similar to: [LLVMdev] polyhedron 2005 results for llvm svn

Dear all, Hi, I have the problems about converting the matrix to adjacency matrix.Here's my example, a b c d e fa 1.0000000 0.4048823682 0.1228531 0.49046991 0.4945158868 0.307443317b 0.4048824 1.0000000000 0.4367475 0.96949219 0.0007378596 0.560747765c 0.1228531 0.4367474719 1.0000000 0.40037341 0.3157538204

Dear members. I am having problems to understand the kmeans- results in R. I am applying kmeans-algorithms to my big data file, and it is producing the results of the clusters. Q1) Does anybody knows how to find out in which cluster (I have fixed numberofclusters = 5 ) which data have been used? COMMAND (kmeans.results <- kmeans(mydata,centers =5, iter.max= 1000, nstart =10000)) Q2) When I

Should I submit this as a bug report? --- reg-tests-3.Rout.save Thu Jul 3 09:55:40 2003 +++ reg-tests-3.Rout Sun Nov 23 13:10:57 2003 @@ -1,17 +1,18 @@ -R : Copyright 2003, The R Development Core Team -Version 1.8.0 Under development (unstable) (2003-07-03) +R : Copyright 2003, The R Foundation for Statistical Computing +Version 1.8.1 (2003-11-21), ISBN 3-900051-00-3 R is free software and

Dear all, May be question seems trivial for most of the R users, but really at least for me, this comes out to be very problematic. Suppose I have the following data: > r [1] -0.0008179960 -0.0277968529 -0.0105731583 -0.0254050262 0.0321847131 0.0328170674 [7] 0.0431894392 -0.0217614918 -0.0218366946 0.0048939739 -0.0012212499 0.0032533579 [13] -0.0081533269 -0.0098725606

Hello, I am trying to compare two histograms using barplot. the idea is to plot the histograms as pairs of columns side by side for each x value. I was able to do it using barplot before but I can't remember now for the life of me now how I did it in the past: > d [,1] [,2] -37.5 0.0000000000 2.789396e-05 -32.5 0.0001394700 5.578801e-05 -27.5 0.0019804742

Dear R experts, I have a Matlab code which I am translating to R in order to examine and enhance it. First of all, I need to reproduce in R the results which were already obtained in Matlab (to make sure that everything is correct). There are some matrix manipulations and '\' operation among them in the code. I have the following data frame > ABS.df Pro syn

Has anyone come across the right combinations to print a limited number of digits? My trial and error approach is taking too much time. Here is what I have tried: > op <- options() > a <- c(1e-10,1,2,3,.5,.25) > names(a) <- c("A", "B", "C", "D", "E", "F") > # default > a A B C D

Hi!, I'm really hoping someone out there will be able to help me. I recently started my MSc dissertation on Population Projection Matrices, which has been going well until now. I am trying to set-up a general script that does a pairwise comparison of all elements in my matrices. So for example, given that I have the following matrix S: > S [,1] [,2] [,3] [1,]

Dear All, I have data some thisng like this : > data <- read.csv(file='ipsample.csv',sep=',' , header=TRUE) > data State Jan Feb Mar Apr May Jun 1 AAA 1 1 0 2 2 0 2 BBB 1298 1195 1212 1244 1158 845 3 CCC 0 0 0 1 2 1 4 DDD 5 11 17 15 10 9 5 EEE 18 28 27 23 23 16 6 FFF 68 152 184 135 111

Dear all, I try to summarise my data per category using aggregate, but for some reason I get the error message "sum" not meaningful for factors even though my vector is numeric. The data set is shown below. Could someone please give a hint. Thanks in advance! Sincerely, Tord > names(test) [1] "ObjektID" "tallstubbyta" > is.factor(test$ObjektID);

Hi, I have the binned data (observed and generated from model)  that I would like to test using the anderson-darling goodness of fit test.  But I'm not sure which package in R to use. I tried ad.test(...) but it does not recognise the test by Vito Ricci in FITTING DISTRIBUTIONS WITH R   > ad.test(hist_hume_beec[,1],hist_hume_beec[,2]) Error: could not find function "ad.test"

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >

Hi, The following works fine: > f [1] 0.08 0.03 0.04 > A2 [,1] [,2] [,3] [1,] 0.000000e+00 0.0000000000 0.000000e+00 [2,] 0.000000e+00 0.0000000000 0.000000e+00 [3,] 2.999651e-03 0.0009094342 1.945708e-03 [4,] 4.124431e-05 0.0001360390 1.203345e-05 [5,] 3.027932e-04 0.0000412920 4.668090e-04 [6,] 0.000000e+00 0.0000000000 0.000000e+00 > b2 [1] 0 0 0 0 0 0

Hi Everyone, I am uncertain that I am writing the contrast statements correctly. Basically, I'm unsure when to use a -1 and a 1 when writing the contrasts. Specifically I am interested in comparing the slopes between different temperature regimes. Temperature is therefore a factor. Time and percent are numerical. Using the gmodels package I made the following model:

I want to get data for a set of ticker symbols and compute the daily return of the adjusted close using quantmod, and then scatterplot returns using pairs(). The following gets data for the list of tickers: tickers <- c("SHY","TLT","SPY","IWM","GLD","IEV","ILF","EWJ","EPP","SAF","ASA")

Dear R-help, I am having a problem with the interpretation of result from validate.cph in the Design package. My purpose is to fit a cox model and validate the Somer's Dxy. I used the hypothetical data given in the help manual with modification to the cox model fit. My research problem is very similar to this example. This is the model without stratification: > library(Design) > f1

Hi, directory<- "/home/arunksa111/data.new" #first function filelist<-function(directory,number,list1){ setwd(directory) filelist1<-dir(directory) direct<-dir(directory,pattern = paste("MSMS_",number,"PepInfo.txt",sep=""), full.names = FALSE, recursive = TRUE) list1<-lapply(direct, function(x) read.table(x,header=TRUE, sep =

I'm trying to use tapply to output means and SD or SE for my data but seem to be limited by how many times I can subset it. Here's a snippet of my data > stems353[1:10,] Time DataSource Plot Elevation Aspect Slope Type Species SizeClass Stems 1 Modern Cameron 70F221 1730 ESE 20 Conifer ABCO Class1 3 2 Modern Cameron 70F221 1730

Hello R-users. I haven't use R for a life time and this might be trivial - I hope you do not mind. I have a questions about arguments in the Glm-function. There seems to be something that I cannot cope. The basics are ok: > y <- as.double(rnorm(20) > .5) > logit.model <- glm(y ~ rnorm(20), family=binomial(link=logit), trace = TRUE) Deviance = 28.34255 Iterations - 1

Dear All, I?m using R package ?msm? to fit a multi state model to infection history data (counts of infections per month upto diagnosis of a particular disease (sink state is state 11). The observed transitions are as follows: to from 1 2 3 4 5 6 7 8 10 11 1 35192 3806 899 233 46 11 3 0 1 534 2 3801 790 249 69 15