similar to: for loop not working

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 R users, i have a matrix with 31 rows and 444 columns and i want to extract every 37th column of that matrix starting from 1. more precisely i want to select columns 1, 38,75, 112 and so on. then doing the same by starting from column number 2(2,39,76,113.......). i know that there is a manual way of doing it but i wanted to make it more quickly as i have fairly large data to dealth with.

Hi, Try this: #mat1 is the data res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i) {new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j) {x1<-rbind(mat1[i,],mat1[j,]); x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2,3])*abs(x1[1,7]-x1[2,7]))+(abs(x1[1,4]-x1[2,4])*abs(x1[1,8]-x1[2,8]))}));new1}))

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,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

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<-

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??

Dear UseRs, MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation. t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T)) Then "1215450" values of "t"(one for

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

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, 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]]

Dear useRs, I have a matrix in the following form [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] 1 1 3 2 3 1 1 2 3 3 2 and following is my desired output (combining the column headers, having same values). a<-1,2,6,7 b<-3,5,9,10 c<-4,8,11 Thanks in advance Elisa [[alternative HTML

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 Elisa, I hope this is what you wanted. dat1<-read.csv("peaks.csv",sep=",") #Subset dat2<-dat1[1:5,] res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i) do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))), function(x) {x1<-rbind(dat2[i,],x);

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

Dear useRs, With the following set of command i managed to create the copula density ================================================ >library(MASS) >library(evd) >X<-matrix(sample(1:3000), ncol=2) >U=cbind(rank(X[,1])/(nrow(X)+1),rank(X[,2])/(nrow(X)+1)) > mat1=kde2d(U[,1],U[,2],n=150) image(mat1$x,mat1$y,mat1$z,col= rev(heat.colors(1500)),xlab="",ylab="")

Dear useRs, my question could be very basic for which i apologize in advance. Each column of a matrix with dimensions 365 rows and 37 columns was drawn against another matrix of dimensions 365 rows and 1 column. with that i was able to draw 37 curves on the same axis. now i want to draw an average curve of these 37 curves on the same axis in such a way that all the curves (average and 37

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]]

Hi all, I want to execute a loop of a program: for (u in Timeframemin:Timeframe){} Imagine that Timeframemin<-10 Timefram<-10000 Is it posible to execute the loop but only proving from 10 to 10000 but jumping 10 each time, for example, execute for 10,20,30.....to Timeframe. Other question is, when a program is "heavy" and has a lot of loops to execute (how can I know where

Dear UseRs, i have a matrix of 365 rows and 444 columns. i drew each column of this matrix against the number of days in a year, which are obviously 365. now i have 444 curves and i want to Use Fourier analysis for the approximation of the average values. does anyone know how to do it? any help in this regards will be deeply appreciated... regards eliza [[alternative HTML version