271605 0.1728395) 2) Start=8.5 62 6 absent (0.9062500 0.0937500) 4) Start=14.5 29 0 absent (1.000 0.000) * 5) Start 14.5 33 6 absent (0.800 0.200) 10) Age 55 12 0 absent (1.000 0.000) * 11) Age=55 21 6 absent (0.600 0.400) 22) Age=111 14 2 absent (0.800 0.200) * 23) Age 111 7 1 present (0.200 0.800) * 3) Start 8.5 19 8 absent (0.5294118 0.4705882) * The trees are dramatically different (the first one is just a root). The predictions are of course different (the first model predicts all cases as absent) but the total number of misclassified observations differs by only 1 (17 vs. 16). Can anyone reproduce this, or is something wrong with my system? Thanks again, Andy PS. rpart version is 3.1-39 rpart results from make fullcheck Testing package rpart Massaging examples into 'rpart-Ex.R' ... Running examples in 'rpart-Ex.R' ... Running specific tests Running `surv_test.R' Running `testall.R' Comparing `testall.Rout' to `testall.Rout.save' ...127c127 g2 22.77 to the right, improve=6.8130, (6 missing) --- g2 22.76 to the right, improve=6.8130, (6 missing) 159c159 g2 22.77 to the right, improve=4.8340, (6 missing) --- g2 22.76 to the right, improve=4.8340, (6 missing) 193c193 grade 3.5 to the left, agree=0.772, adj=0.188, (0 split) --- grade 3.5 to the left, agree=0.772, adj=0.187, (0 split) 199c199 g2 13.47 to the left, improve=3.55300, (0 missing) --- g2 13.48 to the left, improve=3.55300, (0 missing) 241c241 1) root 146 53.420 5.893e-18 --- 1) root 146 53.420 -4.563e-17 275c275 mean=5.893e-18, MSE=0.3659 --- mean=-4.563e-17, MSE=0.3659 346c346 g2 13.47 to the left, improve=4.238e-02, (3 missing) --- g2 13.48 to the left, improve=4.238e-02, (3 missing) 375c375 g2 17.91 to the right, improve=0.1271000, (1 missing) --- g2 17.92 to the right, improve=0.1271000, (1 missing) 515c515 g2 13.47 to the left, improve=1.94600, (3 missing) --- g2 13.48 to the left, improve=1.94600, (3 missing) 555c555 g2 17.91 to the right, improve=3.122000, (1 missing) --- g2 17.92 to the right, improve=3.122000, (1 missing) 647c647 life70.25 to the right, improve=0.25230, (0 missing) --- life70.26 to the right, improve=0.25230, (0 missing) OK Running `usersplits.R' Comparing `usersplits.Rout' to `usersplits.Rout.save' ...174c174 Timing ratio = 3.2 --- Timing ratio = 5.9 OK __ Andy Jaworski 518-1-01 Process Laboratory 3M Corporate Research Laboratory - E-mail: [EMAIL PROTECTED] Tel: (651) 733-6092 Fax: (651) 736-3122 Prof Brian Ripley [EMAIL PROTECTED] ac.uk To [EMAIL PROTECTED] 03/07/2008 03:11 cc AM[EMAIL PROTECTED] R-help@R-project.org
Re: [R] R-Logo in \LaTeX
Gabor,
this is nice, but
1) the logo is a bitmap, it is ugly if you resize it
Sure, it's a bitmap, but the naked eye resolution is only
100 $\mu$m so that a vectorized solution is overkilling
in most common situations IMHO. I have to zoom by a factor
1200% to see some pixellization problems on my screen,
but my eyes are admitedly getting older and older, one more
instance of fortune(75) issue I guess :-(
2) you don't need a pdf version for pdflatex, it
handles jpg (and maybe also png as well), so you can
just use the logos at the R developer site.
Not so sure, I had to convert it a long time ago into a PDF
format for a reason I don't remember. I'm sharing my *.rnw
on a CVS with colleages working under Linux, Unix, Mac and
Windows, so the reason could be that there was a problem
in a given platform. The PDF choice is defensive in the sense
that we are Sweaving with \SweaveOpts{pdf = T, eps = F},
so that including PDF is a pre-condition.
It would be really nice to have a non-bitmap version, though.
If it exists.
There was a not so-old thread about this on R-devel:
http://finzi.psych.upenn.edu/R/R-devel/archive/19448.html
Best,
Jean
--
Jean R. Lobry([EMAIL PROTECTED])
Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - LYON I,
43 Bd 11/11/1918, F-69622 VILLEURBANNE CEDEX, FRANCE
allo : +33 472 43 27 56 fax: +33 472 43 13 88
http://pbil.univ-lyon1.fr/members/lobry/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] LaTeX in R
I finally got everything plotted as I had planned. So, thank you again. Sundars suggestion looks more flexible when it comes to large numbers of categories (i.e. sigmas). In the plot I will be using, there are 3 sigmas... I am keen to learn about that bquote and substitute stuff. I will definitely take a look at your newsletter contribution, uwe. The ';'s were intended to ease the mail2console copy-and-paste, as some mail programs take control of your linebreaks. Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Sundar Dorai-Raj [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 17:22 An: Uwe Ligges Cc: Mario Maiworm; r-help@r-project.org Betreff: Re: [R] LaTeX in R Or my personal favorite if the length of mySigma is variable: mySigma - 2:3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10,sd = mySigma[2]),lty = 2) leg - as.expression(lapply(mySigma, function(x) bquote(sigma == .(x legend(x = topright, lty = c(1,2),legend = leg) Thanks, --sundar Uwe Ligges said the following on 3/7/2008 8:15 AM: You might want to read Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation in Plots. R News 2 (3), 32-34. with an example at the end that meets your requirements: (please note that I removed those ugly ; mySigma[1] - 2 mySigma[2] - 3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10, sd = mySigma[2]), lty = 2) legend1 - substitute(sigma == myS, list(myS = mySigma[1])) legend2 - substitute(sigma == myS, list(myS = mySigma[2])) legend(x = topright, lty = c(1,2), legend = do.call(expression, list(legend1, legend2))) Uwe Ligges Mario Maiworm wrote: Finally, this should work for an array of sigmas. I just realized that the substitute()-command is not evaluated within a c()-environment :( mySigma[1] - 2; mySigma[2] - 3; plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') ; lines(dnorm(1:10,sd = mySigma[2]),lty = 2); legend(x = topright, lty = c(1,2),legend = c(substitute(sigma == myS, list(myS = mySigma[1])),substitute(sigma == myS, list(myS = mySigma[2] Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 16:30 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: AW: [R] LaTeX in R Mario Maiworm wrote: Thank you, uwe and jeremy. I was actually looking exactly for that! But something still doesn't work: I want to plot a symbol in a legend of a plot, lets say \sigma = 2. 2 should be the value of a variable. So, when I try mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=paste(expression(sigma), = ,mySigma),lty=1) , the sigma is not plotted as a symbol. This version: mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=expression(paste(sigma, = ,mySigma)),lty=1) gives me a 'real' sigma but the mySigma variable is not evaluated. Any ideas? Yes: mySigma - 2 plot(1:10, dnorm(1:10, sd = mySigma), type='l') legend(x = topright, lty = 1, legend = substitute(sigma == myS, list(myS = mySigma))) Uwe Ligges Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 15:27 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: [R] LaTeX in R Mario Maiworm wrote: Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? See ?plotmath Uwe Ligges Best, Mario.
