Re: [R] how to multiply a constant to a matrix?
imagine when you have complicated matrix algebra computation using R, you cannot prevent some middle-terms become quadratic and absorbs into one scalar, right? if R cannot intelligently determine this, and you have to manually add drop everywhere, do you think it is reasonable? On 5/23/06, Patrick Burns [EMAIL PROTECTED] wrote: I think drop(B/D) * solve(A) would be a more transparent approach. It isn't that R can not do what you want, it is that it is saving you from shooting yourself in the foot in your attempt. What you are doing is not really a matrix computation. Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) Michael wrote: This is very strange: I want compute the following in R: g = B/D * solve(A) where B and D are quadratics so they are just a scalar number, e.g. B=t(a) %*% F %*% a; I want to multiply B/D to A^(-1), but R just does not allow me to do that and it keeps complaining that nonconformable array, etc. I tried the following two tricks and they worked: as.numeric(B/D) * solve(A) diag(as.numeric(B/D), 5, 5) %*% solve (A) But if R cannot intelligently do scalar and matrix multiplication, it is really problemetic. It basically cannot be used to do computations, since in complicated matrix algebras, you have to distinguish where is scalar, and scalars obtained from quadratics cannot be directly used to multiply another matrix, etc. It is going to a huge mess... Any thoughts? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] conditional replacement
Hi if speed is critical you can use (x=30x=60)*x+(x30)*30+(x60)*60 HTH Petr On 23 May 2006 at 16:34, Rogerio Porto wrote: From: Rogerio Porto [EMAIL PROTECTED] To: Sachin J [EMAIL PROTECTED], R-help@stat.math.ethz.ch Date sent: Tue, 23 May 2006 16:34:04 -0300 Subject:Re: [R] conditional replacement Sachin, there's another slower but more flexible way than Gabor's solution: ifelse(x30,30,ifelse(x60,60,x)) HTH, Rogerio. - Original Message - From: Sachin J [EMAIL PROTECTED] To: R-help@stat.math.ethz.ch Sent: Tuesday, May 23, 2006 3:40 PM Subject: [R] conditional replacement Hi How can do this in R. df 48 1 35 32 80 If df 30 then replace it with 30 and else if df 60 replace it with 60. I have a large dataset so I cant afford to identify indexes and then replace. Desired o/p: 48 30 35 32 60 Thanx in advance. Sachin __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] problem with ad.test
Hi, On Tue, 23 May 2006, Bhismadev Chakrabarti wrote: i am a novice and have been trying to use the anderson-darling test on a simple text file with one column of data. ... EQ is the name ( in the top row of the column, imported with read.table( file, header=TRUE) of the column of the data. Well, unfortunately, this is going to be a case of a novice giving advice to a novice as I'm just learning this too and was recently having problem with the normality tests. The source of your problem is the read.table command and the data structure it creates. I have never used it and am too lazy now to read about it for you. :) I've used the scan method instead, as follows: library(nortest) A - scan(tmp) Read 9 items summary(A) Min. 1st Qu. MedianMean 3rd Qu.Max. 1 3 5 5 7 9 A [1] 1 2 3 4 5 6 7 8 9 ad.test(A) Anderson-Darling normality test data: A A = 0.1368, p-value = 0.9606 Good luck! Ray __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] normality testing with nortest
Hi all, On Mon, 22 May 2006, Rolf Turner wrote: It is worth noting that if the null hypothesis is true, then the p-value is uniformly distributed on [0,1]. This should be kept in mind when assessing the ``instability'' of p-values. Just wanted to thank everyone who participated in this thread for clarifying things for me. I guess I knew bits and pieces but your messages put them together and gave me some R commands (which I am still new to) to check things myself. Thanks a lot! Ray __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Does R have EGARCH modeling function?
I've downloaded fSeries, but looks like it just has an interface to OX(TM) Garch Modeling Software,and that OX(TM) software package is not free. So where can I find an EGARCH function that is truely usable? Thanks a lot! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] ASC/NZSA 2006 - Detailed Program Available via the Website
Australian Statistical Conference / New Zealand Statistical Association Conference 'Statistical Connections' SKYCITY, Auckland, New Zealand 3 - 6 July 2006 LESS THAN TWO MONTHS TO GO Register Online Now! Register now for ASCNZ06 - an international conference with a full and varied scientific program. For up to date and detailed information on the 2006 Program visit: http://www.statsnz2006.com/program.asp http://www.statsnz2006.com/program.asp ASC / NZSA 2006 with less than two months to go has attracted substantial registrations to its outstanding scientific program featuring ; * 4 Plenary Sessions featuring leading international speakers: Professor David Donoho - Stanford University, Title: Sparsity in Estimation and Detection Professor Peter Hall - ANU, Title: PCA For FDA Professor Ray Chambers - University of Southampton Title: Models and Auxiliary Information in Survey Sampling Professor Xiao-Li Meng - Harvard University Title: (Data) Size Does Matter, But You Might Be In for a Surprise ... 25 Invited Speaker Sessions covering leading international and local speakers featuring a wide range of topics in Statistical and Machine Learning, Statistical Inference, Statistics in Ecology and the Environment, Statistics in Biological Sciences and Medicine, Stochastic Processes, Econometrics and Finance, Computational Statistics and High Dimensional Data Modelling. Over 30 contributed paper sessions complementing the Plenary and Invited Sessions A full day theme devoted to Official Statistics starting with the Foreman Lecture by Professor Ray Chambers, followed by an invited session featuring Professor Alastair Scott and Professor Kirk Wolter. After lunch there will be an Official Statistics Forum featuring the New Zealand Government Statistician Mr Brian Pink and the Australian Statistician Mr Dennis Trewin, followed by a contributed session. Other contributed sessions on various aspects of official statistics will be held throughout the conference. Also included in the conference program is a Young Statisticians program Associated workshops on Distance Sampling, Stochastic Processes and the R-Fest Workshops on R Graphics, S programming and Bioconductor. Find out more on the workshops at the following link http://www.statsnz2006.com/workshop.asp http://www.statsnz2006.com/workshop.asp For information on Tours and Social events Please visit the website http://www.statsnz2006.com/sprogram.asp http://www.statsnz2006.com/sprogram.asp We look forward to seeing you in Auckland! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] general Gauss-Newton or support for NSUR: contemporaneously correlated non-linear models
Dear r-Help readers, 1) Is there support for NSUR in some R package yet? 2) Is there a general function of applying the Gauss-Newton or Marquard method, in which the function of calculating the partial derivatives can be specified by the user? Contemporaneously correlated non-linear models (NSUR) is a method to fit a system of non-linear equations. I want to use to fit several non-linear models at the same time, so that one the independent variable is alsway the sum of several other independent variable of other models. This procedure is described in detail by Parresol 2001. He used PROC MODEL by SAS to accomplish the task. If there is no support for NSUR yet, I would try to program a function by myself, although I only used very simple R-functins yet. The essential part is the fit the minimum of complex vector equation. This leads to my second question above. The function (as provided as an parameter) should accept a vector of model parameters and calculate the matrix of partial derivatives at this point in the parameter space. Thanks Thomas Parresol, B.R., 2001. Additivity of nonlinear biomass equations. Canadian Journal of Forest Research-Revue Canadienne De Recherche Forestiere 31, 865-878. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Two functions with the same name
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Re: [R] Two functions with the same name
Function written or loaded last will mask a function written or loaded before. HTH Petr PLEASE do read the posting guide! ^ On 24 May 2006 at 13:47, Ana Conesa wrote: Date sent: Wed, 24 May 2006 13:47:09 +0200 To: r-help@stat.math.ethz.ch From: Ana Conesa [EMAIL PROTECTED] Subject:[R] Two functions with the same name __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] data.frame
