similar to: loop correction needed

Dear useRs, I have a matrix with 38 columns and 365 rows. what i want to do is the following..... 1. subtracting from each column, first itself and then all the remaining columns. More precisely, from column number 1, i will first subtract itself(column 1) and then the remaining 37 columns. Afterwards i will take column number 2 and do the same. In this way i want to proceed till 38th column.

dear useRs, i have a list containing 16 matrices. i want to calculate the column mean of each of them. i tried >sr <- lapply(s,function(x) colMeans(x, na.rm=TRUE)) but i am getting the following error >Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric can it be done in any other way? and why i am getting this error?? thanks in advance.. elisa [[alternative

HI Eliza, You could do this: set.seed(15) mat1<-matrix(sample(1:800,124*12,replace=TRUE),nrow=12) # smaller dataset #Your codes ?list1<-list() ?for(i in 1:ncol(mat1)){ ? list1[[i]]<-t(apply(mat1,1,function(x) x[i]-x)) ? list1} ?x<-list1?? x<-matrix(unlist(x),nrow=12) x<-abs(x) ?y<-colSums(x, na.rm=FALSE) z<-matrix(y,ncol=10) ?z<-as.dist(z) ?z ?# ?? 1?? 2?? 3?? 4?? 5??

Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x)

Dear useRs, i want to split a matrix having 1116rows and 12 columns. i want to split that matrix into 36 small matrices each having 12 columns and 31 rows. The big matrix should be splitted row wise. which means that the first small matrix should copy values which are in first 31 rows and 12 columns of the big matrix. similarly 2nd small matrix should contain values from 32nd to 63rd row of the

Dear useRs, i have vectors of about 27 descriptors, each having 703 elements. what i want to do is the following 1. i want to do regression analysis of these 27 vectors individually, against a dependent vector, say B, having same number of elements.2. i would like to know best 10 regression results, if i do regression analysis of dependent vector against the random combination of any 4

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="")))

Dear UseRs,Extremely sorry for a basic question. I have a matrix of 19 rows and 365 columns. what i want to do is the following...First i want to leave out column number 1 and want to calculate the row wise mean of the remaining columns, which will obviously give me 365 values in one column, and then subtracting these values from the column i left out i.e. col=1 then i want to leave out column 2

Dear useRs, I have the following data "z" of two variables "x"(z[,1]) and "y"(z[,2]). > dput(z) structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66,

Dear R users,Sorry for such a basic question. I really need to know that how can i pick the indices of 5 lowest values from each row of a matrix with dimensions 12*12??Thank you very much in advance Elisa [[alternative HTML version deleted]]

Dear useRs, I have a very basic question. I have a distance matrix and i skipped the upper part of it deliberately. The distance matrix is 1000*1000. Then i used "min" command to extract the lowest value from that matrix. Now i want to know what is the location of that lowest element? More precisely, the row and column number of that lowest element. Thanks in advance elisa

Dear Eliza, Try this: Lines1<-readLines(textConnection("1911.01.01?????? 7.87 1911.01.02?????? 9.26 1911.01.03?????? 8.06 1911.01.04?????? 8.13 1911.01.05????? 12.90 1911.02.06?????? 5.45 1911.02.07?????? 3.26 1911.03.08?????? 5.70 1911.03.09?????? 9.24 1911.04.10?????? 7.60 1911.05.11????? 14.82 1911.05.12????? 14.10 1911.06.13?????? 7.87 1911.06.14?????? 9.26

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)

Dear R family,i am trying to plot and save, simultaneously, about 1000. the name of each plot is contained in "names" file. when i run this loop, i get an error. "Error in plot.new() : Unable to open file 'C:/R/SAVEHERE/myplot_Tak.jpg' for writing". could you please correct the mistake in the loop? >names<-(names(sp)) >for(a in seq_along(names)){ >mypath

Hi Elisa, Try this: date1<-format(seq.Date(as.Date("1991.1.1",format="%Y.%m.%d"),as.Date("1996.12.31",format="%Y.%m.%d"),by="day"),"%Y.%m.%d") ?length(date1) #[1] 2192 mat1<-matrix(c(.314,.314,.273,.273,.236,.236,.236,.236,.273,.314,.403,.314),ncol=1) res1<-

Dear R users, sorry for a very basic question. i wanted to ask that if your column are too much in number and you want to select the remaining columns, starting from column number 117. one way is to use usual command >q[ ,(117:2300)]. is there a way by which i can select the remaining columns starting from column number 117 without using the usual command or without giving the column number

Dear Elisa, Try this: el<- matrix(1:100,ncol=20) ?set.seed(25) ?el1<- matrix(sample(1:100,20,replace=TRUE),ncol=1) In the example you showed, there were no column names.? ?list(el[,sort(el1)[1:3]],sort(el1,index.return=TRUE)$ix[1:3]) #[[1]] ?# ?? [,1] [,2] [,3] #[1,]?? 31?? 61?? 71 #[2,]?? 32?? 62?? 72 #[3,]?? 33?? 63?? 73 #[4,]?? 34?? 64?? 74 #[5,]?? 35?? 65?? 75 # #[[2]] #[1] 9 5 3 A.K.

dear useRs, i created a distance matrix, of certain voltage values. unfortunately, i lost the original values. i am only left with the distance matrix that i created from those values. i wanted to ask that is there a way in R to reverse distance matrix for the original values? thanks in advance eliza [[alternative HTML version deleted]]

Dear R users,[IF THE FORMAT OF MY EMAIL IS NOT CLEAR, I HAVE ATTACHED A TEXT FILE FOR A CLEAR VIEW] I would like to use the R output file in Fortran. my file Is exactly in the following format. ELISA/BOTTO wATER INN FROM 1900 11 1 TO 1996 12 31 1901.11. 1 447.000 1901.11. 2 445.000 1901.11. 3 445.000 1924.11. 4 445.000 1924.11. 5 449.000 1924.11. 6 442.000 1924.11. 7

Dear useRs, 1. how to calculate single median value for two columns of a matrix? 2. i have a matrix of 16 columns and 365 rows, how to calculate median between columns 1 and 16, 2 and 16, 3 and 16, 4 and 16, 5 and 16 till 15th column. is there a loop command to do the said operation? regards eliza [[alternative HTML version deleted]]