[R] confused about CORREP cor.LRtest
After some struggling with the data format, non-standard in BioConductor, I have gotten cor.balance in package CORREP to work. My desire was to obtain maximum-likelihood p-values from the same data object using cor.LRtest, but it appears that this function wants something different, which I can't figure out from the documentation. Briefly, my dataset consists of 36 samples from 12 conditions and I have 497 genes of interest to be correlated. The following works: M - cor.balance(stddata, m = 3, G=497) The following does not: M.p - cor.LRtest(stddata, m1 = 3, m2 = 3) Do I need to do something to stddata between example 1 and 2 or does m stand for something different in the two examples? sessionInfo follows. Thanks, Mark sessionInfo() R version 2.7.0 Under development (unstable) (2008-03-05 r44683) x86_64-unknown-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=en_US.UTF-8;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] grid tools stats graphics grDevices datasets utils [8] methods base other attached packages: [1] rat2302_2.0.1Rgraphviz_1.17.13graph_1.17.17 [4] igraph_0.5 CORREP_1.5.0 e1071_1.5-17 [7] class_7.2-41 affy_1.17.8 preprocessCore_1.1.5 [10] affyio_1.7.13Biobase_1.99.1 loaded via a namespace (and not attached): [1] cluster_1.11.10 -- Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry Indiana University School of Medicine 15032 Hunter Court, Westfield, IN 46074 (317) 490-5129 Work, Mobile VoiceMail (317) 204-4202 Home (no voice mail please) mwkimpelatgmaildotcom __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Time series panel
I have a set of data that consists of a number of biological measurements. The columns are Time that runs from 01/01/2005 to 01/5/2007, Group which has 23 levels and postcode which is nested within group. This is a balanced panel but the number of postcodes differs within groups, from 15 to 400. The rest of this data consists of a number of columns of quantitative measures, largely counts. I would like to set this up as a dataframe but retain the time series element and the structural relations within the data. How can I do this in R? Whenever I try ts I end up with things out of order without the time series element.correctly represented Secondly assuming that I wish to regress temperature against vegetation and type how do I express this as a linear hierarchical model nesting postcode within group but keeping time as non nested (eg group x time) with calendar time as a group level predictor? Any help would be most appreciated. Graham Leask Kind regards Dr Graham Leask Economics and Strategy Group Aston Business School Aston University Aston Triangle Birmingham B4 7ET Tel: Direct line 0121 204 3150 Fax: 0870 759 8408 email [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-Logo in \LaTeX
Jean,
On Fri, Mar 07, 2008 at 06:09:35PM +0100, Jean lobry wrote:
Gabor,
this is nice, but
1) the logo is a bitmap, it is ugly if you resize it
Sure, it's a bitmap, but the naked eye resolution is only
100 $\mu$m so that a vectorized solution is overkilling
in most common situations IMHO.
a vectorized solution can be rendered smoothly at any
desired size. Furthermore, it can be easily edited and the
size of the file is (usually) much smaller than the bitmap
version. No overkill at all, as it is SIMPLER to use the
vector version.
I have to zoom by a factor
1200% to see some pixellization problems on my screen,
but my eyes are admitedly getting older and older, one more
instance of fortune(75) issue I guess :-(
Hmmm, i've downloaded your pdf, and it is clearly ugly at 400%.
I'm talking about the Rlogo.pdf file. E.g. i'm not sure that
it would look good on an A0 poster.
2) you don't need a pdf version for pdflatex, it
handles jpg (and maybe also png as well), so you can
just use the logos at the R developer site.
Not so sure, I had to convert it a long time ago into a PDF
format for a reason I don't remember. I'm sharing my *.rnw
on a CVS with colleages working under Linux, Unix, Mac and
Windows, so the reason could be that there was a problem
in a given platform. The PDF choice is defensive in the sense
that we are Sweaving with \SweaveOpts{pdf = T, eps = F},
so that including PDF is a pre-condition.
Maybe this is different if you use Sweave. pdflatex itself
handles jpg without problems.
It would be really nice to have a non-bitmap version, though.
If it exists.
There was a not so-old thread about this on R-devel:
http://finzi.psych.upenn.edu/R/R-devel/archive/19448.html
Of which the conclusion is that there is no vector version.
Best,
Gabor
Best,
Jean
--
Jean R. Lobry([EMAIL PROTECTED])
Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - LYON I,
43 Bd 11/11/1918, F-69622 VILLEURBANNE CEDEX, FRANCE
allo : +33 472 43 27 56 fax: +33 472 43 13 88
http://pbil.univ-lyon1.fr/members/lobry/
--
Csardi Gabor [EMAIL PROTECTED]UNIL DGM
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[R] Problems installing packages using the inbuilt facility: Error i n gzfile(file, r) : unable to open connection
Hi
I have been trawling the web, FAQs, and R manuals for help on the following
issue, but have failed and was wondering if anyone has a solution to the
following problem:
After having installed R 2.6.2 for Windows (binary), I tried to install various
packages. Every time I try loading a package (any package) via the built-in
menu, I run into the following error message.
utils:::menuInstallPkgs()
trying URL 'http://cran.uk.r-project.org/bin/windows/contrib/2.6/ada_2.0-1.zip
http://cran.uk.r-project.org/bin/windows/contrib/2.6/ada_2.0-1.zip '
Content type 'application/zip' length 971893 bytes (949 Kb)
opened URL
downloaded 949 Kb
Warning in gzfile(file, r) :
cannot open compressed file 'ada/DESCRIPTION', probable reason 'No such file
or directory'
Error in gzfile(file, r) : unable to open connection
traceback()
6: gzfile(file, r)
5: read.dcf(file.path(curPkg, DESCRIPTION), c(Package, Version,
Type))
4: unpackPkg(foundpkgs[okp, 2], foundpkgs[okp, 1], lib, installWithVers)
3: .install.winbinary(pkgs = pkgs, lib = lib, contriburl = contriburl,
method = method, available = available, destdir = destdir,
installWithVers = installWithVers, dependencies = dependencies)
2: install.packages(NULL, .libPaths()[1], dependencies = NA, type = type)
1: utils:::menuInstallPkgs()
I tried saving the ZIP of the package to my C drive, and then installing from
that, but I get the same error message. I then tried calling gzfile with the
filename of the ZIP directly, and I seem to get a file handle without any error
message:
gzfile(C:/Program Files/R/R-2.6.2/library/ada_2.0-1.zip,r)
description
class
C:/Program Files/R/R-2.6.2/library/ada_2.0-1.zip
gzfile
mode
text
rb6
binary
opened
can read
opened
yes
can write
no
It might be worth noting, that I initially I also had problems with the
internet download facility, which I fixed starting R with the flag --internet2,
as indicated in the FAQs.
Can you please help?
Many thanks - Philipp
PS: I did the same on my PC at home, and there everything works without
problems.