Dear all, Does any one knows why should I get the following error message, when trying to do a simple data.frame?? DataF-data.frame(Subject,BiomR,Spp,Capas,Litter,Herbs,LitterD,MaxCanH,DDifS p,DSSp,Slope, CanDens,NearestSp) Erro em data.frame(Subject, BiomR, Spp, Capas, Litter, Herbs, LitterD, : arguments imply differing number of rows: 202, 0 The data I am using is: Subject BiomR Spp Capas Litter Herbs LitterD MaxCanH DDifSp DSSp Slope CanDens NearestSp 1 140.74 Qfa 2 1.460 0 0.778 8 70 3.663 28 2.280 Ear 2 7.26 Aun 2 0.660 0 0.477 8 110 3.988 34 2.476 Ear 3 68.72 Qfa 2 1.019 0 1.000 8 48 2.300 48 2.476 Ear 4 0.13 Vti 2 1.719 0 0.903 4 55 4.141 28 3.258 Qco 5 2.32 Vti 2 1.719 0 0.954 4 65 3.984 32 3.051 Aun 6 2.97 Vti 2 1.719 0 0.954 6 201 3.642 22 2.000 Qfa 7 0.16 Qfa 2 1.238 2 0.778 6 72 3.105 20 1.566 Vti 8 1.76 Qfa 2 1.145 0 0.954 6 67 3.383 22 2.052 Vti 9 0.35 Qfa 3 1.092 2 0.778 6 65 3.557 30 2.106 Vti 10 0.15 Qfa 2 1.092 2 0.778 6 42 3.464 14 2.106 Vti 11 107.33 Qfa 3 1.145 5 0.903 8 147 4.035 24 2.220 Vti 12 12.32 Vti 3 1.145 5 1.000 6 208 3.464 28 2.220 Qfa 13 1.82 Vti 3 1.145 5 0.903 6 251 3.458 20 2.220 Qfa 14 1.84 Vti 1 1.019 5 0.602 6 88 3.960 28 1.363 Aun 15 3.25 Vti 2 1.719 0 0.778 6 70 4.319 14 2.621 Aun 16 3.60 Vti 1 1.460 0 0.954 6 53 4.186 26 1.902 Aun 17 0.69 Vti 2 1.238 5 0.845 4 184 3.856 44 2.547 Aun 18 1.04 Vti 1 1.019 5 0.602 4 144 3.911 32 2.052 Aun 19 0.08 Vti 2 1.460 0 0.699 6 93 3.956 36 2.547 Qco 20 10.43 Vti 2 1.145 5 0.903 6 70 3.919 23 3.258 Qco 21 0.69 Vti 3 1.145 5 0.845 8 120 4.023 30 3.051 Ear 22 13.47 Vti 3 1.460 2 0.811 8 39 3.370 42 3.152 Ear 23 2.83 Vti 2 1.019 5 0.845 6 120 4.102 24 3.258 Ear 24 1.26 Vti 2 1.019 5 0.903 6 153 4.116 30 3.152 Ear 25 0.25 Qfa 2 1.145 5 0.845 6 106 4.210 38 2.699 Qco 26 28.27 Vti 2 0.913 10 0.845 6 203 4.688 32 1.566 Qco 27 0.88 Vti 2 1.460 2 0.845 8 52 4.464 30 3.051 Qfa 28 0.08 Vti 2 1.311 0 0.477 8 86 4.486 48 2.956 Qfa 29 1.13 Vti 3 1.019 5 0.845 8 54 4.291 40 3.492 Qfa 30 11.03 Vti 2 0.821 0 0.903 4 130 4.606 42 2.699 Ear 31 0.63 Vti 2 0.821 5 0.845 4 130 4.464 44 2.699 Ear 32 0.97 Vti 2 0.737 0 0.845 4 123 4.562 34 2.699 Ear 33 4.94 Vti 2 0.737 0 0.903 4 123 4.522 48 2.699 Ear 34 2.26 Vti 2 0.821 5 0.778 4 114 4.478 46 2.699 Ear 35 0.94 Vti 2 0.660 0 0.903 4 144 4.516 31 2.699 Ear 36 0.69 Vti 2 0.589 0 0.699 4 146 4.551 35 2.699 Ear 37 0.13 Vti 3 1.019 0 0.845 8 48 4.362 14 2.343 Qfa 38 2.12 Vti 2 0.737 0 1.041 8 52 4.054 24 2.956 Qfa 39 1.77 Vti 2 0.913 0 0.903 8 56 5.198 6 2.621 Ear 40 0.32 Vti 2 1.145 5 0.954 8 118 5.138 35 3.258 Ear 41 0.11 Vti 2 1.019 0 0.602 8 84 5.237 20 2.621 Ear 42 1.57 Vti 1 0.913 0 0.954 6 137 5.038 12 3.492 Qco 43 1.77 Vti 3 0.737 0 0.954 6 141 5.026 16 3.371 Qco 44 0.17 Vti 2 0.913 0 1.000 8 69 4.901 24 3.258 Qco 45 0.41 Vti 3 1.092 2 0.954 8 81 5.129 44 4.457 Qfa 46 2.62 Vti 1 0.913 0 0.845 4 114 5.047 34 4.457 Ear 47 0.09 Vti 3 0.975 2 0.954 8 98 4.924 20 3.492 Aun 48 0.12 Vti 4 1.311 5 0.845 8 78 4.748 26 3.761 Aun 49 1.70 Vti 3 0.975 2 1.000 8 95 5.146 36 4.457 Qfa 50 14.49 Qfa 1 1.719 0 0.778 6 37 2.962 32 1.766 Pan 51 0.09 Vti 2 1.311 0 0.778 6 56 2.872 20 1.950 Pan 52 2.12 Qfa 2 1.460 0 0.845 6 203 3.097 18 3.152 Vti 53 0.23 Qfa 2 1.238 2 0.778 8 86 2.190 30 2.476 Ear 54 5.36 Qfa 2 0.975 0 0.954 6 134 3.409 42 4.684 Vti 55 0.41 Vti 2 1.145 5 0.845 6 108 3.890 36 4.945 Qco 56 0.19 Vti 1 1.719 0 0.845 6 42 4.130 35 3.492 Qco 57 15.59 Vti 3 1.719 0 1.114 6 91 4.386 16 4.257 Ear 58 3.25 Vti 3 1.719 0 0.903 6 85 4.275 36 4.257 Ear 59 2.32 Vti 3 1.460 0 0.954 6 91 4.223 32 4.257 Ear 60 4.24 Vti 3 1.719 0 1.041 6 36 4.386 42 4.257 Ear 61 31.49 Vti 3 1.719 0 1.176 6 78 4.371 43 4.257 Ear 62 2.97 Vti 3 0.786 2 0.954 6 71 3.988 22 4.076 Qco 63 0.50 Vti 3 0.706 2 0.903 6 76 4.134 20 4.076 Qco 64 20.20 Vti 2 1.719 0 1.079 6 62 3.557 18 5.640 Qfa 65 0.12 Vti 1 1.460 0 0.903 6 75 3.186 20 3.152 Qfa 66 0.10 Vti 1 1.719 0 0.903 6 90 3.154 20 3.152 Qfa 67 0.79 Vti 1 1.719 0 0.954 6 64 3.154 12 3.152 Ear 68 0.17 Vti 1 1.719 0 0.954 6
Re: [R] data.frame
Hi Your arguments has different length and therefore the error message data.frame is an object, which resembles a table from Excel, it has the same number of rows in each column From help page: A data frame, a matrix-like structure whose columns may be of differing types (numeric, logical, factor and character and so on). x1-1:5 x2-1:3 x3-1:10 data.frame(x1,x2,x3) Error in data.frame(x1, x2, x3) : arguments imply differing number of rows: 5, 3, 10 So you can recycle shorter vectors data.frame(cbind(x1,x2,x3)) or you can use list list(x1,x2,x3) I wonder how did you created your arguments? HTH Petr On 24 May 2006 at 13:05, Sara Mouro wrote: From: Sara Mouro [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Date sent: Wed, 24 May 2006 13:05:45 +0100 Subject:[R] data.frame Dear all, Does any one knows why should I get the following error message, when trying to do a simple data.frame?? DataF-data.frame(Subject,BiomR,Spp,Capas,Litter,Herbs,LitterD,MaxCanH ,DDifS p,DSSp,Slope, CanDens,NearestSp) Erro em data.frame(Subject, BiomR, Spp, Capas, Litter, Herbs, LitterD, : arguments imply differing number of rows: 202, 0 The data I am using is: Subject BiomR Spp Capas Litter Herbs LitterD MaxCanH DDifSp DSSp Slope CanDens NearestSp 1 140.74 Qfa 2 1.460 0 0.778 8 70 3.663 28 2.280 Ear 2 7.26 Aun 2 0.660 0 0.477 8 110 3.988 34 2.476 Ear 3 68.72 Qfa 2 1.019 0 1.000 8 48 2.300 48 2.476 Ear 4 0.13 Vti 2 1.719 0 0.903 4 55 4.141 28 3.258 Qco 5 2.32 Vti 2 1.719 0 0.954 4 65 3.984 32 3.051 Aun 6 2.97 Vti 2 1.719 0 0.954 6 201 3.642 22 2.000 Qfa 7 0.16 Qfa 2 1.238 2 0.778 6 72 3.105 20 1.566 Vti 8 1.76 Qfa 2 1.145 0 0.954 6 67 3.383 22 2.052 Vti 9 0.35 Qfa 3 1.092 2 0.778 6 65 3.557 30 2.106 Vti 10 0.15 Qfa 2 1.092 2 0.778 6 42 3.464 14 2.106 Vti 11 107.33 Qfa 3 1.145 5 0.903 8 147 4.035 24 2.220 Vti 12 12.32 Vti 3 1.145 5 1.000 6 208 3.464 28 2.220 Qfa 13 1.82 Vti 3 1.145 5 0.903 6 251 3.458 20 2.220 Qfa 14 1.84 Vti 1 1.019 5 0.602 6 88 3.960 28 1.363 Aun 15 3.25 Vti 2 1.719 0 0.778 6 70 4.319 14 2.621 Aun 16 3.60 Vti 1 1.460 0 0.954 6 53 4.186 26 1.902 Aun 17 0.69 Vti 2 1.238 5 0.845 4 184 3.856 44 2.547 Aun 18 1.04 Vti 1 1.019 5 0.602 4 144 3.911 32 2.052 Aun 19 0.08 Vti 2 1.460 0 0.699 6 93 3.956 36 2.547 Qco 20 10.43 Vti 2 1.145 5 0.903 6 70 3.919 23 3.258 Qco 21 0.69 Vti 3 1.145 5 0.845 8 120 4.023 30 3.051 Ear 22 13.47 Vti 3 1.460 2 0.811 8 39 3.370 42 3.152 Ear 23 2.83 Vti 2 1.019 5 0.845 6 120 4.102 24 3.258 Ear 24 1.26 Vti 2 1.019 5 0.903 6 153 4.116 30 3.152 Ear 25 0.25 Qfa 2 1.145 5 0.845 6 106 4.210 38 2.699 Qco 26 28.27 Vti 2 0.913 10 0.845 6 203 4.688 32 1.566 Qco 27 0.88 Vti 2 1.460 2 0.845 8 52 4.464 30 3.051 Qfa 28 0.08 Vti 2 1.311 0 0.477 8 86 4.486 48 2.956 Qfa 29 1.13 Vti 3 1.019 5 0.845 8 54 4.291 40 3.492 Qfa 30 11.03 Vti 2 0.821 0 0.903 4 130 4.606 42 2.699 Ear 31 0.63 Vti 2 0.821 5 0.845 4 130 4.464 44 2.699 Ear 32 0.97 Vti 2 0.737 0 0.845 4 123 4.562 34 2.699 Ear 33 4.94 Vti 2 0.737 0 0.903 4 123 4.522 48 2.699 Ear 34 2.26 Vti 2 0.821 5 0.778 4 114 4.478 46 2.699 Ear 35 0.94 Vti 2 0.660 0 0.903 4 144 4.516 31 2.699 Ear 36 0.69 Vti 2 0.589 0 0.699 4 146 4.551 35 2.699 Ear 37 0.13 Vti 3 1.019 0 0.845 8 48 4.362 14 2.343 Qfa 38 2.12 Vti 2 0.737 0 1.041 8 52 4.054 24 2.956 Qfa 39 1.77 Vti
Re: [R] Two functions with the same name
Hi you probably created the second function in .Global Environment e.g. directly in console. see ls() and look for the name of the function. If it is there the only way I know about is to remove the function from environment by rm(name.of.the.function) then the masked function shall by usable again provided that you did not detached the package. Probably there is some way through environment() but I never used it and therefore I can not help you other way than to encourage to read its help page. HTH Petr On 24 May 2006 at 14:33, Ana Conesa wrote: To: Petr Pikal [EMAIL PROTECTED] From: Ana Conesa [EMAIL PROTECTED] Subject:Re: [R] Two functions with the same name At 14:04 24/05/2006, you wrote: Function written or loaded last will mask a function written or loaded before. Petr, Thanks for your answer. I know that. What I was asking is the way to go back to the function loaded in the first place, since I need it again. Re-loading the first package does not work for me. Ana HTH Petr PLEASE do read the posting guide! ^ On 24 May 2006 at 13:47, Ana Conesa wrote: Date sent:Wed, 24 May 2006 13:47:09 +0200 To: r-help@stat.math.ethz.ch From:Ana Conesa [EMAIL PROTECTED] Subject:[R] Two functions with the same name __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] O @ Ana Conesa, PhD. @@@ O @@ O @ Centro de Genómica @ O O @ Instituto Valenciano de Investigaciones Agrarias (IVIA) @@@ O Carretera Moncada - Naquera, Km. 4,5 O @ 46113 Moncada (Valencia) SPAIN || Tel. +34 963424000 ext.70161; Fax. +34 963424001 || email: [EMAIL PROTECTED] Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] median of a survfit object
Hello R users ! Here a survfit object : library(survival) essai - aml[aml$x == Maintained,] calc - survfit(Surv(essai$time, 1 - essai$status)) calc Call: survfit(formula = Surv(essai$time, 1 - essai$status)) n events median 0.95LCL 0.95UCL 11 4 103 28 Inf I would like to get the median of the object calc... For example, med - calc$median But it doesn't work... How can I do this ? -- David -- David [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Cross correlation/ bivariate/ mantel
McClatchie, Sam (PIRSA-SARDI mcclatchie.sam at saugov.sa.gov.au writes: Colleagues I have two spatial datasets (latitude, longitude, fish eggs) and (latitude, longitude, fish larvae) at the same 280 stations (i.e. 280 cases). I want to determine if the 2 datasets are spatially correlated. In other words, do high numbers of larvae occur where there are high numbers of eggs? I would like to calculate the cross correlation for these bivariate data and calculate a Mantel statistic as described on pg. 147 of Fortin and Dale 2005 Spatial analysis. Package ade4 has mantel.randtest and package vegan has mantel . Both functions find the Mantel statistic. If you need partial Mantel statistic, you can try partial.mantel in vegan or this code that I contributed some time ago: https://stat.ethz.ch/pipermail/r-help/2003-September/038985.html Marcelino de la Cruz Departamento de Biologia Vegetal Universidad Politecnica de Madrid Spain __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] data transformation
Hi,
I have great problems with my work in R.
