similar to: Reading a Vector into R

Hi, In the example you showed: m1<- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or ?m1[!upper.tri(m1)] <-? rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1<- c(3,4,5) ?m2<- matrix(0,length(vec1),max(vec1)) ?indx <- cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]<- 1 ?m2 #???? [,1] [,2] [,3] [,4] [,5]

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))

Hi, set.seed(45) test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data apply(test1,2,function(x) sprintf("%.1f",median(x))) #columnA columnB # "44.5"? "10.2" par(mfrow=c(1,2)) lapply(test1,function(x) {b<-

Dear Elisa, Try this: vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) vec2<-vec1[1:26] names(vec2)<-LETTERS[1:26] label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i)

Hi, Try: ?vec1<-c(1,1,1,1,1,1,1,1,1) if(all(vec1==1)) "xxxxxxxxx" else? "yyyyyyyyyyy" #[1] "xxxxxxxxx" ?vec2<-c(rep(1,4),2) ?if(all(vec2==1)) "xxxxxxxxx" else? "yyyyyyyyyyy" #[1] "yyyyyyyyyyy" #or if(length(unique(vec1))==1) "xxxxxxxxx" else? "yyyyyyyyyyy" #[1] "xxxxxxxxx" ? if(length(unique(vec2))==1)

Hi, You could try: A<- matrix(unlist(read.table(text=" 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 ",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4

Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1

Hi, I have a following data set: id event time (in sec) 1 add 1373502892 2 add 1373502972 3 delete 1373502995 4 view 1373503896 5 add 1373503996 ... I'd like to add new column "time on task" which is time elapsed between two events (id2 - id1...). What would be the best approach to do that? Thanks, Srecko [[alternative HTML

Hello. Sorry I am not sure this question should go to llvm or mesa3d-dev mailing list, so I post it to both. I am writing a llvm backend for a modern graphics processor which has a ISA very similar to that of Direct 3D. I am reading the code in Gallium-3D driver in a mesa3d branch, which converts the shader programs (TGSI tokens) to LLVM IR. For the shader instruction also found in LLVM IR,

Alex wrote: > Hello. > > Sorry I am not sure this question should go to llvm or mesa3d-dev mailing > list, so I post it to both. > > I am writing a llvm backend for a modern graphics processor which has a ISA > very similar to that of Direct 3D. > > I am reading the code in Gallium-3D driver in a mesa3d branch, which > converts the shader programs (TGSI tokens) to

I would like to know whether there is a faster way to do the below operation (updating vec1). My objective is to update the elements of a vector (vec1), where a particular element i is dependent on the previous one. I need to do this on vectors that are 1 million or longer and need to repeat that process several hundred times. The for loop works but is slow. If there is a faster way, please let

Hi, Try: vec1<- "mue#d/sjbijk at ruepvnvbnceiicrpgxkgcyl@keduhqvqi/ubudvxopddpfddgitrynzshzdcwgneyffrkpbxwilwqngrsals#geqmtkcpkp/qecgdfa#uag" library(seqinr) ?res<-lapply(0:4,function(i) lapply(2:5,function(j) splitseq(s2c(gsub("[#@/]","",vec1)),word=j,frame=i))) #or library(stringr) res1<-lapply(0:4,function(i) lapply(2:5,function(j)

Dear expeRts, I know I can't paste expressions in the normal way, but I just couldn't figure out how to get the following (I want to paste a character vector to an expression vector) right with bquote() or substitute. vec1 <- c("a", expression(tilde(b)), "c") vec2 <- c("1", "2", "3") main <- as.expression(paste(vec1, vec2))

I'm not sure this list is the right place for this thing. I noticed some erratic behaviour in sammon(). Running sammon on two nearly identical sets of data results in very different results. Below is an example. I create an initial configuration with cmdscale() and store it into 'vec1'. I write this to file, and read it back in again to 'vec2'. According to cor() on the three

Hello R specialists, I have a base vector called vBase with 102 elements. I have another vector vec1 which elements are only part of vBase but is shorter. I transform vec1 so I get a vector with the same size as vBase and with each common element on the same indexed row. If a value is missing in vec1 then I put a Na like this: vec1 vBase Amsterdam Amsterdam Na

Dear all, It was a pleasure to meet you at Iowa State University. Two days ago I submitted two experimental packages to CRAN (hope it will be there soon): rindex: quick indexing of large objects (currently only character, see ?index) regtest: some first support for automated regression testing (heavily used in \dontshow{} section of ?index) With rindex you can for example i <-

Hello R-users, I need to create a vector inserting an 1 after each value of another vector. For example: vec1<-c(2,3,4) I need to create a vector with the values 2,1,3,1,4 Does anyone know how create this vector without loops (vec1 could have 1000 elements) Thank you, Juan -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hallo all Please help me. I am lost and do not know what is the problem. I have a factor called kvartaly. > attributes(kvartaly) $levels [1] "1Q.04" "2Q.04" "3Q.04" "4Q.04" "1Q.05" "2Q.05" "3Q.05" "4Q.05" $class [1] "factor" > mode(kvartaly) [1] "numeric" > str(kvartaly) Factor w/ 8

Dear all, I have a big problem: - I got two matrices, A and B - A shows identifies the value of B, however the values of B must be summed - For instance, 1 1 2 2 2 2 1 1 gives matrix a 3 4 2 1 1 1 2 2 gives matrix b Now the result for the value 1 would be 7 4 which are the rowsums of the values of matrix B given that matrix A has the value 1. How can I do this automatically? I

There seems to be an issue in computing rank correlations with missing values present. I think this comes from the way rank() function works but I am not sure how to go about this. Rank function places missing values at the end by default thus skewing the rank relationship between two vectors: Example: R : Copyright 2003, The R Foundation for Statistical Computing Version 1.8.1 (2003-11-21),