Re: [R] How to Store the executed values in a dataframe rle function
Thanks,It does work for the sample data. When I use it for my actual data it is throwing this error Error in data.frame(Sample = .samp, Chr = .set$Chr[1L], Start = min(.set$Start), : arguments imply differing number of rows: 1, 0 I am not able to understand Why I am getting this? waiting for your reply, Thanks, Suji On Wed, Sep 28, 2011 at 4:15 PM, jim holtman jholt...@gmail.com wrote: I only used textConnection for the sample data. Just put your file name in the read.table; e.g., x-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric')) as you have in your email. I used 'x' in my code, so I replaced your 'm' with 'x'. Try it and see if it works; no reason it shouldn't. On Wed, Sep 28, 2011 at 3:03 PM, viritha k virith...@gmail.com wrote: Hi Jim, Thanks for the reply, ok but I dont want to use textConnection and paste each line but want the input to be read from a file like m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric'). So how do I incorporate that in your code. Thanks, Suji On Wed, Sep 28, 2011 at 2:40 PM, jim holtman jholt...@gmail.com wrote: The solution that I sent will handle the 150 different samples; just list the column names in the argument to the top 'lapply'. You don't need the 'rle' in my approach. On Wed, Sep 28, 2011 at 2:13 PM, viritha k virith...@gmail.com wrote: Hi, This is the code that I wrote for 3 samples: code: m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric','numeric')) s-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) names(s)=c(Values,Probes) c-data.frame(Sample=character(s$Probes),Chr=character(s$Probes),Start=numeric(s$Probes),End=numeric(s$Probes),Values=numeric(s$Probes),Probes=numeric(s$Probes),stringsAsFactors=FALSE) G=1 n=4 for(i in 1:length(s$Probes)){ + if(G==1){c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else if((G-1) length(m$Sample1)) { + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:(G+s$Probes[i]-1)]) + c[i,3]-min(m$Start[G:(G+s$Probes[i]-1)]) + c[i,4]-max(m$End[G:(G+s$Probes[i]-1)]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else { + G=1 + n=n+1 + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])}} c Sample ChrStart End Values Probes 1 Sample1 chr2 9896633 14404502 0 4 2 Sample1 chr2 14421718 16048724 -0.43 4 3 Sample1 chr2 37491676 37703009 0 2 4 Sample2 chr2 9896633 9896690 0 2 5 Sample2 chr2 14314039 16048724 -0.35 6 6 Sample2 chr2 37491676 37703009 0 2 7 Sample3 chr2 9896633 14314098 0 3 8 Sample3 chr2 14404467 16031769 0.32 3 9 Sample3 chr2 16036178 37491735 0.45 3 10 Sample3 chr2 37702947 37703009 0 1 The problem that I am facing is for expanding rle function for values and probes. Defintely your code looks simpler, but I would like to read the file by just giving the name of the file as written in my code because my original file contains 150 samples,but how to use lapply or rle function for 150 such samples, if my file contain 150 samples similiar to sample1 and sample2. waiting for your reply, Thanks, Suji On Wed, Sep 28, 2011 at 11:37 AM, jim holtman jholt...@gmail.com wrote: Here one approach: x - read.table(textConnection(Chr start end sample1 sample2 + chr2 9896633 9896683 0 0 + chr2 9896639 9896690 0 0 + chr2 14314039 14314098 0 -0.35 + chr2 14404467 14404502 0 -0.35 + chr2 14421718 14421777 -0.43 -0.35 + chr2 16031710 16031769 -0.43 -0.35 + chr2 16036178 16036237 -0.43 -0.35 + chr2 16048665 16048724 -0.43 -0.35 + chr2 37491676 37491735 0 0 + chr2 37702947 37703009 0 0), header = TRUE, as.is = TRUE) closeAllConnections() result - lapply(c('sample1', 'sample2'), function(.samp){ + # split by breaks in the values + .grps - split(x, cumsum(c(0, diff(x[[.samp]]) != 0))) + + # combine the list of dataframes + .range - do.call(rbind, lapply(.grps, function(.set){ + # create a dataframe of the results + data.frame(Sample = .samp +, Chr = .set$Chr[1L] +, Start =
