similar to: call a variable from outside of for loop

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 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: 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 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

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: #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}))

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 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,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 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)

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 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

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 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 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, i have certain commands for some operations in R. They are good if you have a small dataset but my dataset, apart from what i used in the recent past, is prety large. I want to convert these massive sets of commands into a simple loop. Your help is required on it thanks in advance eliza kindly note: "e" is matrix whose each column has to be executed into a distance vector