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Hi, #working directory data1 #changed name data to data1.? Added some files in each of sub directories a1, a2, etc. ?indx1<- indx[indx!=""] lapply(indx1,function(x) list.files(x)) #[[1]] #[1] "a1.txt"??????? "mmmmm11kk.txt" #[[2]] #[1] "a2.txt"??????? "mmmmm11kk.txt" #[[3]] #[1] "a3.txt"??????? "mmmmm11kk.txt" #[[4]] #[1]

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

How can i replicate this in Linux: source(file.choose()) I've tried source(tkgetOpenFile()) but with no luck

Dear all, I am having a for loop that iterates a given number of measurements that I would like to split over 16 available cores. The code is in the following format inputForFunction<-expand.grid(caseList,filterList) for (i in c(1:length(inputForFunction$Var1))){#           FileList<-GetFileList(flag=as.vector(inputForFunction$Var1[i]));        print(sprintf("Calling the plotsCreate

Hi, Please dput() your example dataset. dat1<- read.table(text="a?? a?? a?? a??? a??? b?? b??? b??? c??? c?? c??? c 2? 4??? 5?? 2??? 7??? 2?? 2??? 6??? 3??? 7?? 9??? 3 3? 3?? 4?? 1???? 6??? 8?? 1??? 3??? 5??? 2??? 6??? 3",sep="",header=FALSE,stringsAsFactors=FALSE) library(reshape) ?dat2<-melt(as.data.frame(t(dat1)),id.var="V1")[,-2]

There are two line 216 and 218 Three development stages 5 WEEK (5W), 7W, 9W. Three tissue: Ca, Co, Pa each with 2 biological replicate. With two biological replicate. I want to do differential gene expression analysis using DESeq2 so I tried these codes after reading about DESeq2: ,my aim is to do the pairwise comparison. how to make colData and design formula. library("DESeq2")

Hello, I have RNAseq data, which I am trying to analyze with DESeq. My file (tab delimited .txt) appears to be correct: >head(myfile) VZ_w13 VZ_w14a VZ_w14b VZ_w15a VZ_w15b VZ_w16a ENSG00000253101 0 0 0 0 0 0 ENSG00000223972 0 0 0 0 0 0... However, when I try to analyze the data with >cds <-

Hello R users, Doing My PhD I collected count data which I believe is zero inflated. I have run a statistical model with lmer and family=poisson and got summary(model)@sigma=1 so I believe there is no overdispertion. I would like to use the fmr function from the 'gnlm' library but I just cannot figure out from the examples in the help page and some forums out there how to convert the lmer

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: >

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

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

If you wanted to do a t.test res1<-do.call(cbind,lapply(seq_len(nrow(m)),function(i) do.call(rbind,lapply(split(rbind(m[i,-1],n),1:nrow(rbind(m[i,-1],n))), function(x) {x1<- rbind(x,m[i,-1]); t.test(x1[1,],x1[2,])$p.value})))) ?res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i) c(c(tail(res1[seq(1,i,1),i],-1),1),res1[-c(1:i),i]))) ?attr(res2,"dimnames")<-NULL ?res2

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

Should I be concerned? [root@fs1 ~]# cobbler import --name=F18 --path=rsync://mirrors.kernel.org/fedora/releases/18/Fedora/x86_64/os --arch=x86_64 task started: 2013-01-18_062009_import task started (id=Media import, time=Fri Jan 18 06:20:09 2013) Found a redhat compatible signature: Packages adding distros creating new distro: F18-x86_64 creating new profile: F18-x86_64 associating repos

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,]

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

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