Re: [R] How to Store the executed values in a dataframe rle function
Here one approach: x - read.table(textConnection(Chr start end sample1 sample2 + chr2 9896633 9896683 0 0 + chr2 9896639 9896690 0 0 + chr2 14314039 14314098 0 -0.35 + chr2 14404467 14404502 0 -0.35 + chr2 14421718 14421777 -0.43 -0.35 + chr2 16031710 16031769 -0.43 -0.35 + chr2 16036178 16036237 -0.43 -0.35 + chr2 16048665 16048724 -0.43 -0.35 + chr2 37491676 37491735 0 0 + chr2 37702947 37703009 0 0), header = TRUE, as.is = TRUE) closeAllConnections() result - lapply(c('sample1', 'sample2'), function(.samp){ + # split by breaks in the values + .grps - split(x, cumsum(c(0, diff(x[[.samp]]) != 0))) + + # combine the list of dataframes + .range - do.call(rbind, lapply(.grps, function(.set){ + # create a dataframe of the results + data.frame(Sample = .samp +, Chr = .set$Chr[1L] +, Start = min(.set$start) +, End = max(.set$end) +, Values = .set[[.samp]][1L] +, Probes = nrow(.set) +) + })) + }) # put the list of dataframes together result - do.call(rbind, result) result Sample ChrStart End Values Probes 0 sample1 chr2 9896633 14404502 0.00 4 1 sample1 chr2 14421718 16048724 -0.43 4 2 sample1 chr2 37491676 37703009 0.00 2 01 sample2 chr2 9896633 9896690 0.00 2 11 sample2 chr2 14314039 16048724 -0.35 6 21 sample2 chr2 37491676 37703009 0.00 2 On Mon, Sep 26, 2011 at 10:30 AM, sujitha virith...@gmail.com wrote: Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0 This is the output that I am expecting: Sample Chr Start End Values Probes sample1 chr2 9896633 14404502 0 4 sample1 chr2 14421718 16048724 -0.43 4 sample1 chr2 37491676 37703001 0 2 sample2 chr2 9896633 9896690 0 2 sample2 chr2 14314039 16048724 -0.35 6 sample2 chr2 37491676 37703009 0 2 Here the Chr value is same but can be any other value aswell so unique among the similar values. The Start for the first line would be the least value until values are similiar (4) then the end would be highest value. The values is the unique value among the common values and probes is number of similar values. Code: m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric')) #reading the test file s-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]])) # to get the last 2 columns names(s)=c(Values,Probes) G=1 for(i in 1:length(s$Probes)){ + if(G==1){first-unique(m$Chr[G:s$Probes[i]]) + second-min(m$Start[G:s$Probes[i]]) + third-max(m$End[G:s$Probes[i]]) + c-cbind(first,second,third,s$Values[i],s$Probes[i]) + print (c) + G=(G+s$Probes[i])} + else if((G-1) length(m$Sample1)) { + first-unique(m$Chr[G:(G+s$Probes[i]-1)]) + second-min(m$Start[G:(G+s$Probes[i]-1)]) + third-max(m$End[G:(G+s$Probes[i]-1)]) + c-cbind(first,second,third,s$Values[i],s$Probes[i]) + print (c) + G=(G+s$Probes[i])} + else { + G=1 + first-unique(m$Chr[G:s$Probes[i]]) + second-min(m$Start[G:s$Probes[i]]) + third-max(m$End[G:s$Probes[i]]) + c-cbind(first,second,third,s$Values[i],s$Probes[i]) + print (c) + G=(G+s$Probes[i])} + } so the output is: first second third [1,] chr2 9896633 14404502 0 4 first second third [1,] chr2 14421718 16048724 -0.43 4 first second third [1,] chr2 37491676 37703009 0 2 first second third [1,] chr2 9896633 9896690 0 2 first second third [1,] chr2 14314039 16048724 -0.35 6 first second third [1,] chr2 37491676 37703009 0 2 I get almost the required output but just need 3 modifications to this code: 1) Since this is just a small part of the file (with 2 samples), but my actual file has 150 samples, so how do I write rle function for that? 2) How do I store all the executed c values as a dataframe (here I am just printing the values)? 3) How do I include sample name in execution? Waiting for your reply , Thanks, Suji -- View this message in context: http://r.789695.n4.nabble.com/How-to-Store-the-executed-values-in-a-dataframe-rle-function-tp3843944p3843944.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve?