==
Please access the attached hyperlink for an important el...{{dropped:8}}
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[R] Trouble with R CMD check
Friends, I changed one line of a package at the source level and then rebuilt it. When I run R CMD check, I get an error: installing R.css in C:/polsplineRS.Rcheck -- Making package polsplineRS adding build stamp to DESCRIPTION making DLL ... making hareall.d from hareall.c making heftall.d from heftall.c making lsdall.d from lsdall.c making lspecall.d from lspecall.c making nlsd.d from nlsd.c making polyall.d from polyall.c making polymars.d from polymars.c gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c hareall.c -o hareall.o hareall.c: In function `gremdim': hareall.c:2127: warning: unused variable `x' hareall.c: In function `compall': hareall.c:1656: warning: 'lala' might be used uninitialized in this function hareall.c: In function `newton': hareall.c:1876: warning: 'rall' might be used uninitialized in this function hareall.c: In function `search': hareall.c:646: warning: 'iloc' might be used uninitialized in this function hareall.c:646: warning: 'lloc' might be used uninitialized in this function hareall.c:646: warning: 'uloc' might be used uninitialized in this function hareall.c: In function `adders': hareall.c:529: warning: 'crit1' might be used uninitialized in this function hareall.c: In function `share': hareall.c:2126: warning: 'bb1' might be used uninitialized in this function hareall.c:2126: warning: 'bt1' might be used uninitialized in this function hareall.c:2126: warning: 'bb2' might be used uninitialized in this function hareall.c:2126: warning: 'bw' might be used uninitialized in this function hareall.c:2127: warning: 'wald' might be used uninitialized in this function hareall.c: In function `sphare': hareall.c:3417: warning: 's' might be used uninitialized in this function hareall.c: At top level: hareall.c:132: warning: 'iigvector' declared `static' but never defined gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c heftall.c -o heftall.o heftall.c: In function `sheft': heftall.c:301: warning: 'nintx' might be used uninitialized in this function heftall.c:301: warning: 'addi' might be used uninitialized in this function heftall.c:1848: warning: 'loloc' might be used uninitialized in this function heftall.c:1848: warning: 'uploc' might be used uninitialized in this function heftall.c:1848: warning: 'nowloc1' might be used uninitialized in this function gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c lsdall.c -o lsdall.o lsdall.c: In function `knotplace': lsdall.c:1890: warning: 'ia' might be used uninitialized in this function lsdall.c:1895: warning: 'u3' might be used uninitialized in this function lsdall.c:1895: warning: 'u4' might be used uninitialized in this function lsdall.c: In function `liter': lsdall.c:369: warning: 'zerror' might be used uninitialized in this function lsdall.c:370: warning: 'newlikelihood' might be used uninitialized in this function lsdall.c: In function `logcensor': lsdall.c:51: warning: 'xnknots' might be used uninitialized in this function lsdall.c:55: warning: 'xczheta' might be used uninitialized in this function lsdall.c: In function `pqlsd': lsdall.c:2908: warning: 'j' might be used uninitialized in this function lsdall.c: At top level: lsdall.c:1458: warning: 'fun2' defined but not used lsdall.c:1471: warning: 'fun48' defined but not used gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c lspecall.c -o lspecall.o lspecall.c: In function `tsadd': lspecall.c:626: warning: 'b2' might be used uninitialized in this function gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c nlsd.c -o nlsd.o nlsd.c: In function `ludcmp': nlsd.c:695: warning: 'imax' might be used uninitialized in this function nlsd.c: In function `setupspace': nlsd.c:1717: warning: 'm' might be used uninitialized in this function nlsd.c: In function `startspace': nlsd.c:1838: warning: 'l' might be used uninitialized in this function nlsd.c: In function `pompall': nlsd.c:1158: warning: 'f' might be used uninitialized in this function nlsd.c: In function `nlsd': nlsd.c:747: warning: 'nowloc1' might be used uninitialized in this function nlsd.c:747: warning: 'loloc' might be used uninitialized in this function nlsd.c:747: warning: 'uploc' might be used uninitialized in this function gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c polyall.c -o polyall.o polyall.c: In function `poly': polyall.c:3121: warning: 'okd' might be used uninitialized in this function polyall.c:2777: warning: 'aj' might be used uninitialized in this function polyall.c:2777: warning: 'ak' might be used uninitialized in this function polyall.c:2777: warning: 'sj' might be used uninitialized in this function polyall.c:2777: warning: 'sk' might be used uninitialized in this function polyall.c:2777: warning: 'bbi' might be used uninitialized in this function polyall.c:2777: warning: 'bbj' might be used uninitialized in this function polyall.c:2777: warning: 'bbk' might be used uninitialized in
[R] Numerical Integration in 1D
Dear UseRs,
I'm curious about the derivative of n!.
We know that Gamma(n+1)=n! So when on takes the derivative of
Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
I've tried code like
integrand-function(x) {log(x)*exp(x)*x^n}
integrate(integrand,lower=0,upper=Inf)
It seems that R doesn't like to integrate for any n, and I was
wondering if anyone knew a way around this?
-Max
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Re: [R] Trouble with R CMD check
Thanks! That's seemed to partially work, but now when I try to load it in R I get: library(polsplineRS) Error in library.dynam(polspline, pkg, lib) : shared library 'polspline' not found Error in library(polsplineRS) : .First.lib failed for 'polsplineRS' -Original Message- From: Duncan Murdoch [mailto:[EMAIL PROTECTED] Sent: Friday, March 07, 2008 1:36 PM To: Shewcraft, Ryan Cc: r-help@r-project.org Subject: Re: [R] Trouble with R CMD check On 3/7/2008 1:21 PM, Shewcraft, Ryan wrote: Friends, I changed one line of a package at the source level and then rebuilt it. When I run R CMD check, I get an error: You're getting lots of warnings from the compiler; presumably those have nothing to do with you. Then at the end you're getting an error because you haven't set up your R toolset completely: hhc: not found This is the Microsoft Help Workshop help compiler. Either you didn't install it, or didn't put it in the right place. But you can avoid the error by just not building compiled help, using Rcmd install --docs=normal polsplineRS Duncan Murdoch installing R.css in C:/polsplineRS.Rcheck -- Making package polsplineRS adding build stamp to DESCRIPTION making DLL ... making hareall.d from hareall.c making heftall.d from heftall.c making lsdall.d from lsdall.c making lspecall.d from lspecall.c making nlsd.d from nlsd.c making polyall.d from polyall.c making polymars.d from polymars.c gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c hareall.c -o hareall.o hareall.c: In function `gremdim': hareall.c:2127: warning: unused variable `x' hareall.c: In function `compall': hareall.c:1656: warning: 'lala' might be used uninitialized in this function hareall.c: In function `newton': hareall.c:1876: warning: 'rall' might be used uninitialized in this function hareall.c: In function `search': hareall.c:646: warning: 'iloc' might be used uninitialized in this function hareall.c:646: warning: 'lloc' might be used uninitialized in this function hareall.c:646: warning: 'uloc' might be used uninitialized in this function hareall.c: In function `adders': hareall.c:529: warning: 'crit1' might be used uninitialized in this function hareall.c: In function `share': hareall.c:2126: warning: 'bb1' might be used uninitialized in this function hareall.c:2126: warning: 'bt1' might be used uninitialized in this function hareall.c:2126: warning: 'bb2' might be used uninitialized in this function hareall.c:2126: warning: 'bw' might be used uninitialized in this function hareall.c:2127: warning: 'wald' might be used uninitialized in this function hareall.c: In function `sphare': hareall.c:3417: warning: 's' might be used uninitialized in this function hareall.c: At top level: hareall.c:132: warning: 'iigvector' declared `static' but never defined gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c heftall.c -o heftall.o heftall.c: In function `sheft': heftall.c:301: warning: 'nintx' might be used uninitialized in this function heftall.c:301: warning: 'addi' might be used uninitialized in this function heftall.c:1848: warning: 'loloc' might be used uninitialized in this function heftall.c:1848: warning: 'uploc' might be used uninitialized in this function heftall.c:1848: warning: 'nowloc1' might be used uninitialized in this function gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c lsdall.c -o lsdall.o lsdall.c: In function `knotplace': lsdall.c:1890: warning: 'ia' might be used uninitialized in this function lsdall.c:1895: warning: 'u3' might be used uninitialized in this function lsdall.c:1895: warning: 'u4' might be used uninitialized in this function lsdall.c: In function `liter': lsdall.c:369: warning: 'zerror' might be used uninitialized in this function lsdall.c:370: warning: 'newlikelihood' might be used uninitialized in this function lsdall.c: In function `logcensor': lsdall.c:51: warning: 'xnknots' might be used uninitialized in this function lsdall.c:55: warning: 'xczheta' might be used uninitialized in this function lsdall.c: In function `pqlsd': lsdall.c:2908: warning: 'j' might be used uninitialized in this function lsdall.c: At top level: lsdall.c:1458: warning: 'fun2' defined but not used lsdall.c:1471: warning: 'fun48' defined but not used gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c lspecall.c -o lspecall.o lspecall.c: In function `tsadd': lspecall.c:626: warning: 'b2' might be used uninitialized in this function gcc -Ic:/PROGRA~1/R/R-24~1.1/include -Wall -O2 -std=gnu99 -c nlsd.c -o nlsd.o nlsd.c: In function `ludcmp': nlsd.c:695: warning: 'imax' might be used uninitialized in this function nlsd.c: In function `setupspace': nlsd.c:1717: warning: 'm' might be used uninitialized in this function nlsd.c: In function `startspace': nlsd.c:1838: warning: 'l' might be
Re: [R] Numerical Integration in 1D
Hi Max,
The analytic integral \int _0 ^\Inf exp(-t) t^n log(t) might not converge
because the integrand tends to -Inf as t - 0.