I look for to model the growth of fish.
I have Longitudinal data, a serie of repeated
measures for each individual.
Using the corresponding packages nlme in R.
I treat to fit to the data different growth functions,
wich were entered by me.
Unfortunately for no it was arrived at the
convergence, several error messages appeared.
I am going to display the growth functions so as they
were entered and a nlme call example:
# differents growth function
#vonBertlalanffy
vonBert- function(x, Linf, K, t0)
Linf*(1-exp(-K*(x-t0)))
size ~ vonBert(age, Linf, K,t0)
vonBert -deriv(~ Linf*(1-exp(-K*(x-t0))),
c(Linf,K,t0),function(x,Linf,K,t0){})
vonBertInit - function(mCall, LHS, data)
{
xy - sortedXyData(mCall[[x]], LHS, data)
Linf - 900
if (Linf != max(xy[,y])) Linf - -Linf
K - 0.3
t0-0
value - c(Linf, K, t0)
names(value) - mCall[c(Linf, K,t0)]
value
}
vonBert - selfStart(vonBert, initial = vonBertInit)
class(vonBert)
#Richards
Rich - function(x, Linf, K, t0, m)
Linf*(1-exp(-K*(x-t0)))^(1/(1-m))
size ~ Rich(age, Linf, K, t0, m)
Rich -deriv(~ Linf*(1-exp(-K*(x-t0)))^(1/(1-m)),
c(Linf,K,t0,m),function(x,Linf,K,t0,m){})
RichInit - function(mCall, LHS, data)
{
xy - sortedXyData(mCall[[x]], LHS, data)
Linf - 900
if (Linf != max(xy[,y])) Linf - -Linf
K - 0.3
t0-0
m - 0.3
value - c(Linf, K, t0, m)
names(value) - mCall[c(Linf, K,t0,m)]
value
}
Rich - selfStart(Rich, initial = RichInit)
class(Rich)
#call
Rich.nlme - nlme(size ~ Rich(age, Linf, K, t0, m),
data = L.gd,
fixed = Linf + K +t0 +m~ 1,
start = list(fixed = c(900, 0.3,-3,0.9)))
#error message
Error: Singularity in backsolve at level 0, block 1
In addition: Warning message:
NaNs produced in: log(x)
What is the problem? I do not understand that it is
what is bad: the data, the entered growth functions,
some specification...
I will thank for any contribution of information.
Lic. Gabriela Escati Peñaloza
Biología y Manejo de Recursos Acuáticos
Centro Nacional Patagónico(CENPAT).
CONICET
Bvd. Brown s/nº.
(U9120ACV)Pto. Madryn
Chubut
Argentina
Tel: 54-2965/451301/451024/451375/45401 (Int:277)
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Re: [R] median of a survfit object
see: [R] How to access results of survival analysis Xiaochun Li (06 May 2006) At 15:03 24.05.2006 +0200, David Hajage wrote: Hello R users ! Here a survfit object : library(survival) essai - aml[aml$x == Maintained,] calc - survfit(Surv(essai$time, 1 - essai$status)) calc Call: survfit(formula = Surv(essai$time, 1 - essai$status)) n events median 0.95LCL 0.95UCL 11 4 103 28 Inf I would like to get the median of the object calc... For example, med - calc$median But it doesn't work... How can I do this ? -- David -- David [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] median of a survfit object
It works ! thank you ! 2006/5/24, Heinz Tuechler [EMAIL PROTECTED]: see: [R] How to access results of survival analysis Xiaochun Li (06 May 2006) At 15:03 24.05.2006 +0200, David Hajage wrote: Hello R users ! Here a survfit object : library(survival) essai - aml[aml$x == Maintained,] calc - survfit(Surv(essai$time, 1 - essai$status)) calc Call: survfit(formula = Surv(essai$time, 1 - essai$status)) n events median 0.95LCL 0.95UCL 11 4 103 28 Inf I would like to get the median of the object calc... For example, med - calc$median But it doesn't work... How can I do this ? -- David -- David [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- David [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] dendrogram plotting problem
j.joshua thomas researchjj at gmail.com writes: Dear List RGui Version : 2.3.0 User : 1 month I am having the *dendrogram plotting problem * The code i tried: library(cluster) DD-DataSetS01022 # 575 x 2 matrix VC-hclust(dist(DD),ave) *Warning message: NAs introduced by coercion* ( what does it mean? Is that the problem?) plot(VC,hang=-2) Output: http://roughjade.blogspot.com I think the problem is ploting too many branches (575) in the same dendrogram. You may consider reading help(dendrogram) to find alternative ways of visualizing such huge amount of data. I would try something like VCd-as.dendrogram(VC) plot(VCd, nodePar=list(pch = c(NA,NA), cex=0.8, lab.cex = 0.01)) and use a magnifying glass to se the labels ;) or better VCd2 - cut(VCd, h=5) plot(VCd2$upper) #to visualize the general structure of the dendrogram and then plot(VCd 2$lower[[1]]) plot(VCd 2$lower[[2]]) plot(VCd 2$lower[[3]]) ... etc, to visualize every branch. Marcelino de la Cruz Departamento de Biologia Vegetal Universidad Politecnica de Madrid __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] data.frame
Sara, You didn't read your data into R correctly. If your data are really in the form you posted (one long column of mixed data types and lots of blank lines), then I would: 1. remove the blank lines with my text editor and save (say, mydata.txt) 2. scan() the variable names into a vector nm - scan(mydata.txt, what = , nlines = 13) 3. scan() the data values into a list x - scan(mydata.txt, what = list(0,0,,0,0,0,0,0,0,0,0,0,), skip = 13) 4. attach the names names(x) - nm 5. convert to data frame DF - data.frame(x) (You really shouldn't post your whole data set; a couple of records would have been sufficient.) Peter Ehlers Sara Mouro wrote: Dear all, Does any one knows why should I get the following error message, when trying to do a simple data.frame?? DataF-data.frame(Subject,BiomR,Spp,Capas,Litter,Herbs,LitterD,MaxCanH,DDifS p,DSSp,Slope, CanDens,NearestSp) Erro em data.frame(Subject, BiomR, Spp, Capas, Litter, Herbs, LitterD, : arguments imply differing number of rows: 202, 0 The data I am using is: Subject BiomR Spp Capas Litter Herbs LitterD MaxCanH DDifSp DSSp Slope CanDens NearestSp 1 140.74 Qfa 2 1.460 0 0.778 8 70 3.663 28 2.280 Ear 2 7.26 Aun 2 0.660 0 0.477 8 110 3.988 34 2.476 Ear __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] glmmADMB and the GPL -- formerly-- How to buy R.
On Tue May 23 18:01:03 CEST 2006 Dave Fourier wrote: Dear List, Some of you have been following the discussion of the GPL and its inclusion in the glmmADMB package we created for R users. I would like to provide a bit of background and include an email we received from Prof. Ripley so that everyone can be aware of how some might use the GPL to try to force access to proprietary software. I think this is interesting because many have voiced the opinion about the benign nature of the GPL and that commercial enterprises who avoid it do so mainly out of ignorance. I have noticed two things: Users of the R-help list appear to rely largely on the advice of a rather small number of statistical experts. Second, the R users regard R as being more cutting edge and up to date than lists devoted to commercial statistical packages like SAS. For these reasons I was surprised to see the following post on the web in reply to a question on negative binomial mixed models. https://stat.ethz.ch/pipermail/r-help/2005-February/066146.html I thought that this was bad advice as certainly our ADMB-RE software could handle this problem easily. However one never knows exactly what sort of data people might use in a particular example that could lead to difficulties so I decided to code up a program that R users could test for this problem. However R users are used to a different approach for model formulation so that it was difficult for the average R user to access the program. I approached Anders Nielsen who is both an experienced ADMB user and R user and asked him to write an interface in R which would make the program more accessible to R users. He created a package and the whole thing seems to have had some success with at least one PhD thesis based on calculations using it. The R code that Anders wrote is simply an interface which takes the R specification for the model and outputs a data file in the format the an ADMB program expects. The ADMB program is a stand alone exe. The R script then reads the ADMB output files and presents the results to the user in a more familiar R format. Now it appears at some revision someone put a GPL notice on this package although Anders states that he did not do so, and and he is certain that it was not originally included by him. In any event the R script is easily extracted from the package by those who know how to do so and we have no problem with making the ADMB-RE source to the exe (TPL file) available. In fact the original was on our web site but was modified as we made to program more robust to deal with difficult data sets. The compiled TPL file links with our proprietary libraries and we have no intention of providing the source for these, but that is exactly what Prof. Ripley seems to be demanding since he claims that he wants the program to run on his computer which it apparently does not do at present. Prof. Ripley seems to feel that he is a qualified spokesman for the open source community. I have no idea what the community at large feels about this. What follows is Hans Skaug's post with Prof. Ripley's reply. On Mon, 22 May 2006, H. Skaug wrote: About glmmADMB and GPL: We were not very cautious when we put in the GPL statement. What we wanted to say was that the use of glmmADMB is free, and does not require a license for AD Model Builder. But that is not what you said, and you are legally and morally bound to fulfill the promise you made. Am I correct in interpreting this discussion so that all we have to do is to remove the License: GPL statement from the DESCRIPTION file (and everywhere else it may occur), and there will be no conflict between glmmADMB and the rules of the R community? I have made a request under the GPL. `All' you have to do is to fulfill it. We have temporarily withdrawn glmmADMB until this question has been settled. You can withdraw the package, but it has already been distributed under GPL, and those who received it under GPL have the right to redistribute it under GPL, including the sources you are obliged to give them. That's part of the `freedom' that GPL gives. hans Brian Ripley wrote: The issue in the glmmADMB example is not if they were required to release it under GPL (my reading from the GPL FAQ is that they probably were not, given that communication is between processes and the R code is interpreted). Rather, it is stated to be under GPL _but_ there is no source code offer for the executables (and the GPL FAQ says that for anonymous FTP it should be downloadable via the same site, and the principles apply equally to HTTP sites). As the executables are not for my normal OS and I would like to exercise my freedom to try the GPLed code, I have requested the sources
Re: [R] multiple plots with par mfg
On 23 May 2006, at 21:47, Romain Francois wrote: Hi, An other possibility might be to use two devices and use dev.set to go from one to another : Thanks. Actually I did try that, but there are quite a lot of points to plot, and the switching between plots slowed the whole simulation down a bit. Thanks too to Henrik Bengtsson for the code snippet. I've managed to get it working now, using par(mfg). Yan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] converting hexadecimal
Hi, do you know, a method to convert an hexadecimal value to the corresponding integer value (decimal) ? thinks for help. Romain -- Lorrillière Romain UMR 8079 Laboratoire Ecologie, Systématique et Evolution Bât. 362 Université Paris-Sud 91405 Orsay cedex France tel : 01 69 15 56 85 fax : 01 69 15 56 96 mobile : 06 81 70 90 70 email : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to make attributes persist after indexing?
