similar to: Pairwise comparison matrix elements

Update to KSR[T] Advisory #002 Our advisory contained one serious piece of mis-information. The latest version of ld.so that we tested (1.9.2) still appeared to be vulnerable to this overflow. We strongly recommend that anyone running linux install the patch distributed with the advisory, or wait for your vendor to release an updated ld.so package and install that as soon as possible. The

I am getting html output in my email body with spurious characters that cannot be rendered, so that my links are not handled properly. What am I not getting here? My email body contains: -----------------clip------------- <A href=mailto:&quot;<%=h(''support-t1rxLZ7CIXjQT0dZR+AlfA@public.gmane.org'')%>? Support&quot;>Support</A><br/> with any

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: >

----- KSR[T] Website : http://www.dec.net/ksrt E-mail: ksrt@dec.net ----- KSR[T] Advisory #005 Date: Dec 6, 1997 ID #: lin-dcrn-005 Operating System(s): Slackware 3.4 Affected Program: dillon crontab / crond ( dcron

----- KSR[T] Website : http://www.dec.net/ksrt E-mail: ksrt@dec.net ----- KSR[T] Advisory #003 Date: Aug 05, 1997 ID #: lin-cron-003 Operating System(s): Redhat linux 4.1, SuSE Linux 5.0, Slackware 3.3 Affected Program:

----- KSR[T] Website : http://www.dec.net/ksrt E-mail: ksrt@dec.net ----- KSR[T] Advisory #004 Date: Oct 6, 1997 ID

KSR[T] Advisory #002 Date: Jul 16, 1997 ID #: lin-ldso-002 Operating System(s): Linux Affected Program: ld.so / ld-linux.so Problem Description: ld.so is the run-time linker used by dynamically linked executables(a.out). Inside the error reporting

I am finding with the patch that all of the Polyhedron 2005 benchmarks pass on i686-apple-darwin9. Could someone clarify the regression rules for releases? Not building a secondary language on a primary target is usually considered a P1 regression for FSF gcc. Not doing so here gives one the impression that llvm.org isn't playing by the same rules. No one is ever going to want to use these

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

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 =

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, I have problem here : I'm using garchFit from fSeries package, here is part of the script : > data <- read.table("d:/data.txt") > a <- garchFit(~garch(1,1),ts(data)) I also attached the file here. In my experience, I got my R not responding. I also tried with > a <- garchFit(~garch(1,1),ts(data*10)) and it's worked. I

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

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

I'm trying to plot the spatial distribution of the precipitation within an area. However, I think some thing wrong with the way to create the matrix for the precipitation data, so it came out lots of stripes. Could somebody help to see what's the problem? Here's the code: library(maptools);gpclibPermit() KSR <-

I need to completely remove row.names from a dataframe. Are there other ways to remove them (and not anything else) besides: mydataframe<-data.frame(mydataframe, row.names=NULL) I realize that this doesn't really remove the row.names; it merely replaces the current row.names vector with the numbers 1..nrow (in quotes). ===================== Dr. Marc R. Feldesman Professor and

I need to completely remove row.names from a dataframe. Are there other ways to remove them (and not anything else) besides: mydataframe<-data.frame(mydataframe, row.names=NULL) I realize that this doesn't really remove the row.names; it merely replaces the current row.names vector with the numbers 1..nrow (in quotes). ===================== Dr. Marc R. Feldesman Professor and

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 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

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