similar to: alternative to leaps command

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 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

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 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, 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 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 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

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 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,

[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,

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 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 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, 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 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 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 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

Dear useRs, While using print command in "for" loop, i designated a variable being printed by "e". Although the output was shown inside the loop but when i tried to call the variable outside the loop it only gave the first row, where as it should have 35 rows as it showed inside loop.The command which i used in the loop is e<-print(sum(abs(b-m[,i]))) Kindly help me on it..

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)

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}))