So, here is a numerical approach to estimating the derivative of the gamma
function:
library(numDeriv)
fx - function(x, n) exp(-x) * x^n
gf - function(n) {integrate(fx, lower=0, upper=Inf, n=n)$val}
grad(x=3, func=gf)
[1] 7.536706
grad(x=10, func=gf)
[1] 8534040
Best,
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Max
Sent: Friday, March 07, 2008 1:41 PM
To: [EMAIL PROTECTED]
Subject: [R] Numerical Integration in 1D
Dear UseRs,
I'm curious about the derivative of n!.
We know that Gamma(n+1)=n! So when on takes the derivative of
Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
I've tried code like
integrand-function(x) {log(x)*exp(x)*x^n}
integrate(integrand,lower=0,upper=Inf)
It seems that R doesn't like to integrate for any n, and I was
wondering if anyone knew a way around this?
-Max
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Numerical Integration in 1D
Prof Brian Ripley formulated on Friday :
On Fri, 7 Mar 2008, Max wrote:
Dear UseRs,
I'm curious about the derivative of n!.
We know that Gamma(n+1)=n! So when on takes the derivative of
Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
I've tried code like
integrand-function(x) {log(x)*exp(x)*x^n}
integrate(integrand,lower=0,upper=Inf)
It seems that R doesn't like to integrate for any n, and I was
wondering if anyone knew a way around this?
ln(x) e^x x^n is not integrable on (0, Inf). You presumably slipped over
a minus sign, but your definition of gamma(n) is wrong -- see ?gamma.
integrate(function(x) exp(-x)*x^n, lower=0, upper=Inf)
will work for gamma(n+1).
I did miss a minus sign in the integration, which explains part of my
problems. The function of interest is the derivative of Gamma(n+1) with
respect to n, but obviously integrated over x from 0 to Infinity.
What happens now is:
integrand-function(x) {log(x)*exp(-x)*x^n}
integrate(integrand,lower=0,upper=Inf)
Error in f(x, ...) : object n not found
Any ideas on how to get around this error?
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Re: [R] Numerical Integration in 1D
Hi max,
Prof. Ripley is right. Your problem is that you missed a (-) sign in the
exponential. Here is a demonstration showing the agreement between
numerical and analytical results:
gx - function(x, n) exp(-x) * x^n * log(x)
df - function(n) {integrate(gx, lower=0, upper=Inf, n=n)$val}
library(numDeriv)
fx - function(x, n) exp(-x) * x^n
gf - function(n) {integrate(fx, lower=0, upper=Inf, n=n)$val}
grad(x=6, func=gf)
[1] 1348.405
df(6)
[1] 1348.405
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Max
Sent: Friday, March 07, 2008 1:41 PM
To: [EMAIL PROTECTED]
Subject: [R] Numerical Integration in 1D
Dear UseRs,
I'm curious about the derivative of n!.
We know that Gamma(n+1)=n! So when on takes the derivative of
Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
I've tried code like
integrand-function(x) {log(x)*exp(x)*x^n}
integrate(integrand,lower=0,upper=Inf)
It seems that R doesn't like to integrate for any n, and I was
wondering if anyone knew a way around this?
-Max
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and provide commented, minimal, self-contained, reproducible code.
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[R] error in random forest
Hi, I get the following error when I try to predict the probabilities of a test sample: Error in predict.randomForest(fit.EBA.OM.rf.50, x.OM, type = prob) : New factor levels not present in the training data I have about 630 predictor variables in the dataset x.OM (25 factor variables and the remaining are continuous variables). Any ideas on how to trace it? Thank you, Nagu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] zoo object won't plot
1. This is not reproducible.
Lines was not provided in reproducible form.
Please look at my prior emails and use that
form so that one can copy from your post and
paste it directly into R and observe the error.
2. What do you mean by does not plot? Do you get
an error or does nothing appear? If its the latter
its because there are no lines in the plot, only
disjoint points separated by NAs so one would
not expect there to be any lines.
plot(na.omit(z1))
plot(na.approx(z1, na.rm = FALSE))
plot(z1, type = p)
would all create plots with lines (or points) if that's the problem.
On Fri, Mar 7, 2008 at 1:53 PM, stephen sefick [EMAIL PROTECTED] wrote:
DateTimeRM61
11/30/2006 12:31NA
11/30/2006 12:46NA
11/30/2006 13:012646784125
11/30/2006 13:16NA
11/30/2006 13:31NA
11/30/2006 13:46NA
11/30/2006 14:012666435177
11/30/2006 14:16NA
11/30/2006 14:31NA
11/30/2006 14:46NA
11/30/2006 15:012653041914
11/30/2006 15:16NA
11/30/2006 15:31NA
11/30/2006 15:46NA
11/30/2006 16:012693126189
11/30/2006 16:16NA
11/30/2006 16:31NA
11/30/2006 16:46NA
11/30/2006 17:012658366411
11/30/2006 17:16NA
11/30/2006 17:31NA
11/30/2006 17:46NA
11/30/2006 18:012705885426
11/30/2006 18:16NA
11/30/2006 18:31NA
11/30/2006 18:46NA
11/30/2006 19:012707635675
11/30/2006 19:16NA
11/30/2006 19:31NA
11/30/2006 19:46NA
11/30/2006 20:012721482049
11/30/2006 20:16NA
11/30/2006 20:31NA
11/30/2006 20:46NA
11/30/2006 21:012712886423
11/30/2006 21:16NA
11/30/2006 21:31NA
11/30/2006 21:46NA
11/30/2006 22:012720426598
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11/30/2006 22:31NA
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11/30/2006 23:012724468912
#read data in as zoo object
# chron
library(chron)
fmt.chron - function(x) {
chron(sub( .*, , x), gsub(.* (.*), \\1:00, x))
}
z1 - read.zoo(textConnection(Lines), sep = ,, header = TRUE, FUN =
fmt.chron)
the above data will not plot
#rplot
plot(z1)
--
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.
-K. Mullis
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] [R-pkgs] bayesm version 2.2-0
bayesm version 2.2-0 is now available on CRAN. Major changes include: 1. general density estimation using a Dirichlet Process Prior and a normal base 2. linear instrumental variable models with unknown error distributions (the Bayesian analogue of IV methods). Achieved via DP priors. peter r Peter E. Rossi Joseph T. and Bernice S. Lewis Professor of Marketing and Statistics Director, Kilts Center for Marketing Editor, Quantitative Marketing and Economics Rm 353, Graduate School of Business, U of Chicago 5807 S. Woodlawn Ave, Chicago IL 60637 Tel: (773) 702-7513 | Fax: (773) 834-2081 WWW: http://ChicagoGsb.edu/fac/peter.rossi SSRN: http://ssrn.com/author=22862 ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] triple integral: adapt package question
Dear All,
I have a function f(x,y,z)=exp(x^3+y^4+x^2*y+x*z^2+y/z) over D, where is D={
(x,y,z)| 0 zInf, 0yc1*z, 0xc2*/y}. x,y,z are all vectors and c1 and c2
are constants. I tried the adapt package and I get some error. This is the
error message:
Error in function (z, y, x) : argument x is missing, with no default
I included my R code. Can anyone please let me know how I can calculate the
numerical integral of such a function where some of the boundaries are
functions of the other variables?
require(adapt)
fn-function(z,y,x) {exp(x^3+y^4+x^2*y+x*z^2+y/z)}
x-runif(200);y-runif(200);z-runif(200);
c1-.5;c2-5;M-100;#M to represent infinity
i1-adapt(3,lo=c(.0001,0,0),up=c(M,c1*z,c2/y),functn=fn)$value
print(i1);
Thanks,
--
Davood Tofighi
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Re: [R] training svm
Also, see the nearZeroVar function in the caret package.