Dear All! For descriptive purposes I would like to add attributes to objects. These attributes should be kept, even if by indexing only part of the object is used. I noted that some attributes like levels and class of a factor exist also after indexing, while others, like comment or label vanish. Is there a way to make an arbitrary attribute to be kept after indexing? This would be especially useful when indexing a data.frame. ## example for loss of attributes fx - factor(1:5, ordered=TRUE) attr(fx, 'comment') - 'Comment for fx' attr(fx, 'label') - 'Label for fx' attr(fx, 'testattribute') - 'just for fun' attributes(fx) # all attributes are shown attributes(fx[])# all attributes are shown attributes(fx[1:5]) # only levels and class are shown attributes(fx[1]) # only levels and class are shown Thanks, Heinz Tüchler __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to make attributes persist after indexing?
You could create your own child class with its own [ method.
[.myfactor - function(x, ...) {
attr - attributes(x)
x - NextMethod([)
attributes(x) - attr
x
}
gx - structure(fx, class = c(myfactor, class(fx)))
attributes(gx[1])
On 5/24/06, Heinz Tuechler [EMAIL PROTECTED] wrote:
Dear All!
For descriptive purposes I would like to add attributes to objects. These
attributes should be kept, even if by indexing only part of the object is
used.
I noted that some attributes like levels and class of a factor exist also
after indexing, while others, like comment or label vanish.
Is there a way to make an arbitrary attribute to be kept after indexing?
This would be especially useful when indexing a data.frame.
## example for loss of attributes
fx - factor(1:5, ordered=TRUE)
attr(fx, 'comment') - 'Comment for fx'
attr(fx, 'label') - 'Label for fx'
attr(fx, 'testattribute') - 'just for fun'
attributes(fx) # all attributes are shown
attributes(fx[])# all attributes are shown
attributes(fx[1:5]) # only levels and class are shown
attributes(fx[1]) # only levels and class are shown
Thanks,
Heinz Tüchler
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Re: [R] converting hexadecimal
On 5/24/2006 11:01 AM, Romain Lorrilliere wrote: Hi, do you know, a method to convert an hexadecimal value to the corresponding integer value (decimal) ? as.double (which is called by as.numeric) automatically converts hex to numeric values. You need to make sure the hex strings have a 0x prefix, e.g. as.numeric(0x0A) [1] 10 You can use paste() or gsub() to put the prefix on existing strings. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to make attributes persist after indexing?
On Wed, 2006-05-24 at 17:20 +0100, Heinz Tuechler wrote: Dear All! For descriptive purposes I would like to add attributes to objects. These attributes should be kept, even if by indexing only part of the object is used. I noted that some attributes like levels and class of a factor exist also after indexing, while others, like comment or label vanish. Is there a way to make an arbitrary attribute to be kept after indexing? This would be especially useful when indexing a data.frame. ## example for loss of attributes fx - factor(1:5, ordered=TRUE) attr(fx, 'comment') - 'Comment for fx' attr(fx, 'label') - 'Label for fx' attr(fx, 'testattribute') - 'just for fun' attributes(fx) # all attributes are shown attributes(fx[])# all attributes are shown attributes(fx[1:5]) # only levels and class are shown attributes(fx[1]) # only levels and class are shown Thanks, Heinz Tüchler Non-standard attributes do not survive the use of [. You could create a new class and subset method for the objects where you require this type of functionality. Frank Harrell has done that in the Hmisc package to support the use of the labeling attributes, which in turn are used by some of his functions such as latex(). You might want to review what he has done in ?label where the [.labelled method has been added. Alternatively, you could create your own function to save the attributes, do the subsetting and then restore the attributes to the resultant object. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] multiple destinations in duration (survival) analysis
Hi, I'm trying to estimate a (parametric) competing risks model in the context of duration (or survival, if you wish) analysis. That is, instead of studying the transition of subjects to death, I wish to study the transitions to multiple *destinations* (which is different from studying multiple *durations*, or recurrent events). I am more interested in the hazard function rather than the survival, but that would do as some manipulations in the estimates will give me what I want. Does any one know were I can find functions/routines to perform such estimation with R? I'd appreciate any tips on this. Thanks, Dimitri [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R Reference Card (especially useful for Newbies)
Newbies (and others!) may find useful the R Reference Card made available by Tom Short and Rpad at http://www.rpad.org/Rpad/Rpad-refcard.pdf or through the Contributed link on CRAN (where some other reference cards are also linked). It categorizes and organizes a bunch of R's basic, most used functions so that they can be easily found. For example, paste() is under the Strings heading and expand.grid() is under Data Creation. For newbies struggling to find the right R function as well as veterans who can't quite remember the function name, it's very handy. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] glmmADMB and the GPL -- formerly-- How to buy R.
I promise the list that this will be my last posting regarding this matter. The package glmmADMB is not and never was the property or a product of Otter Research Ltd. It was simply a probram I wrote using ADMB-RE and gave away. I do this all the time to help various people. I don't monitor or try to control what they do with it. The executable nbmm.exe was produced by using the ADMB-RE package, but under our license executables produced using ADMB are freely distributable. See for example the software Coleraine or stock synthesis. As I stated in my earlier post I was motivated to produce software for negative binomial mixed models by a post of Prof Ripleys, that no such good software existed. I assumed that to mean anywhere not just as a currently existing R package. I originally intended this to be just a stand alone program that anyone could use after they put the data into the proper format. I soon realized, however that R users had a different way of thinking about the methods for specifying these models. Since I have extremely limited R skills I asked Anders Nielsen and Hans Skaug if they would be interested in producing something that could be used more easily by R users. They were not paid for this and acted simply out of a wish to provide a service to the R community. I had no idea how that would go and at that point my involvement in the development (except for modifying the nbmm.exe to deal with difficult data) came to an end. Neither Anders or Hans are employees of or own any part of Otter Research Ltd. I am its sole employee/owner. When the package was finished I agreed to host it on my web site. I also put up a forum because I wanted to see how the software worked for people. I appears that Hans put the GPL on the package. He can speak for himself, but I believe he intended the GPL to apply to the R package which he had taken over from Anders and not the nbmm.exe Certainly he would have no authority to bind Otter Research Ltd in this matter. Now as to flames and Lawyers. As I said Otter Research Ltd is a company of one person. I have successfully sold software for 16 years without a lawyer, mostly to the fisheries management community. I never felt the need for a lawyer until I gave something away for free to the R community. I removed the link that Prof Ripley referred to. If there is another offending link I will be happy to remove it as well if you tell me where it is. For those who want to use the software I will put it up in its original form which means that you will need to put your data into the format that an AD Model Builder program requires. It will still have the same functionality but unfortunately will be more difficult for R users to access. Of course if Hans or Anders or anyone else wants to they are welcome to take the nbmm.exe and make it available with their own R package if they can satisfy whatever the requirements are -- just not on my website. Cheers, Dave -- David A. Fournier P.O. Box 2040, Sidney, B.C. V8l 3S3 Canada Phone/FAX 250-655-3364 http://otter-rsch.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] glmmADMB and the GPL -- formerly-- How to buy R.
- Original Message - From: dave fournier [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Tuesday, May 23, 2006 9:01 AM Subject: [R] glmmADMB and the GPL -- formerly-- How to buy R. Dear List, Some of you have been following the discussion of the GPL and its inclusion in the glmmADMB package we created for R users. I would like to provide a bit of background and include an email we received from Prof. Ripley so that everyone can be aware of how some might use the GPL to try to force access to proprietary software. I think this is interesting because many have voiced the opinion about the benign nature of the GPL and that commercial enterprises who avoid it do so mainly out of ignorance. I have noticed two things: Users of the R-help list appear to rely largely on the advice of a rather small number of statistical experts. Second, the R users regard R as being more cutting edge and up to date than lists devoted to commercial statistical packages like SAS. For these reasons I was surprised to see the following post on the web in reply to a question on negative binomial mixed models. https://stat.ethz.ch/pipermail/r-help/2005-February/066146.html I thought that this was bad advice as certainly our ADMB-RE software could handle this problem easily. However one never knows exactly what sort of data people might use in a particular example that could lead to difficulties so I decided to code up a program that R users could test for this problem. However R users are used to a different approach for model formulation so that it was difficult for the average R user to access the program. I approached Anders Nielsen who is both an experienced ADMB user and R user and asked him to write an interface in R which would make the program more accessible to R users. He created a package and the whole thing seems to have had some success with at least one PhD thesis based on calculations using it. The R code that Anders wrote is simply an interface which takes the R specification for the model and outputs a data file in the format the an ADMB program expects. The ADMB program is a stand alone exe. The R script then reads the ADMB output files and presents the results to the user in a more familiar R format. Now it appears at some revision someone put a GPL notice on this package although Anders states that he did not do so, and and he is certain that it was not originally included by him. In any event the R script is easily extracted from the package by those who know how to do so and we have no problem with making the ADMB-RE source to the exe (TPL file) available. In fact the original was on our web site but was modified as we made to program more robust to deal with difficult data sets. The compiled TPL file links with our proprietary libraries and we have no intention of providing the source for these, but that is exactly what Prof. Ripley seems to be demanding since he claims that he wants the program to run on his computer which it apparently does not do at present. Prof. Ripley seems to feel that he is a qualified spokesman for the open source community. I have no idea what the community at large feels about this. What follows is Hans Skaug's post with Prof. Ripley's reply. On Mon, 22 May 2006, H. Skaug wrote: About glmmADMB and GPL: We were not very cautious when we put in the GPL statement. What we wanted to say was that the use of glmmADMB is free, and does not require a license for AD Model Builder. But that is not what you said, and you are legally and morally bound to fulfill the promise you made. Am I correct in interpreting this discussion so that all we have to do is to remove the License: GPL statement from the DESCRIPTION file (and everywhere else it may occur), and there will be no conflict between glmmADMB and the rules of the R community? I have made a request under the GPL. `All' you have to do is to fulfill it. We have temporarily withdrawn glmmADMB until this question has been settled. You can withdraw the package, but it has already been distributed under GPL, and those who received it under GPL have the right to redistribute it under GPL, including the sources you are obliged to give them. That's part of the `freedom' that GPL gives. hans Brian Ripley wrote: The issue in the glmmADMB example is not if they were required to release it under GPL (my reading from the GPL FAQ is that they probably were not, given that communication is between processes and the R code is interpreted). Rather, it is stated to be under GPL _but_ there is no source code offer for the executables (and the GPL FAQ says that for anonymous FTP it should be downloadable via the same site, and the principles apply equally to HTTP sites). As the executables are not for my normal OS and I would like to exercise my freedom to try
