similar to: column selection

Displaying 20 results from an estimated 20000 matches similar to: "column selection"

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
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="")))
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)
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 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 Feb 11
2
FORMAT EDITING
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
2013 Apr 15
2
nearest stations in distance matrix
Dear R-user, Is there a way in R to locate the nearest 5 indices to a station, based on distances in a distance matrix. In other words i want to have nearest stations based on the distances in the matrix. The distance matrix, i have, has dimension 44*44. Thankyou very much in advance Elisa [[alternative HTML version deleted]]
2013 Feb 19
1
data format
Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x)
2013 Jan 03
5
splitting matrices
Dear useRs, i want to split a matrix having 1116rows and 12 columns. i want to split that matrix into 36 small matrices each having 12 columns and 31 rows. The big matrix should be splitted row wise. which means that the first small matrix should copy values which are in first 31 rows and 12 columns of the big matrix. similarly 2nd small matrix should contain values from 32nd to 63rd row of the
2013 Apr 15
3
Indices of lowest values in matrix
Dear R users,Sorry for such a basic question. I really need to know that how can i pick the indices of 5 lowest values from each row of a matrix with dimensions 12*12??Thank you very much in advance Elisa [[alternative HTML version deleted]]
2013 Apr 25
2
connecting matrices
Dear Elisa, Try this: el<- matrix(1:100,ncol=20) ?set.seed(25) ?el1<- matrix(sample(1:100,20,replace=TRUE),ncol=1) In the example you showed, there were no column names.? ?list(el[,sort(el1)[1:3]],sort(el1,index.return=TRUE)$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.
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
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 Jan 11
3
locating element in distance matrix
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
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 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 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 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
2012 Oct 05
3
loop for column substraction of a matrix
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.