MAx
On Fri, Mar 7, 2008 at 7:41 AM, Charilaos Skiadas [EMAIL PROTECTED] wrote:
On Mar 7, 2008, at 2:17 AM, Oldrich Kruza wrote:
Hello Soumyadeep,
if you store the data in a tabular file, then I suggest using standard
text-editing tools like cut (say your file is called data.csv, fields
are separated with commas and you want to get rid of the third and
sixth column):
$ cut --complement --delimiter=, --fields=3,6 data.csv
data_cut.csv
If you're not in an Unix environment but have perl, then you may use a
script like:
open SRC, data.csv or die(couldn't open source);
open DST, data_cut.csv or die(couldn't open destination);
while (SRC) {
chomp;
@fields = split /,/;#substitute the comma for the
delimiter you use
splice @fields, 2, 1;#get rid of third column (they're
zero-based, thus 2 instead of 3)
splice @fields, 5, 1;#get rid of sixth column
print DST join(,, @fields), \n;
}
If you need to do the selection within R, then you can do it by
indexing the data structure. Suppose you have the data in a data.frame
called data. Then:
data - data[,-6]
data - data[,-3]
might do the trick (but since I'm not much of an R hacker, this is
without guarantee). I think it might be better however to do the
preprocessing before the data get into R because then you avoid
loading the columns to discard into memory.
I am guessing that the data is already in R, so it should be easier
to do it in R, especially if he doesn't know which columns are the
ones with all identical values. For instance, suppose the data set is
called x. Then the following would return TRUE for the columns that
have all values the same:
allsame - sapply(x,function(y) length(table(y))==1)
and then the following will take them out
newdata - x[,!allsame]
Hope this helps
~ Oldrich
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
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--
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Re: [R] Numerical Integration in 1D
On Fri, 7 Mar 2008, Max wrote:
Prof Brian Ripley formulated on Friday :
On Fri, 7 Mar 2008, Max wrote:
Dear UseRs,
I'm curious about the derivative of n!.
We know that Gamma(n+1)=n! So when on takes the derivative of
Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
I've tried code like
integrand-function(x) {log(x)*exp(x)*x^n}
integrate(integrand,lower=0,upper=Inf)
It seems that R doesn't like to integrate for any n, and I was
wondering if anyone knew a way around this?
ln(x) e^x x^n is not integrable on (0, Inf). You presumably slipped over
a minus sign, but your definition of gamma(n) is wrong -- see ?gamma.
integrate(function(x) exp(-x)*x^n, lower=0, upper=Inf)
will work for gamma(n+1).
I did miss a minus sign in the integration, which explains part of my
problems. The function of interest is the derivative of Gamma(n+1) with
respect to n, but obviously integrated over x from 0 to Infinity.
And you said n!, so n must be integer and you cannot differentiate a
function of a integer argument.
If you are interested in the derivative of gamma(x), check out ?digamma.
What happens now is:
integrand-function(x) {log(x)*exp(-x)*x^n}
integrate(integrand,lower=0,upper=Inf)
Error in f(x, ...) : object n not found
Any ideas on how to get around this error?
Set 'n' in the evaluation environment.
E.g.
n - 3
integrate(integrand, lower=0, upper=Inf)
7.536706 with absolute error 4.7e-06
digamma(4)*gamma(4)
[1] 7.536706
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] Combine two columns
Is there a way to combine two columns within a data frame? Example data: id snp AL1 AL2 150030 A B 151030 A A 152030 A B This is what I would like: indvsnp AL1AL2 150030 AB 151030 AA 152030 AB Any help is greatly appreciated. Alysta __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combine two columns
Try: transform(x, AL1Al2 = paste(AL1, AL2, sep=''))[-c(3:4)] On 07/03/2008, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Is there a way to combine two columns within a data frame? Example data: id snp AL1 AL2 150030 A B 151030 A A 152030 A B This is what I would like: indvsnp AL1AL2 150030 AB 151030 AA 152030 AB Any help is greatly appreciated. Alysta __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing function to tapply as a string
Or perhaps:
myfun - function(fname, ...)match.fun(fname)(...)
On 07/03/2008, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Was wondering if it is possible to pass function name as a parameter
Yes. This isn't exactly what you wanted, but it demonstrates the
principle.
x = rnorm(5)
[1] -0.6510448 0.4591730 1.3225205 1.2314391 -0.0888139
myfun - function(fname, x) eval(parse(text=paste(fname,(x),sep=)))
myfun('max',x)
[1] 1.322521
myfun('min',x)
[1] -0.6510448
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Combine two columns
Depending on your purpose you might want to look at ?interaction or its synonym : (i.e. a colon) On Fri, Mar 7, 2008 at 4:09 PM, [EMAIL PROTECTED] wrote: Is there a way to combine two columns within a data frame? Example data: id snp AL1 AL2 150030 A B 151030 A A 152030 A B This is what I would like: indvsnp AL1AL2 150030 AB 151030 AA 152030 AB Any help is greatly appreciated. Alysta __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding Interaction and main effects contrasts for two-way ANOVA
Dale,
Other than the first SAS contrast, does the following demonstrate what
your asking for?
summary(twoway)
material temp voltage
1:12 50:12 Min. : 20
2:12 65:12 1st Qu.: 70
3:12 80:12 Median :108
Mean :106
3rd Qu.:142
Max. :188
contrasts(twoway$material)
2 3
1 0 0
2 1 0
3 0 1
contrasts(twoway$temp)
65 80
50 0 0
65 1 0
80 0 1
fit - aov(voltage ~ material*temp, data=twoway)
summary.aov(fit)
Df Sum Sq Mean Sq F value Pr(F)
material 2 1068453427.91 0.0020 **
temp 2 39119 19559 28.97 1.9e-07 ***
material:temp 4 961424033.56 0.0186 *
Residuals 27 18231 675
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# setting (partial) contrasts
contrasts(twoway$material) - c(1,-1,0) # ignoring the second
available df
contrasts(twoway$temp) - c(0,1,-1) # ignoring the second available df
contrasts(twoway$material)
[,1] [,2]
11 -0.41
2 -1 -0.41
30 0.82
contrasts(twoway$temp)
[,1] [,2]
500 -0.82
651 0.41
80 -1 0.41
summary.aov(fit, split=list(material=list('m1-m2'=1), temp=list('t50 -
t80'=1)))
Df Sum Sq Mean Sq F value Pr(F)
material 2 1068453427.91 0.00198 **
material: m1-m2 1 380038005.63 0.02506 *
temp 2 39119 19559 28.97 1.9e-07 ***
temp: t50 - t80 1 11310 11310 16.75 0.00035 ***
material:temp 4 961424033.56 0.01861 *
material:temp: m1-m2.t50 - t80 1 497049707.36 0.01146 *
Residuals27 18231 675
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# other examples of setting contrasts
# compare m1 vs m2 and m2 vs m3
contrasts(twoway$material) - matrix(c(1,-1,0,1,1,-2), nrow=3)
contrasts(twoway$material)
[,1] [,2]
110
2 -11
30 -1
# compare m1 vs m2 and m1+m2 vs m3
contrasts(twoway$material) - matrix(c(1,-1,0,1,1,-2), nrow=3)
contrasts(twoway$material)
[,1] [,2]
111
2 -11
30 -2
I'm not sure if 'summary.aov' is the only lm-family summary method with
the split argument.
DaveT.