Re: [R] normality testing with nortest
see inline Raymond Wan wrote: Hi all, On Mon, 22 May 2006, Rolf Turner wrote: It is worth noting that if the null hypothesis is true, then the p-value is uniformly distributed on [0,1]. SG: This is true in many but not all cases, and I think is a reasonable approximation even when not exactly true. Example: Testing any null hypothesis about a discrete distribution with only N possible outcomes must produce N possible outcomes in the distribution of the p-value. hope this helps, Spencer Graves This should be kept in mind when assessing the ``instability'' of p-values. Just wanted to thank everyone who participated in this thread for clarifying things for me. I guess I knew bits and pieces but your messages put them together and gave me some R commands (which I am still new to) to check things myself. Thanks a lot! Ray __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Statistical Power
On Tue, 23 May 2006, Christopher Brown wrote: How can I compute a power analysis on a multi-factor within-subjects design? If you are capable of installing source packages and if you know what a general linear hypothesis test is, you can: download 'hpower' to your computer http://gd.tuwien.ac.at/languages/R/src/contrib/hpower_0.1-0.tar.gz and install it. Then library( hpower ) ?glhpwr should get you started. [ Part 3.28: Included Message ] Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] UBTA - Wall Street Journal Agree - Ref. wj277963
Kelsey, P2P Sports Betting Software May Change Casino Sports Betting VANCOUVER, BRITISH COLUMBIA - UBA Technology Inc. (UBTA), has entered into initial negotiations to install its proprietary betting exchange software in traditional land-based casinos. Read the whole story: http://geocities.yahoo.com.br/incidentless63143 Alma Lyles, Acct. Rep. go820972 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Regression line limited by the rage of values
Hi In R, using plot(x,y) followed by abline(lm(y~x)) produces a graph with a regression line spanning the whole plot . This means that the line extends beyond the swarm of data points to the defined of default plot region. With par(xpd=T) it will span the entire figure region. But how can I limit a regression line to the data range, i.e between (xmin,ymin) and (xmax,ymax)? Sorry for not knowing the lingo here. If you don't understand the question, please run the following script: x1-c(1,2,3,4) x2-c(5,6,7,8) y1-c(2,4,5,8) y2-c(10,11,12,16) plot(x1,y1,xlim=c(0,10),ylim=c(0,20),col=blue) points(x2,y2,col=red) abline(lm(y1~x1),col=blue) abline(lm(y2~x2),col=red) The resulting plot isn't very informative. There is no overlap in the two groups of data, yet the two ablines overlap. I instead want the blue line to go from (1,2) to (4,8) and the red line from (5,10) to (8,16). So, how can I constrain the abline to the relevant region, i.e stop abline from extrapolating beyond the actual range of data. Or should I use a function line 'lines' to do this? Cheers Andres __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Regression line limited by the rage of values
On Wed, 2006-05-24 at 18:51 +0200, Andreas Svensson wrote: Hi In R, using plot(x,y) followed by abline(lm(y~x)) produces a graph with a regression line spanning the whole plot . This means that the line extends beyond the swarm of data points to the defined of default plot region. With par(xpd=T) it will span the entire figure region. But how can I limit a regression line to the data range, i.e between (xmin,ymin) and (xmax,ymax)? Sorry for not knowing the lingo here. If you don't understand the question, please run the following script: x1-c(1,2,3,4) x2-c(5,6,7,8) y1-c(2,4,5,8) y2-c(10,11,12,16) plot(x1,y1,xlim=c(0,10),ylim=c(0,20),col=blue) points(x2,y2,col=red) abline(lm(y1~x1),col=blue) abline(lm(y2~x2),col=red) The resulting plot isn't very informative. There is no overlap in the two groups of data, yet the two ablines overlap. I instead want the blue line to go from (1,2) to (4,8) and the red line from (5,10) to (8,16). So, how can I constrain the abline to the relevant region, i.e stop abline from extrapolating beyond the actual range of data. Or should I use a function line 'lines' to do this? Cheers Andres Try this instead of the two abline()'s: lines(x1, fitted(lm(y1 ~ x1)), col = blue) lines(x2, fitted(lm(y2 ~ x2)), col = red) The function fitted() will extract the model fitted y values from the lm() model object. See ?lines and ?fitted HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Statistical Power
On Wed, 2006-05-24 at 09:22 -0700, Charles C. Berry wrote: On Tue, 23 May 2006, Christopher Brown wrote: How can I compute a power analysis on a multi-factor within-subjects design? If you are capable of installing source packages and if you know what a general linear hypothesis test is, you can: download 'hpower' to your computer http://gd.tuwien.ac.at/languages/R/src/contrib/hpower_0.1-0.tar.gz and install it. Then library( hpower ) ?glhpwr Downloading, unzipping the package on to FC4 running R 2.3.0 gives an error message about a bad description file when trying to install hpower. R CMD INSTALL hpower Rick B. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] changing font size in plot(effect())
I can't seem to be able to change the font size in an effect display. I've tried the following: par(cex.lab=4) plot(effect (alti,reg8), ylab=detection probability) and plot(effect (alti,reg8), ylab=detection probability, cex=4) but nothing changes. Can anyone help me? thanks. Emilie Berthiaume graduate student Sherbrooke University 2500 boul. de l'Université Sherbrooke, Québec J1K 2R1 CANADA Phone: 1-819-821-8000 poste 2059 Fax: 1-819-821-8049 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] FastICA and matching variance of inputs
I have been playing around with the FastICA algorithm in R, and I have come across the following issue. I'm new to this topic, so I'm hoping someone will be able to shed some insight. We know that the variance of the Independent Components is constrained to be 1 and they are uncorrelated, so the sum of squares of the coefficients of the mixing matrix (a$A) will give the variance of the reconstructed inputs (a$X). It seems to me that this should match the variance of the actual inputs (X). But the former seems to be smaller than the latter, by varying amounts. Does anyone know why that might be? # results are close S - matrix(runif(1), 500, 20) A - matrix(runif(4000), 20, 200) X - S%*%A a - fastICA(X, 20) as.vector(apply(X,2,sd))[1:10] apply(a$A,2,function(x) sqrt(sum(x^2)))[1:10] -- as.vector(apply(X,2,sd))[1:10] [1] 0.6815156 0.7160899 0.8237067 0.6777511 0.8045670 0.7584224 0.7818801 [8] 0.8015249 0.7252270 0.6337887 apply(a$A,2,function(x) sqrt(sum(x^2)))[1:10] [1] 0.6808337 0.7153735 0.8228825 0.6770730 0.8037620 0.7576636 0.7810978 [8] 0.8007229 0.7245014 0.6331546 -- # results are not close S - matrix(runif(1), 500, 20) A - matrix(runif(4000), 20, 200) X - S%*%A a - fastICA(X, 2) as.vector(apply(X,2,sd))[1:10] apply(a$A,2,function(x) sqrt(sum(x^2)))[1:10] -- as.vector(apply(X,2,sd))[1:10] [1] 0.7400437 0.7775435 0.6862873 0.7890104 0.7516134 0.8099342 0.8095056 [8] 0.7653593 0.6743729 0.7875858 apply(a$A,2,function(x) sqrt(sum(x^2)))[1:10] [1] 0.6463631 0.7131771 0.5987285 0.6979883 0.6664734 0.7300787 0.7179024 [8] 0.6571067 0.5549075 0.6684818 -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] changing font size in plot(effect())
Try this: trellis.par.set(par.ylab.text = list(cex = 4)) plot(effect (alti,reg8), ylab=detection probability) ?trellis.par.set Emilie Berthiaume wrote: I can't seem to be able to change the font size in an effect display. I've tried the following: par(cex.lab=4) plot(effect (alti,reg8), ylab=detection probability) and plot(effect (alti,reg8), ylab=detection probability, cex=4) but nothing changes. Can anyone help me? thanks. Emilie Berthiaume graduate student Sherbrooke University 2500 boul. de l'Université Sherbrooke, Québec J1K 2R1 CANADA Phone: 1-819-821-8000 poste 2059 Fax: 1-819-821-8049 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Statistical Power
On Wed, May 24, 2006 20:22, Rick Bilonick wrote: Downloading, unzipping the package on to FC4 running R 2.3.0 gives an error message about a bad description file when trying to install hpower. R CMD INSTALL hpower Indeed, it does so on both my Ubuntu and Win XP boxes. The problem is that the DESCRIPTION file does not have a proper Maintainer line. If you simply add the line: Maintainer: no-one [EMAIL PROTECTED] and follow the regular procedures, it should install. However, I am not so sure that this is the right thing to do with regards to permissions etc. Anyone any advice? Ioannis -- Ioannis C. Dimakos, Ph.D. University of Patras Department of Elementary Education Patras, GR-26500 GREECE http://www.elemedu.upatras.gr/dimakos/ http://yannishome.port5.com/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] problem-nlme
Hi,
I have great problems with my work in R.
I look for to model the growth of fish.
I have Longitudinal data, a serie of repeated
measures for each individual.
Using the corresponding packages nlme in R.
I treat to fit to the data different growth functions,
wich were entered by me.
Unfortunately for no it was arrived at the
convergence, several error messages appeared.
I am going to display the growth functions so as they
were entered and a nlme call example:
# differents growth function
#vonBertlalanffy
vonBert- function(x, Linf, K, t0)
Linf*(1-exp(-K*(x-t0)))
size ~ vonBert(age, Linf, K,t0)
vonBert -deriv(~ Linf*(1-exp(-K*(x-t0))),
c(Linf,K,t0),function(x,Linf,K,t0){})
vonBertInit - function(mCall, LHS, data)
{
xy - sortedXyData(mCall[[x]], LHS, data)
Linf - 900
if (Linf != max(xy[,y])) Linf - -Linf
K - 0.3
t0-0
value - c(Linf, K, t0)
names(value) - mCall[c(Linf, K,t0)]
value
}
vonBert - selfStart(vonBert, initial = vonBertInit)
class(vonBert)
#Richards
Rich - function(x, Linf, K, t0, m)
Linf*(1-exp(-K*(x-t0)))^(1/(1-m))
size ~ Rich(age, Linf, K, t0, m)
Rich -deriv(~ Linf*(1-exp(-K*(x-t0)))^(1/(1-m)),
c(Linf,K,t0,m),function(x,Linf,K,t0,m){})
RichInit - function(mCall, LHS, data)
{
xy - sortedXyData(mCall[[x]], LHS, data)
Linf - 900
if (Linf != max(xy[,y])) Linf - -Linf
K - 0.3
t0-0
m - 0.3
value - c(Linf, K, t0, m)
names(value) - mCall[c(Linf, K,t0,m)]
value
}
Rich - selfStart(Rich, initial = RichInit)
class(Rich)
#call
Rich.nlme - nlme(size ~ Rich(age, Linf, K, t0, m),
data = L.gd,
fixed = Linf + K +t0 +m~ 1,
start = list(fixed = c(900, 0.3,-3,0.9)))
#error message
Error: Singularity in backsolve at level 0, block 1
In addition: Warning message:
NaNs produced in: log(x)
What is the problem? I do not understand that it is
what is bad: the data, the entered growth functions,
some specification...
