similar to: loop for column substraction of a matrix

Displaying 20 results from an estimated 2000 matches similar to: "loop for column substraction of a matrix"

2013 Jan 10
1
merging command
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??
2012 Oct 26
3
regression analysis in R
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
2013 Feb 19
1
data format
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)
2012 Nov 27
3
loop command to matrix
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
2013 Jan 03
5
splitting matrices
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
2013 Jan 02
4
list of matrices
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
2013 Apr 15
3
Indices of lowest values in matrix
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]]
2012 Dec 25
2
for loop not working
dear R family,i have a matrix of 444 columns. what i want to do is the following. 1. starting from column 1 i want to select every 37th column on the way. more precisely i want to select column 1, 38,75,112,149 and so on. 2.starting from column 2, i again want to select every 37th column. which means 2,39,76,113,150 and so on. similarly starting from 3 till 37th column. i have tried following loop
2013 Feb 13
3
date and matrices
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<-
2013 Feb 17
6
histogram
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="")))
2012 Oct 15
2
median between matrix column
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]]
2013 Apr 24
2
Distance matrices Combinations
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
2013 Feb 27
2
matrix multiplication
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}))
2013 Jan 02
2
loop correction needed
Dear useRs, i have a data frame with 16 lists in it. each list has variable number of lines. i want to create a loop which will start deleting every 32nd line in each list, till the end of each list. more precisely if a list has 200 rows i want to delete row number 32, 64, 96 and so on...for that i created the followng loop. >e<-lapply(seq(1),function(i)
2013 Mar 06
3
combining column having same values
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
2013 Feb 17
1
addition in the initial question
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)
2012 Dec 24
2
colmeans not working
[text file is also attached in case you find the format of email difficult to understand] Dear useRs,You must all the planning for the christmas, but i am stucked in my office on the following issue i had a file containg information about station name, year, month, day, and discharge information. i opened it by using following command > dat1<-read.table("EL.csv",header=TRUE,
2013 Mar 22
3
Distance calculation
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);
2013 Oct 19
3
bold dot size and name in plot
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,
2013 Feb 15
2
data formatting
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