*
Silviculture Data Analyst
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
[EMAIL PROTECTED]
http://ofri.mnr.gov.on.ca
*
-Original Message-
From: Steele [mailto:[EMAIL PROTECTED]
Sent: March 6, 2008 09:08 PM
To: [EMAIL PROTECTED]
Subject: [R] Finding Interaction and main effects contrasts
for two-way ANOVA
I've tried without success to calculate interaction and main effects
contrasts using R. I've found the functions C(), contrasts(),
se.contrasts() and fit.contrasts() in package gmodels. Given the url
for a small dataset and the two-way anova model below, I'd like to
reproduce the results from appended SAS code. Thanks. --Dale.
## the dataset (from Montgomery)
twoway - read.table(http://dsteele.veryspeedy.net/sta501/twoway.txt;,
col.names=c('material', 'temp','voltage'),colClasses=c('factor',
'factor', 'numeric'))
## the model
fit - aov(voltage ~ material*temp, data=twoway)
/* SAS code */
proc glm data=twoway;
class material temp;
model voltage = material temp material*temp;
contrast '21-22-31+32' material*temp 0 0 0 1 -1 0 -1 1 0;
estimate '21-22-31+32' material*temp 0 0 0 1 -1 0 -1 1 0;
contrast 'material1-material2' material 1 -1 0;
estimate 'material1-material2' material 1 -1 0;
contrast 'temp50 - temp80' temp 1 0 -1;
estimate 'temp50 - temp80' temp 1 0 -1;
run;
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[R] Warning: matrix by vector division
Dear list, I just made a very simple mistake, but it was hard to spot. And I think that I should warn other people, because it is probably so simple to make... === R code === # Let us create a matrix: (a - cbind(c(0,1,1), rep(1,3))) # [,1] [,2] # [1,]01 # [2,]11 # [3,]11 # That is a MISTAKE: a/colSums(a) # [,1] [,2] # [1,] 0.000 0.333 # [2,] 0.333 0.500 # [3,] 0.500 0.333 # I just wonder if some R warning should be issued here? # That is what I actually needed (column-wise frequencies): t(t(a)/colSums(a)) # [,1] [,2] # [1,] 0.0 0.333 # [2,] 0.5 0.333 # [3,] 0.5 0.333 === end of R code === With best wishes and regards, Alexey Shipunov __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Irregular Time Series Issue
Hello, I have an irregular time series of the form : Time Data Time1 Data1 1 b1 e 7 g 4i NA NA 5 k NA NA NA NA ... (the columns have varying length of NAs after a certain point) Converting this to regular time series with Pastecs does not seem to work, when I see the entire data as a single series. So I remove the NAs and deal with one series at a time in a loop. Any suggestions ? Thanks A. Mani -- A. Mani Member, Cal. Math. Soc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Warning: matrix by vector division
and you might want to check ?prop.table prop.table(a, 2) [,1] [,2] [1,] 0.0 0.333 [2,] 0.5 0.333 [3,] 0.5 0.333 or even ?sweep (which will be useful for more complex situations) sweep(a, 2, colSums(a), /) [,1] [,2] [1,] 0.0 0.333 [2,] 0.5 0.333 [3,] 0.5 0.333 b On Mar 7, 2008, at 5:32 PM, Alexey Shipunov wrote: Dear list, I just made a very simple mistake, but it was hard to spot. And I think that I should warn other people, because it is probably so simple to make... === R code === # Let us create a matrix: (a - cbind(c(0,1,1), rep(1,3))) # [,1] [,2] # [1,]01 # [2,]11 # [3,]11 # That is a MISTAKE: a/colSums(a) # [,1] [,2] # [1,] 0.000 0.333 # [2,] 0.333 0.500 # [3,] 0.500 0.333 # I just wonder if some R warning should be issued here? # That is what I actually needed (column-wise frequencies): t(t(a)/colSums(a)) # [,1] [,2] # [1,] 0.0 0.333 # [2,] 0.5 0.333 # [3,] 0.5 0.333 === end of R code === With best wishes and regards, Alexey Shipunov __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to Estimate Covariance by Week based on a linear regression model
Hi all: I have always used SPSS to estimate weekly covariance based on a linear regression model but have to hard code the model Std. Error and the Mean-Square and then execute one week a the time. I was wondering if someone could give me an idea on how to estimate weekly(WK) covariance using the summary and anova of dfr(lineal model below). I have to do this for 52 weeks(WK) but I am providing a dataset with only two weeks below. The first week(WK 38 is missing values) dfr - read.table(textConnection(percentQ Efficiency 1.5650.0125 1.94 0.0213 0.8760.003736 1.0270.006 1.5360.0148 1.5360.0162 2.6070.02 1.4560.0157 2.16 0.0103 1.6980.0196 1.64 0.0098684 1.8140.0183 2.3940.0107 2.4690.0221 3.6110.0197 3.4660.0155 1.8770.0283 2.8930.0189 1.8510.009772 2.8340.0285 1.9230.022 2.5810.0159 2.3610.0053591 2.43 0.0185 1.66 0.0151 2.2850.0084034 2.2850.0124 2.37 0.0122 2.3920.0146 2.2440.0175), header=TRUE) # Linear model Reg-lm(Efficiency~percentQ,data=dfr) summary(Reg) # Coefficients standard error Std=Betas[,Std. Error] Std[1]^2 Std[1]^2 # Analysis of Variance (ANOVA) MS - anova(lm(Efficiency~percentQ,data=dfr)) MS # value of the Residual Mean-Square MS$Mean Sq[2] #I want to estimate weekly(WK) covariance of the dataset below using the linear model above. temp53 - read.table(textConnection(XD TD PD WK 38 38 38 38 3.0259 0.022522163299 38 2.2316 0.01724 120315 38 2.3374 0.017944137874 38 2.2024 0.017046160524 39 2.4216 0.018504163565 39 1.4672 0.012157143973 39 1.4817 0.012253111956 39 1.4959 0.0123488967739 1.4431 0.0119979526939 1.5676 0.0128258155839), header=TRUE) # I read about the cov function and tried it with my data but couldn't get the desired results..I would really appreciate any hints..Thanks Felipe D. Carrillo Fishery Biologist US Fish Wildlife Service California, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Warning: matrix by vector division
R is working exactly as documented. If you look at how the matrix is stored (column-wise) and then what you are dividing by, you will see that it is doing what you asked (when recycling values): as.vector(a) [1] 0 1 1 1 1 1 as.vector(a)/c(2,3) [1] 0.000 0.333 0.500 0.333 0.500 0.333 On Fri, Mar 7, 2008 at 5:32 PM, Alexey Shipunov [EMAIL PROTECTED] wrote: Dear list, I just made a very simple mistake, but it was hard to spot. And I think that I should warn other people, because it is probably so simple to make... === R code === # Let us create a matrix: (a - cbind(c(0,1,1), rep(1,3))) # [,1] [,2] # [1,]01 # [2,]11 # [3,]11 # That is a MISTAKE: a/colSums(a) # [,1] [,2] # [1,] 0.000 0.333 # [2,] 0.333 0.500 # [3,] 0.500 0.333 # I just wonder if some R warning should be issued here? # That is what I actually needed (column-wise frequencies): t(t(a)/colSums(a)) # [,1] [,2] # [1,] 0.0 0.333 # [2,] 0.5 0.333 # [3,] 0.5 0.333 === end of R code === With best wishes and regards, Alexey Shipunov __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Warning: matrix by vector division
Dear all, Thank you very much for all your valuable suggestions! Best wishes, Alexey Shipunov __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problems installing packages using the inbuilt facility: Error i n gzfile(file, r) : unable to open connection
Works fine for me on the same setup. Try this and compare (especially
the size of the downloaded file):
url - http://cran.uk.r-project.org/bin/windows/contrib/2.6/ada_2.0-1.zip;;
download.file(url, basename(url), mode=wb) # Note wb!!!