I will thank for any contribution of information.
Lic. Gabriela Escati Peñaloza
Biología y Manejo de Recursos Acuáticos
Centro Nacional Patagónico(CENPAT).
CONICET
Bvd. Brown s/nº.
(U9120ACV)Pto. Madryn
Chubut
Argentina
Tel: 54-2965/451301/451024/451375/45401 (Int:277)
Fax: 54-29657451543
__
Yahoo! Autos. Más de 100 vehículos vendidos por día.
¿Qué esperás para vender el tuyo?
Hacelo ahora y ganate un premio de Yahoo!
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[R] Joining variables
Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Joining variables
On Wed, 2006-05-24 at 14:45 -0400, Guenther, Cameron wrote: Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1 Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. See ?paste and note the 'sep' argument: NewVar - paste(Var1, Var2, sep = ) NewVar [1] SA10005511319851113 HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] LARS error
Hi All, when i run cv.lars(x3,y3) it runs fine. but when i run cv.lars(x3,y3,fraction=seq(0,0.1,100)) I get the following error. Error in apply((y[omit] - fit)^2, 2, mean) : dim(X) must have a positive length Any help/suggestions will be appreciated. Thanks. Harsh - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Joining variables
Thanks to all who responded. The paste function is what I was looking for. Thanks again. Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] -Original Message- From: Berton Gunter [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 24, 2006 2:54 PM To: Guenther, Cameron; r-help@stat.math.ethz.ch Subject: RE: [R] Joining variables What does combination of both mean exactly. I can think of two interpretations that have two different answers. If you give a small example (as the posting guide suggests) it would certainly help me provide an answer. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Guenther, Cameron Sent: Wednesday, May 24, 2006 11:46 AM To: r-help@stat.math.ethz.ch Subject: [R] Joining variables Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1 Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] multiple destinations in duration (survival) analysis
I have been working on a similar project and here is how I approached it (though I would be very happy to hear other ideas): Our situation was that we wanted to predict length of stay in the emergency room of the hospital, there are multiple competing places to go from the ER: released to go home, admitted to the main hospital, transferred to another specialty hospital, leave without seeing the doctor, Not suprisingly the predictors have opposite effects on time till admission and time till released to home. There is a function aj in package changeLOS that computes the Aalen-Johansen estimator for competing risks, but it does not fit anything with covariates. You can also do multiple cox model regressions using each destination as the event and all other destinations as censoring, but this does not give some of the nice interpretations of the Aelen-Johansen estimates, so here is how we combined the 2 approaches: 1. fit a Cox ph model for each competing risk (using all others as censoring). 2. Choose a baseline set of covariates (means of all, or typical values, ...). 3. Compute the predicted survival distribution for this baseline individual (using survfit on the coxph model) for each of the coxph models. 4. Simulate 1,000 observations from each of the predicted survival distributions, put these into a 1,000 by (number of competing risks) matrix. 5. take the minimum of each row as the time to event and which row has that minimum as the event that occurred and use that as input to the aj function (actually I think the trans function is used in there somewhere, then the aj function). 6. Create a plot based on the output of AJ showing at each time period what the probabilities of being still in the system or any of the destinations is. 7. Choose a 2nd set of covariates (generally just change 1 value from the baseline) and repeate the whole process, compare the 2 graphs to get a feel for the effect of changing that covariate. 8. Repeate, but changing different covariates to see the effect. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Dimitri Szerman Sent: Wednesday, May 24, 2006 9:47 AM To: R-Help Subject: [R] multiple destinations in duration (survival) analysis Hi, I'm trying to estimate a (parametric) competing risks model in the context of duration (or survival, if you wish) analysis. That is, instead of studying the transition of subjects to death, I wish to study the transitions to multiple *destinations* (which is different from studying multiple *durations*, or recurrent events). I am more interested in the hazard function rather than the survival, but that would do as some manipulations in the estimates will give me what I want. Does any one know were I can find functions/routines to perform such estimation with R? I'd appreciate any tips on this. Thanks, Dimitri [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Joining variables
Guenther, Cameron wrote: Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1 Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. See ?paste or ?sprintf Var1 - factor(SA100055113) Var2 - 19851113 NewVar - paste(Var1, Var2, sep = ) NewVar - sprintf(%s%s, Var1, Var2) HTH, --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Joining variables
What does combination of both mean exactly. I can think of two interpretations that have two different answers. If you give a small example (as the posting guide suggests) it would certainly help me provide an answer. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Guenther, Cameron Sent: Wednesday, May 24, 2006 11:46 AM To: r-help@stat.math.ethz.ch Subject: [R] Joining variables Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1 Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Problem with Statistics::R Save plots
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Re: [R] Joining variables
look at ?paste(). Best, Dimitris Dimitris Rizopoulos Ph.D. Student 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 Quoting Guenther, Cameron [EMAIL PROTECTED]: Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1 Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Decimal places
Hello, I would like to force R to print 2 decimal places. I use options(digits=2) but as far as I know it prints significant digits so in certain cases is more then 2. Is there other way to do it? Best, Robert [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Regression line limited by the range of values
Thankyou very much Marc for that nifty little script. When I use it on my real dataset though, the lines are fat in the middle and thinner towards the ends. I guess it's because lines draw one fitted line for each x, and if you have hundreds of x, this turns into a line that is thicker that it should be (due to rounding errors perhaps). I got the tip to use segments, and draw one line from min(x),min(y) to max(x),max(y) but with real data with a bunch of na.rm and na.actions this becomes very long and bulky. For a regression with two data swarms and two trend lines it becomes this long mess: plot(TCgonad[Period==1], ABelly[Period==1],xlim=c(0,20),ylim=c(130,160), col=”blue”) points(TCgonad[Period==2], ABelly[Period==2],col=”red”) segments( min(TCgonad[Period==1],na.rm=T), min(fitted(lm(ABelly[Period==1]~TCgonad[Period==1], na.action=na.exclude)),na.rm=T), max(TCgonad[Period==1],na.rm=T), max(fitted(lm(ABelly[Period==1]~TCgonad[Period==1], na.action=na.exclude)),na.rm=T),col=”blue”) segments( min(TCgonad[Period==2],na.rm=T), min(fitted(lm(ABelly[Period==2]~TCgonad[Period==2], na.action=na.exclude)),na.rm=T), max(TCgonad[Period==2],na.rm=T), max(fitted(lm(ABelly[Period==2]~TCgonad[Period==2], na.action=na.exclude)),na.rm=T),col=”red”) I just think it's strange that abline has as a nonadjustable default to extrapolate the line to outside the data - a mortal sin in my field. Cheers Andreas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Joining variables
Hello, Although I cannot speak for the original individual posting this question but I would like to see an example, if possible, of how to construct a factor of several factors. I think this would be more efficient than paste if such a procedure were possible and, of course, would also answer the postee's original question. Many thanks, Mark LoPresti Thomson Financial/Research Group -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Berton Gunter Sent: Wednesday, May 24, 2006 2:54 PM To: 'Guenther, Cameron'; r-help@stat.math.ethz.ch Subject: Re: [R] Joining variables What does combination of both mean exactly. I can think of two interpretations that have two different answers. If you give a small example (as the posting guide suggests) it would certainly help me provide an answer. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Guenther, Cameron Sent: Wednesday, May 24, 2006 11:46 AM To: r-help@stat.math.ethz.ch Subject: [R] Joining variables Hello, If I have two variables that are factors or characters and I want to create a new variable that is the combination of both what function can I use to accomplish this? Ex. Var1 Var2 SA100055113 19851113 And I want NewVar SA10005511319851113 Thanks in advance. Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Regression line limited by the range of values
On Wed, 2006-05-24 at 21:53 +0200, Andreas Svensson wrote: Thankyou very much Marc for that nifty little script. When I use it on my real dataset though, the lines are fat in the middle and thinner towards the ends. I guess it's because lines draw one fitted line for each x, and if you have hundreds of x, this turns into a line that is thicker that it should be (due to rounding errors perhaps). I got the tip to use segments, and draw one line from min(x),min(y) to max(x),max(y) but with real data with a bunch of na.rm and na.actions this becomes very long and bulky. For a regression with two data swarms and two trend lines it becomes this long mess: plot(TCgonad[Period==1], ABelly[Period==1],xlim=c(0,20),ylim=c(130,160), col=”blue”) points(TCgonad[Period==2], ABelly[Period==2],col=”red”) segments( min(TCgonad[Period==1],na.rm=T), min(fitted(lm(ABelly[Period==1]~TCgonad[Period==1], na.action=na.exclude)),na.rm=T), max(TCgonad[Period==1],na.rm=T), max(fitted(lm(ABelly[Period==1]~TCgonad[Period==1], na.action=na.exclude)),na.rm=T),col=”blue”) segments( min(TCgonad[Period==2],na.rm=T), min(fitted(lm(ABelly[Period==2]~TCgonad[Period==2], na.action=na.exclude)),na.rm=T), max(TCgonad[Period==2],na.rm=T), max(fitted(lm(ABelly[Period==2]~TCgonad[Period==2], na.action=na.exclude)),na.rm=T),col=”red”) I just think it's strange that abline has as a nonadjustable default to extrapolate the line to outside the data - a mortal sin in my field. Cheers Andreas Andreas, What is likely happening is that your x values are not sorted in increasing order as they were in the dummy data set, thus there is probably some movement of the lines back and forth from one x value to the next in the order that they appear in the data set. Here is a more generic approach that should work. It is based upon the examples in ?predict.lm. We create the two models and then use range(x?) for the new min,max x values to be used for the prediction of the fitted y values. Thus: # Create model1 using random data set.seed(1) x1 - rnorm(15) y1 - x1 + rnorm(15) model1 - lm(y1 ~ x1) # Create model2 using random data set.seed(2) x2 - rnorm(15) y2 - x2 + rnorm(15) model2 - lm(y2 ~ x2) # Plot the first set of points # set the axis ranges based upon the common ranges of the # two sets of data plot(x1, y1, col = red, xlim = range(x1, x2), ylim = range(y1, y2)) # Add the second set of points points(x2, y2, col = blue) # Create the first fitted line # Create two x values for the prediction based upon the # range of x1 lines(range(x1), predict(model1, data.frame(x1 = range(x1))), col = red) # Create the second fitted line # Same here for the x2 values lines(range(x2), predict(model2, data.frame(x2 = range(x2))), col = blue) Note that in both cases for the fitted lines, the lines will be constrained by the range of the actual x? data values. This should be easier to apply in general. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Survey proportions... Can I use population as denominator?