trying URL 'http://cran.uk.r-project.org/bin/windows/contrib/2.6/ada_2.0-1.zip'
Content type 'application/zip' length 971893 bytes (949 Kb)
opened URL
downloaded 949 Kb
file.info(basename(url))
size isdir mode mtime ctime
ada_2.0-1.zip 971893 FALSE 666 2008-03-07 15:43:52 2008-03-07 15:43:28
atime exe
ada_2.0-1.zip 2008-03-07 15:43:52 no
install.packages(basename(url), repos=NULL)
package 'ada' successfully unpacked and MD5 sums checked
updating HTML package descriptions
library(ada)
Loading required package: rpart
/Henrik
On Fri, Mar 7, 2008 at 10:02 AM, Kauer, Philipp
[EMAIL PROTECTED] wrote:
Hi
I have been trawling the web, FAQs, and R manuals for help on the following
issue, but have failed and was wondering if anyone has a solution to the
following problem:
After having installed R 2.6.2 for Windows (binary), I tried to install
various packages. Every time I try loading a package (any package) via the
built-in menu, I run into the following error message.
utils:::menuInstallPkgs()
trying URL
'http://cran.uk.r-project.org/bin/windows/contrib/2.6/ada_2.0-1.zip
http://cran.uk.r-project.org/bin/windows/contrib/2.6/ada_2.0-1.zip '
Content type 'application/zip' length 971893 bytes (949 Kb)
opened URL
downloaded 949 Kb
Warning in gzfile(file, r) :
cannot open compressed file 'ada/DESCRIPTION', probable reason 'No such
file or directory'
Error in gzfile(file, r) : unable to open connection
traceback()
6: gzfile(file, r)
5: read.dcf(file.path(curPkg, DESCRIPTION), c(Package, Version,
Type))
4: unpackPkg(foundpkgs[okp, 2], foundpkgs[okp, 1], lib, installWithVers)
3: .install.winbinary(pkgs = pkgs, lib = lib, contriburl = contriburl,
method = method, available = available, destdir = destdir,
installWithVers = installWithVers, dependencies = dependencies)
2: install.packages(NULL, .libPaths()[1], dependencies = NA, type = type)
1: utils:::menuInstallPkgs()
I tried saving the ZIP of the package to my C drive, and then installing
from that, but I get the same error message. I then tried calling gzfile with
the filename of the ZIP directly, and I seem to get a file handle without any
error message:
gzfile(C:/Program Files/R/R-2.6.2/library/ada_2.0-1.zip,r)
description
class
C:/Program Files/R/R-2.6.2/library/ada_2.0-1.zip
gzfile
mode
text
rb6
binary
opened
can read
opened
yes
can write
no
It might be worth noting, that I initially I also had problems with the
internet download facility, which I fixed starting R with the flag
--internet2, as indicated in the FAQs.
Can you please help?
Many thanks - Philipp
PS: I did the same on my PC at home, and there everything works without
problems.
==
Please access the attached hyperlink for an important el...{{dropped:8}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] boxcox.fit error
As the error message says: the transformation requires positive data Your data has a minimum value at zero. That means it is merely non-negative. Positive means all the values must be greater than zero. Strictly. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Alexys Herleym Rodriguez Avellaneda Sent: Saturday, 8 March 2008 1:40 AM To: r-help@r-project.org Subject: [R] boxcox.fit error Hi, Thakns all for your help I am doing the next in my dataframe tabla, column pend1, because the Lilliefors (Kolmogorov-Smirnov) test give me a pvalue alfa. (data no normal distribution). I need do a transformation with box-cox or other: bc - boxcox.fit(tabla$pend1) R send to me: Error in boxcox.fit(tabla$pend1) : Transformation requires positive data The summary for my data is: summary(tabla$pend1) Min. 1st Qu. MedianMean 3rd Qu.Max. 0.012.819.820.227.257.9 length(tabla$pend1) [1] 4408 Someone can help me?. What is the error? Thanks Alexys H __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error in random forest
The error message is pretty clear, really. To spell it out a bit more, what you have done is as follows. Your training set has factor variables in it. Suppose one of them is f. In the training set it has 5 levels, say. Your test set also has a factor f, as it must, but it appears that in the test set it has 6 levels, or more, or levels that do not agree with those for f in the training set. This mismatch measn that the predict method for randomForest cannot use this test set. What you have to do is make sure that the factor levels agree for every factor in both test and training set. One way to do this is to put the test and training set together with rbind(...) say, and then separate them again. But even this will still have a problem for you. Because you training set will have some factor levels empty, which are not empty in the test set. The error will most likely be more subtle, though. You really need to sort this out yourself. It is not particularly an R problem, but a confusion over data. To be useful, your training set need to cover the field for all levels of every factor. Think about it. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Nagu Sent: Saturday, 8 March 2008 5:37 AM To: r-help@r-project.org; [EMAIL PROTECTED] Subject: [R] error in random forest Hi, I get the following error when I try to predict the probabilities of a test sample: Error in predict.randomForest(fit.EBA.OM.rf.50, x.OM, type = prob) : New factor levels not present in the training data I have about 630 predictor variables in the dataset x.OM (25 factor variables and the remaining are continuous variables). Any ideas on how to trace it? Thank you, Nagu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Irregular Time Series Issue
Have you considered using 'corCAR1' with 'lme' in the 'nlme' package? Hope this helps. Spencer Graves p.s. If you are not familiar with this, I highly recommend Pinheiro, J.C., and Bates, D.M. (2000) Mixed-Effects Models in S and S-PLUS (Springer). The ~R\library\nlme\scripts directory contains ch01.R, ch02.R, etc., which contain R scripts to reproduce virtually all the examples in the text. These are especially valuable for R, because the book was written before 'nlme' was ported to R, and there are a few changes for which the version in the book gives the wrong answer, while following the script gives the correct answer. A Mani wrote: Hello, I have an irregular time series of the form : Time Data Time1 Data1 1 b1 e 7 g 4i NA NA 5 k NA NA NA NA ... (the columns have varying length of NAs after a certain point) Converting this to regular time series with Pastecs does not seem to work, when I see the entire data as a single series. So I remove the NAs and deal with one series at a time in a loop. Any suggestions ? Thanks A. Mani __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error in random forest
Thank you very much. I'll jump in to the data and verify the consistency between the training and testing variables and their levels. On Fri, Mar 7, 2008 at 5:14 PM, [EMAIL PROTECTED] wrote: The error message is pretty clear, really. To spell it out a bit more, what you have done is as follows. Your training set has factor variables in it. Suppose one of them is f. In the training set it has 5 levels, say. Your test set also has a factor f, as it must, but it appears that in the test set it has 6 levels, or more, or levels that do not agree with those for f in the training set. This mismatch measn that the predict method for randomForest cannot use this test set. What you have to do is make sure that the factor levels agree for every factor in both test and training set. One way to do this is to put the test and training set together with rbind(...) say, and then separate them again. But even this will still have a problem for you. Because you training set will have some factor levels empty, which are not empty in the test set. The error will most likely be more subtle, though. You really need to sort this out yourself. It is not particularly an R problem, but a confusion over data. To be useful, your training set need to cover the field for all levels of every factor. Think about it. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Nagu Sent: Saturday, 8 March 2008 5:37 AM To: r-help@r-project.org; [EMAIL PROTECTED] Subject: [R] error in random forest Hi, I get the following error when I try to predict the probabilities of a test sample: Error in predict.randomForest(fit.EBA.OM.rf.50, x.OM, type = prob) : New factor levels not present in the training data I have about 630 predictor variables in the dataset x.OM (25 factor variables and the remaining are continuous variables). Any ideas on how to trace it? Thank you, Nagu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ask for help on nonlinear fitting
I have a table like the following. I want to fit Cm to Vm like this: Cm ~ Cl+Q1*b1*38.67*exp(-b1*(Vm-Vp1)*0.03867)/(1+exp(-b1*(Vm-Vp1)*0.03867))^2+Q2*b2*38.67*exp(-b2*(Vm-Vp2)*0.03867)/(1+exp(-b2*(Vm-Vp2)*0.03867))^2 I use nls, with start=list(Q1=2e-3, b1=1, Vp1=-25, Q2=3e-3, b2=1, Vp2=200). But I always get 'singlular gradient' error like this. But in SigmaPlot I can get the result. How can I get with R. Thanks! The table: Vm Cm Ih -147.715 8.15 -0.107 -146.944 8.081 -0.106 -146.173 8.089 -0.106 -145.409 8.114 -0.108 -144.638 8.105 -0.107 -143.873 8.085 -0.105 -143.102 8.151 -0.102 -142.338 8.084 -0.1 -141.567 8.079 -0.101 -140.796 8.077 -0.101 -140.032 8.077 -0.102 -139.261 8.126 -0.101 -138.497 8.153 -0.099 -137.726 8.129 -0.097 -136.962 8.073 -0.096 -136.191 8.088 -0.097 -135.426 8.119 -0.098 -134.655 8.121 -0.098 -133.885 8.062 -0.096 -133.12 8.135 -0.093 -132.349 8.089 -0.091 -131.585 8.065 -0.091 -130.814 8.064 -0.092 -130.05 8.071 -0.094 -129.279 8.139 -0.092 -128.515 8.15 -0.09 -127.744 8.119 -0.087 -126.973 8.074 -0.086 -126.208 8.083 -0.087 -125.438 8.084 -0.088 -124.673 8.102 -0.089 -123.902 8.083 -0.087 -123.138 8.132 -0.084 -122.367 8.108 -0.082 -121.596 8.072 -0.083 -120.832 8.07 -0.083 -120.061 8.083 -0.085 -119.297 8.134 -0.084 -118.532 8.14 -0.081 -117.762 8.119 -0.078 -116.997 8.078 -0.077 -116.226 8.102 -0.079 -115.455 8.112 -0.08 -114.691 8.104 -0.08 -113.92 8.093 -0.078 -113.156 8.126 -0.075 -112.385 8.109 -0.073 -111.621 8.066 -0.073 -110.85 8.054 -0.074 -110.085 8.077 -0.076 -109.315 8.134 -0.075 -108.544 8.133 -0.073 -107.779 8.131 -0.069 -107.008 8.093 -0.068 -106.244 8.113 -0.069 -105.473 8.094 -0.071 -104.709 8.106 -0.071 -103.938 8.1 -0.069 -103.174 8.135 -0.066 -102.403 8.115 -0.064 -101.632 8.061 -0.065 -100.868 8.05 -0.066 -100.097 8.091 -0.067 -99.332 8.122 -0.066 -98.561 8.137 -0.064 -97.797 8.123 -0.061 -97.026 8.098 -0.061 -96.255 8.117 -0.061 -95.491 8.108 -0.063 -94.72 8.107 -0.063 -93.956 8.108 -0.062 -93.185 8.144 -0.059 -92.421 8.115 -0.057 -91.65 8.078 -0.057 -90.885 8.053 -0.058 -90.114 8.093 -0.06 -89.344 8.134 -0.059 -88.579 8.147 -0.057 -87.815 8.122 -0.054 -87.044 8.08 -0.053 -86.28 8.128 -0.054 -85.509 8.094 -0.055 -84.745 8.114 -0.056 -83.974 8.119 -0.054 -83.203 8.135 -0.051 -82.438 8.122 -0.049 -81.667 8.062 -0.049 -80.903 8.072 -0.05 -80.132 8.094 -0.052 -79.368 8.142 -0.051 -78.597 8.131 -0.049 -77.826 8.122 -0.046 -77.062 8.097 -0.045 -76.291 8.138 -0.046 -75.527 8.094 -0.048 -74.756 8.124 -0.048 -73.991 8.116 -0.046 -73.221 8.137 -0.044 -72.456 8.118 -0.042 -71.685 8.082 -0.042 -70.914 8.051 -0.043 -70.15 8.089 -0.044 -69.379 8.143 -0.044 -68.615 8.14 -0.042 -67.844 8.113 -0.039 -67.08 8.099 -0.038 -66.309 8.139 -0.039 -65.544 8.084 -0.041 -64.774 8.119 -0.041 -64.003 8.144 -0.039 -63.238 8.142 -0.037 -62.467 8.142 -0.035 -61.703 8.076 -0.035 -60.932 8.069 -0.036 -60.168 8.093 -0.037 -59.397 8.133 -0.037 -58.626 8.131 -0.035 -57.862 8.107 -0.032 -57.098 8.106 -0.031 -56.327 8.146 -0.032 -55.562 8.091 -0.033 -54.791 8.12 -0.034 -54.027 8.141 -0.032 -53.256 8.138 -0.03 -52.485 8.143 -0.028 -51.721 8.071 -0.028 -50.95 8.058 -0.029 -50.186 8.097 -0.03 -49.415 8.146 -0.03 -48.651 8.128 -0.028 -47.88 8.125 -0.025 -47.115 8.116 -0.024 -46.344 8.15 -0.025 -45.573 8.092 -0.026 -44.809 8.142 -0.027 -44.038 8.159 -0.026 -43.274 8.164 -0.023 -42.503 8.161 -0.021 -41.739 8.078 -0.021 -40.968 8.079 -0.022 -40.204 8.105 -0.024 -39.433 8.157 -0.023 -38.662 8.15 -0.021 -37.897 8.131 -0.018 -37.127 8.122 -0.017 -36.362 8.149 -0.018 -35.591 8.092 -0.02 -34.827 8.138 -0.02 -34.056 8.16 -0.019 -33.285 8.152 -0.017 -32.521 8.164 -0.014 -31.75 8.092 -0.014 -30.986 8.074 -0.015 -30.215 8.094 -0.017 -29.45 8.148 -0.016 -28.68 8.151 -0.014 -27.915 8.121 -0.012 -27.144 8.125 -0.01 -26.38 8.162 -0.011 -25.609 8.094 -0.012 -24.845 8.132 -0.013 -24.074 8.17 -0.011 -23.31 8.158 -0.01 -22.539 8.153 -0.007 -21.774 8.081 -0.007 -21.004 8.09 -0.008 -20.233 8.103 -0.01 -19.468 8.151 -0.009 -18.697 8.124 -0.007 -17.933 8.132 -0.005 -17.162 8.122 -0.003 -16.398 8.149 -0.004 -15.627 8.081 -0.005 -14.863 8.13 -0.006 -14.092 8.157 -0.004 -13.321 8.175 -0.003 -12.557 8.157 0 -11.786 8.078 0 -11.021 8.085 -0.001 -10.25 8.106 -0.002 -9.486 8.145 -0.002 -8.715 8.13 0 -7.944 8.136 0.003 -7.18 8.108 0.005 -6.409 8.132 0.004 -5.645 8.099 0.002 -4.874 8.134 0.002 -4.11 8.158 0.003 -3.339 8.165 0.005 -2.574 8.162 0.008 -1.803 8.082 0.008 -1.033 8.087 0.007 -0.268 8.096 0.005 0.503 8.137 0.006 1.267 8.121 0.008 2.038 8.122 0.01 2.802 8.12 0.012 3.566 8.124 0.011 4.337 8.091 0.01 5.108 8.125 0.009 5.873 8.147 0.011 6.643 8.172 0.013 7.408 8.146 0.015 8.179 8.081 0.016 8.943 8.086 0.015 9.714 8.114 0.013 10.485 8.131 0.014 11.249 8.103 0.016 12.02 8.127 0.018 12.784 8.108 0.02 13.555 8.107 0.019 14.32 8.087 0.018 15.09 8.106 0.017 15.855 8.142 0.019 16.626 8.164 0.021 17.397 8.138 0.024 18.161 8.068 0.024 18.932 8.075 0.023 19.696 8.088 0.022 20.467 8.126 0.022 21.231 8.103 0.025 22.002 8.127 0.027