On Tue, 23 May 2006, David L. Van Brunt, Ph.D. wrote: Just giving the survey package a spin... I'm accustomed to stata, and it seems very similar in many respects. One thing is throwing me, however. I've gotten my data in, and specified the design. Looks like the weighting is right (based on published population estimates from these data), but now I'd like to check my marginal means for proportions against those that have been published. No, actually you get means with svymean(). Proportions for factor levels come by svymean() with a factor variable, eg: data(api) dclus1-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) svymean(~stype,dclus1) mean SE stypeE 0.786885 0.0463 stypeH 0.076503 0.0268 stypeM 0.136612 0.0296 svymean(~sch.wide,dclus1) mean SE sch.wideNo 0.12568 0.0204 sch.wideYes 0.87432 0.0204 -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] biplot with grouped observations
Hi, I'd like to plot the results of PCA in biplots, grouping the observations by using different symbols. With xyplot in lattice I can easily differentiate the groups (see for example DAAG, p 285) but it seems difficult (or impossible?) to plot the vectors for the variables as well as xyplot cannot handle secondary axis very well. Using the basic plot I can combine observations and variables in one plot but cannot differentiate the groups of observations as easily as in xyplot. What could I do to get grouped observations in a biplot? Thanks already, Kim __ Kim Milferstedt University of Illinois at Urbana-Champaign Department of Civil and Environmental Engineering 4125 Newmark Civil Engineering Building 205 North Mathews Avenue MC 250 Urbana, IL 61801 USA phone: (001) 217 333-9663 fax: (001) 217 333-6968 email: [EMAIL PROTECTED] http://cee.uiuc.edu/research/morgenroth __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] legend title in effects plot
Could someone tell how I can change/remove the legend title in a (multiline) effect plot? Thanks V In general, how is one supposed to find out answers to such questions? It is virtually impossible to keep track of what graphical parameters are passed on to which low level function and which don't. Every funciton calls legend labels a different name (title, key, legend.lab, etc.). I tried the help pages of par,plot,effect,legend, without an effect :-) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Decimal places
On Wed, 24 May 2006, Robert Mcfadden wrote: Hello, I would like to force R to print 2 decimal places. I use options(digits=2) but as far as I know it prints significant digits so in certain cases is more then 2. Is there other way to do it? ?round, possibly format(rounf(x, 2), nsmall = 2) if you want exactly 2 decimal places. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] problem-nlme
Gabriela,
it may be that the model that you are trying to fit is vey complex.
Try something simpler. Note that the default setting for the random
effects is the same as the fixed effects. In my experience this is a
very difficult model to fit, especially for the Richards and von
Bertalanffy models. So, try a simpler model, and build from there.
Eg
Rich.nlme.linf - nlme(size ~ Rich(age, Linf, K, t0, m),
data = L.gd,
fixed = Linf + K + t0 + m ~ 1,
random = Linf ~ 1,
start = list(fixed = c(900, 0.3,-3,0.9)))
Good luck!
Andrew
On Wed, May 24, 2006 at 06:30:43PM +, gabriela escati pe?aloza wrote:
Hi,
I have great problems with my work in R.
I look for to model the growth of fish.
I have Longitudinal data, a serie of repeated
measures for each individual.
Using the corresponding packages nlme in R.
I treat to fit to the data different growth functions,
wich were entered by me.
Unfortunately for no it was arrived at the
convergence, several error messages appeared.
I am going to display the growth functions so as they
were entered and a nlme call example:
# differents growth function
#vonBertlalanffy
vonBert- function(x, Linf, K, t0)
Linf*(1-exp(-K*(x-t0)))
size ~ vonBert(age, Linf, K,t0)
vonBert -deriv(~ Linf*(1-exp(-K*(x-t0))),
c(Linf,K,t0),function(x,Linf,K,t0){})
vonBertInit - function(mCall, LHS, data)
{
xy - sortedXyData(mCall[[x]], LHS, data)
Linf - 900
if (Linf != max(xy[,y])) Linf - -Linf
K - 0.3
t0-0
value - c(Linf, K, t0)
names(value) - mCall[c(Linf, K,t0)]
value
}
vonBert - selfStart(vonBert, initial = vonBertInit)
class(vonBert)
#Richards
Rich - function(x, Linf, K, t0, m)
Linf*(1-exp(-K*(x-t0)))^(1/(1-m))
size ~ Rich(age, Linf, K, t0, m)
Rich -deriv(~ Linf*(1-exp(-K*(x-t0)))^(1/(1-m)),
c(Linf,K,t0,m),function(x,Linf,K,t0,m){})
RichInit - function(mCall, LHS, data)
{
xy - sortedXyData(mCall[[x]], LHS, data)
Linf - 900
if (Linf != max(xy[,y])) Linf - -Linf
K - 0.3
t0-0
m - 0.3
value - c(Linf, K, t0, m)
names(value) - mCall[c(Linf, K,t0,m)]
value
}
Rich - selfStart(Rich, initial = RichInit)
class(Rich)
#call
Rich.nlme - nlme(size ~ Rich(age, Linf, K, t0, m),
data = L.gd,
fixed = Linf + K +t0 +m~ 1,
start = list(fixed = c(900, 0.3,-3,0.9)))
#error message
Error: Singularity in backsolve at level 0, block 1
In addition: Warning message:
NaNs produced in: log(x)
What is the problem? I do not understand that it is
what is bad: the data, the entered growth functions,
some specification...
I will thank for any contribution of information.
Lic. Gabriela Escati Pe?aloza
Biolog?a y Manejo de Recursos Acu?ticos
Centro Nacional Patag?nico(CENPAT).
CONICET
Bvd. Brown s/n?.
(U9120ACV)Pto. Madryn
Chubut
Argentina
Tel: 54-2965/451301/451024/451375/45401 (Int:277)
Fax: 54-29657451543
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--
Andrew Robinson
Department of Mathematics and StatisticsTel: +61-3-8344-9763
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au
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Re: [R] Decimal places
Thank you very much. Round(x, digids=2) works perfect. Rob -Original Message- From: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 24, 2006 10:47 PM To: Robert Mcfadden Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Decimal places On Wed, 24 May 2006, Robert Mcfadden wrote: Hello, I would like to force R to print 2 decimal places. I use options(digits=2) but as far as I know it prints significant digits so in certain cases is more then 2. Is there other way to do it? ?round, possibly format(rounf(x, 2), nsmall = 2) if you want exactly 2 decimal places. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Regression line limited by the rage of values
For single lines the usual method is to just use the lines or segments
function, however I have been thinking that there may be some other uses
for clipping within a plot region when adding info (abline, but others
as well). So here is a first stab at a function to clip within the
region, it needs the TeachingDemos package to work. I'll probably put
in some more error checking, document it and include it in a future
TeachingDemos release, but you are welcome to use it in the meantime.
For your case you would create your scatterplot the same as you already
did, then add the lines like:
clipplot( abline(lm(y1~x1),col=blue), xlim=c(1,4) )
clipplot( abline(lm(y2~x2),col=red), xlim=c(5,8) )
Here is the function definition:
clipplot - function(fun, xlim=par('usr')[1:2],
ylim=par('usr')[3:4] ){
old.par - par(c('plt','xpd'))
if( length(xlim) 2 ) stop('xlim must be a vector with at least 2
elements')
if( length(ylim) 2 ) stop('ylim must be a vector with at least 2
elements')
if( !require(TeachingDemos) ) stop('TeachingDemos package needed')
xl - range(xlim)
yl - range(ylim)
pc - cnvrt.coords(xl,yl)$fig
par(plt=c(pc$x,pc$y),xpd=FALSE)
fun
par(old.par)
box() # need to plot something to reset
}
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Andreas Svensson
Sent: Wednesday, May 24, 2006 10:52 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Regression line limited by the rage of values
Hi
In R, using plot(x,y) followed by abline(lm(y~x)) produces a graph
with a regression line spanning the whole plot . This means that the
line extends beyond the swarm of data points to the defined of default
plot region. With par(xpd=T) it will span the entire figure region. But
how can I limit a regression line to the data range, i.e between
(xmin,ymin) and (xmax,ymax)?
Sorry for not knowing the lingo here. If you don't understand the
question, please run the following script:
x1-c(1,2,3,4)
x2-c(5,6,7,8)
y1-c(2,4,5,8)
y2-c(10,11,12,16)
plot(x1,y1,xlim=c(0,10),ylim=c(0,20),col=blue)
points(x2,y2,col=red)
abline(lm(y1~x1),col=blue)
abline(lm(y2~x2),col=red)
The resulting plot isn't very informative. There is no overlap in the
two groups of data, yet the two ablines overlap.
I instead want the blue line to go from (1,2) to (4,8) and the red line
from (5,10) to (8,16).
So, how can I constrain the abline to the relevant region, i.e stop
abline from extrapolating beyond the actual range of data.
Or should I use a function line 'lines' to do this?
Cheers
Andres
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[R] Logistic Regression - Results?
Hi, I use SPSS at work and have R installed both at work and on my home machine. I've been trying to do some logistic regressions in R and SPSS, but the results I'm getting are different. I've followed a few R tutorials, and with most of them, I get the following instructions: result - glm(z ~ x + y, family=binomial(logit)) In the case above, with three variables (z being dependent). In SPSS, I'm told to use Analyze - Regression - Binary Logistic, where I put x, y in Covariates and z in Dependent. Note that my values for x and y are either 1 or 0. The results I get from these two tests are different, however, and I was wondering why. Am I choosing the wrong commands? If not, why are the results different? Any help would be greatly appreciated, and please note that I have a limited amount of stats knowledge. Thanks, Wojciech -- Five Minutes to Midnight: Youth on human rights and current affairs http://www.fiveminutestomidnight.org/ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to make attributes persist after indexing?
Thank you for your answer, Gabor. I will see, if I understood it.
Heinz
At 11:31 24.05.2006 -0400, Gabor Grothendieck wrote:
You could create your own child class with its own [ method.
[.myfactor - function(x, ...) {
attr - attributes(x)
x - NextMethod([)
attributes(x) - attr
x
}
gx - structure(fx, class = c(myfactor, class(fx)))
attributes(gx[1])
On 5/24/06, Heinz Tuechler [EMAIL PROTECTED] wrote:
Dear All!
For descriptive purposes I would like to add attributes to objects. These
attributes should be kept, even if by indexing only part of the object is
used.
I noted that some attributes like levels and class of a factor exist also
after indexing, while others, like comment or label vanish.
Is there a way to make an arbitrary attribute to be kept after indexing?
This would be especially useful when indexing a data.frame.
## example for loss of attributes
fx - factor(1:5, ordered=TRUE)
attr(fx, 'comment') - 'Comment for fx'
attr(fx, 'label') - 'Label for fx'
attr(fx, 'testattribute') - 'just for fun'
attributes(fx) # all attributes are shown
attributes(fx[])# all attributes are shown
attributes(fx[1:5]) # only levels and class are shown
attributes(fx[1]) # only levels and class are shown
Thanks,
Heinz Tüchler
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[R] plot xy data with error bars
Dear All, I have x,y data with errors associated with both x and y. What would be the easiest way of creating plots with both x and y ? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] plot xy data with error bars
Dear All, I have x,y data with errors associated with both x and y. What would be the easiest way of creating plots with both x and y error bars? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Does R have EGARCH modeling function?
I just got 39 hits for RSiteSearch(egarch). The one that looks to me like it comes the closest to answering your question is http://finzi.psych.upenn.edu/R/Rhelp02a/archive/49245.html;. I don't know the definition of egarch, but I would expect that it would not be too hard to program if you know the definition. You could read the source code for some of the other GARCH-type functions, find one that's moderately close, and modify it. hope this helps. Spencer Graves Michael wrote: I've downloaded fSeries, but looks like it just has an interface to OX(TM) Garch Modeling Software,and that OX(TM) software package is not free. So where can I find an EGARCH function that is truely usable? Thanks a lot! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] general Gauss-Newton or support for NSUR: contemporaneously correlated non-linear models
Have you looked at nlme and the book Pinheiro and Bates (2000) Mixed-Effects Models in S and S-Plus (Springer)? I'm not familiar with NSUR, but it sound like nlme might handle it very well. hope this helps, Spencer Graves Thomas Wutzler wrote: Dear r-Help readers, 1) Is there support for NSUR in some R package yet? 2) Is there a general function of applying the Gauss-Newton or Marquard method, in which the function of calculating the partial derivatives can be specified by the user? Contemporaneously correlated non-linear models (NSUR) is a method to fit a system of non-linear equations. I want to use to fit several non-linear models at the same time, so that one the independent variable is alsway the sum of several other independent variable of other models. This procedure is described in detail by Parresol 2001. He used PROC MODEL by SAS to accomplish the task. If there is no support for NSUR yet, I would try to program a function by myself, although I only used very simple R-functins yet. The essential part is the fit the minimum of complex vector equation. This leads to my second question above. The function (as provided as an parameter) should accept a vector of model parameters and calculate the matrix of partial derivatives at this point in the parameter space. Thanks Thomas Parresol, B.R., 2001. Additivity of nonlinear biomass equations. Canadian Journal of Forest Research-Revue Canadienne De Recherche Forestiere 31, 865-878. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to make attributes persist after indexing?
On Thu, 2006-05-25 at 00:43 +0100, Heinz Tuechler wrote:
Thank you for your answer, Gabor. I will see, if I understood it.
Heinz
At 11:31 24.05.2006 -0400, Gabor Grothendieck wrote:
You could create your own child class with its own [ method.
[.myfactor - function(x, ...) {
attr - attributes(x)
x - NextMethod([)
attributes(x) - attr
x
}
gx - structure(fx, class = c(myfactor, class(fx)))
attributes(gx[1])
Heinz,
What Gabor has proposed is essentially what Frank Harrell does in the
Hmisc package (which I referenced), though in a more generic fashion
with respect to the attributes that are saved and reset.
By using:
gx - structure(fx, class = c(myfactor, class(fx)))
you are taking an object 'fx' and adding a child class attribute called
myfactor to it. So it is in effect an object that retains the
attributes of the original class of 'fx', plus the new class attribute
'myfactor'.
As a parallel, for example, consider that a square is a child class of a
rectangle. A square inherits all of the attributes of a rectangle, plus
the additional attribute that all four sides are of equal length. So for
example, if 'x' is a square, you might see:
x
[1] 4 4 4 4
attr(,class)
[1] squarerectangle
as compared to a rectangle 'y':
y
[1] 4 2 4 2
attr(,class)
[1] rectangle
Note the two class attributes for 'x', with 'square' preceding
'rectangle' in the order. The order is important, because in R, function
methods are dispatched based upon the class attributes in the order that
they appear in the vector.
So...back to 'gx'.
When the generic [ function is called with gx (ie. gx[1]), the first
class attribute 'myfactor' is picked up from 'gx'. Then, the method that
Gabor has presented, [.myfactor, is dispatched (executed). It is the
generic function [ with the specific class method defined by
.myfactor.
In that function, the first thing that takes place is that the
attributes of the 'x' argument are saved in 'attr'.
attr - attributes(x)
This would include any new attributes that you have defined and added,
such as labels and comments.
The next thing that happens is that the generic [ is now called again:
x - NextMethod([)
but this time, the method dispatched is based upon the Next entry in
the class vector. In this case, whatever the original class of 'fx' was,
which could be an atomic vector, a factor, a matrix or a data frame, for
example.
You can see the methods available for [ by using:
methods([)
The appropriate method for [ is then executed, resulting in 'x', which
is the subset version of 'gx'.
Then:
attributes(x) - attr
restores the original attributes saved in 'attr' to 'x'.
Then, finally, the 'x' object is returned to the calling environment.
So, in effect, you have created a new subset function [ that retains
your new attributes. The great thing about this, is that once the method
is defined in your working environment, all you have to do is to add the
newly associated class attribute to the objects you want to subset and R
does the rest transparently.
Thus:
# Add the new method
[.myfactor - function(x, ...) {
attr - attributes(x)
x - NextMethod([)
attributes(x) - attr
x
}
# Create fx
fx - factor(1:5, ordered = TRUE)
attr(fx, 'comment') - 'Comment for fx'
attr(fx, 'label') - 'Label for fx'
attr(fx, 'testattribute') - 'just for fun'
attributes(fx)
$levels
[1] 1 2 3 4 5
$class
[1] ordered factor
$comment
[1] Comment for fx
$label
[1] Label for fx
$testattribute
[1] just for fun
# Create gx, which is identical to fx, with the
# additional class attribute
gx - structure(fx, class = c(myfactor, class(fx)))
attributes(gx)
$levels
[1] 1 2 3 4 5
$class
[1] myfactor ordered factor # - Note change here
$comment
[1] Comment for fx
$label
[1] Label for fx
$testattribute
[1] just for fun
# Show the structure of fx[1]
# Note that other attributes are lost
str(fx[1])
Ord.factor w/ 5 levels 1234..: 1
# Show the structure of gx[1]
# Note that attributes are retained
str(gx[1])
Ord.factor w/ 5 levels 1234..: 1
- attr(*, comment)= chr Comment for fx
- attr(*, label)= chr Label for fx
- attr(*, testattribute)= chr just for fun
HTH,
Marc Schwartz
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Re: [R] general Gauss-Newton or support for NSUR: contemporaneously correlated non-linear models
On Wed, 24 May 2006, Spencer Graves wrote: Have you looked at nlme and the book Pinheiro and Bates (2000) Mixed-Effects Models in S and S-Plus (Springer)? I'm not familiar with NSUR, but it sound like nlme might handle it very well. hope this helps, Spencer Graves I don't think this is what Thomas needs. NSUR is I think nonlinear seemingly unrelated regressions. The SUR bit means that the response is multivariate. If he hasn't already, Thomas should look at systemfit: systemfit: Simultaneous Equation Estimation Package This package contains functions for fitting simultaneous systems of linear and nonlinear equations using Ordinary Least Squares (OLS), Weighted Least Squares (WLS), Seemingly Unrelated Regressions (SUR), Two-Stage Least Squares (2SLS), Weighted Two-Stage Least Squares (W2SLS), Three-Stage Least Squares (3SLS), and Weighted Three-Stage Least Squares (W3SLS). David Scott _ David Scott Department of Statistics, Tamaki Campus The University of Auckland, PB 92019 AucklandNEW ZEALAND Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000 Email: [EMAIL PROTECTED] Graduate Officer, Department of Statistics ASC/NZSA 2006: Statistical Connections The joint conference of the Statistical Society of Australia Inc. and the New Zealand Statistical Association, July 3--6, 2006 in Auckland. Go to: http://www.statsnz2006.com/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Ploting two datasets
Dears, I need your help. I have two sets A and B each one containing n lines and being composed by the atributes x and y. Three questions: 1) How can I plot A and B both on the same graphic? I need to show that the classes A and B are separable. In this way, I need to plot the points of A as *, and the points of B as +, for example. This would result two clouds. 2) How can I plot the result of question one using multi-levels, that is, a set of ellipsis that start on the mean of A and on the mean of B? 3) How can I trace an hiperplane separating the classes A and B just depicted in question 1? I appreciate any help. Thanks for all. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] boosting
Hi I am using boosting for a classification and prediction problem. For some reason it is giving me an outcome that doesn't fall between 0 and 1 for the predictions. I have tried type=response but it made no difference. Can anyone see what I am doing wrong? Screen output shown below: boost.model - gbm(as.factor(train$simNuance) ~ ., # formula + data=train, # dataset + # +1: monotone increase, + # 0: no monotone restrictions + distribution=gaussian, # bernoulli, adaboost, gaussian, + # poisson, and coxph available + n.trees=3000,# number of trees + shrinkage=0.005, # shrinkage or learning rate, + # 0.001 to 0.1 usually work + interaction.depth=3, # 1: additive model, 2: two-way interactions, etc. + bag.fraction = 0.5, # subsampling fraction, 0.5 is probably best + train.fraction = 0.5,# fraction of data for training, + # first train.fraction*N used for training + n.minobsinnode = 10, # minimum total weight needed in each node + cv.folds = 5,# do 5-fold cross-validation + keep.data=TRUE, # keep a copy of the dataset with the object + verbose=FALSE)# print out progress best.iter = gbm.perf(boost.model,method=cv) pred = predict.gbm(boost.model, test, best.iter) summary(pred) Min. 1st Qu. MedianMean 3rd Qu.Max. 0.4772 1.5140 1.6760 1.5100 1.7190 1.9420 Checked by AVG Free Edition. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] exporting long character vectors to dbf
I am trying to use RODBC, but still run into some problems. If I
understood correctly, I have to change the default type for char
vectors to text (or varchar(3000)). But when I try something like
library(RODBC)
channel - odbcConnect(supremo, uid=R, case=tolower)
setSqlTypeInfo(MySQL,
list(double=double,
integer=integer,
character=varchar(3000),
logical=varchar(5)))
Warning message:
number of items to replace is not a multiple of replacement length
I think there is a pair of brackets missing in setSqlTypeInfo here
typesR2DBMS[driver] - value[c(double, integer, character,
logical)]
by the way, is there any easy way to debug packages that does not
involve reapeatedly modifying, compiling and installing them again??
On 5/24/06, Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Wed, 24 May 2006, Mulholland, Tom wrote:
I assume this is (or was) a specification issue. I think write.dbf uses
the shapefile library (C not R library) so it applies to the use of
shapefiles and just happens to have been included in the foreign package
because it has a generic usefullness. (Is that a word?)
Actually, the width of the dbf field comes from the lines in write.dbf:
else if (is.character(x)) {
mf - max(nchar(x[!is.na(x)]))
precision[i] - min(max(nlen, mf), 254)
scale[i] - 0
so that's the limit (254, not 255). This is stated as a limitation of the
dbf format at
http://www.clicketyclick.dk/databases/xbase/format/data_types.html
so I don't think you can do what you want with .dbf.
Since I very rarely care about the elegance of my solutions, just that
they work, I would try saving the file in another format that you can
get open office to read and let it do the conversion rather than trying
to get R to do it. I'm sure OpenOffice can deal with straightforward
text files if that's a last resort.
Indeed. Using an ODBC driver (and RODBC) to write to the database might
be a good option.
Tom
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Eduardo Leoni
Sent: Wednesday, 24 May 2006 7:13 AM
To: r-help@stat.math.ethz.ch
Subject: [R] exporting long character vectors to dbf
Hi -
I need to export data to openoffice base, where one of the elements is
a long character vector (255 characters.) write.dbf exports it as
varchar, truncating the data. Any idea how to do this?
thanks,
-eduardo
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--
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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