Re: [R] How to Store the executed values in a dataframe rle function
The solution that I sent will handle the 150 different samples; just list the column names in the argument to the top 'lapply'. You don't need the 'rle' in my approach. On Wed, Sep 28, 2011 at 2:13 PM, viritha k virith...@gmail.com wrote: Hi, This is the code that I wrote for 3 samples: code: m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric','numeric')) s-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) names(s)=c(Values,Probes) c-data.frame(Sample=character(s$Probes),Chr=character(s$Probes),Start=numeric(s$Probes),End=numeric(s$Probes),Values=numeric(s$Probes),Probes=numeric(s$Probes),stringsAsFactors=FALSE) G=1 n=4 for(i in 1:length(s$Probes)){ + if(G==1){c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else if((G-1) length(m$Sample1)) { + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:(G+s$Probes[i]-1)]) + c[i,3]-min(m$Start[G:(G+s$Probes[i]-1)]) + c[i,4]-max(m$End[G:(G+s$Probes[i]-1)]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else { + G=1 + n=n+1 + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])}} c Sample Chr Start End Values Probes 1 Sample1 chr2 9896633 14404502 0 4 2 Sample1 chr2 14421718 16048724 -0.43 4 3 Sample1 chr2 37491676 37703009 0 2 4 Sample2 chr2 9896633 9896690 0 2 5 Sample2 chr2 14314039 16048724 -0.35 6 6 Sample2 chr2 37491676 37703009 0 2 7 Sample3 chr2 9896633 14314098 0 3 8 Sample3 chr2 14404467 16031769 0.32 3 9 Sample3 chr2 16036178 37491735 0.45 3 10 Sample3 chr2 37702947 37703009 0 1 The problem that I am facing is for expanding rle function for values and probes. Defintely your code looks simpler, but I would like to read the file by just giving the name of the file as written in my code because my original file contains 150 samples,but how to use lapply or rle function for 150 such samples, if my file contain 150 samples similiar to sample1 and sample2. waiting for your reply, Thanks, Suji On Wed, Sep 28, 2011 at 11:37 AM, jim holtman jholt...@gmail.com wrote: Here one approach: x - read.table(textConnection(Chr start end sample1 sample2 + chr2 9896633 9896683 0 0 + chr2 9896639 9896690 0 0 + chr2 14314039 14314098 0 -0.35 + chr2 14404467 14404502 0 -0.35 + chr2 14421718 14421777 -0.43 -0.35 + chr2 16031710 16031769 -0.43 -0.35 + chr2 16036178 16036237 -0.43 -0.35 + chr2 16048665 16048724 -0.43 -0.35 + chr2 37491676 37491735 0 0 + chr2 37702947 37703009 0 0), header = TRUE, as.is = TRUE) closeAllConnections() result - lapply(c('sample1', 'sample2'), function(.samp){ + # split by breaks in the values + .grps - split(x, cumsum(c(0, diff(x[[.samp]]) != 0))) + + # combine the list of dataframes + .range - do.call(rbind, lapply(.grps, function(.set){ + # create a dataframe of the results + data.frame(Sample = .samp + , Chr = .set$Chr[1L] + , Start = min(.set$start) + , End = max(.set$end) + , Values = .set[[.samp]][1L] + , Probes = nrow(.set) + ) + })) + }) # put the list of dataframes together result - do.call(rbind, result) result Sample Chr Start End Values Probes 0 sample1 chr2 9896633 14404502 0.00 4 1 sample1 chr2 14421718 16048724 -0.43 4 2 sample1 chr2 37491676 37703009 0.00 2 01 sample2 chr2 9896633 9896690 0.00 2 11 sample2 chr2 14314039 16048724 -0.35 6 21 sample2 chr2 37491676 37703009 0.00 2 On Mon, Sep 26, 2011 at 10:30 AM, sujitha virith...@gmail.com wrote: Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0 This is the output that I am expecting: Sample Chr Start End Values Probes sample1 chr2 9896633 14404502 0 4 sample1 chr2 14421718 16048724 -0.43 4 sample1 chr2 37491676 37703001 0 2 sample2 chr2 9896633 9896690 0 2 sample2 chr2 14314039 16048724 -0.35 6 sample2 chr2 37491676 37703009 0 2 Here the Chr value is same but
Re: [R] How to Store the executed values in a dataframe rle function
Hi, This is the code that I wrote for 3 samples: code: m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric','numeric')) s-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) names(s)=c(Values,Probes) c-data.frame(Sample=character(s$Probes),Chr=character(s$Probes),Start=numeric(s$Probes),End=numeric(s$Probes),Values=numeric(s$Probes),Probes=numeric(s$Probes),stringsAsFactors=FALSE) G=1 n=4 for(i in 1:length(s$Probes)){ + if(G==1){c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else if((G-1) length(m$Sample1)) { + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:(G+s$Probes[i]-1)]) + c[i,3]-min(m$Start[G:(G+s$Probes[i]-1)]) + c[i,4]-max(m$End[G:(G+s$Probes[i]-1)]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else { + G=1 + n=n+1 + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])}} c Sample ChrStart End Values Probes 1 Sample1 chr2 9896633 14404502 0 4 2 Sample1 chr2 14421718 16048724 -0.43 4 3 Sample1 chr2 37491676 37703009 0 2 4 Sample2 chr2 9896633 9896690 0 2 5 Sample2 chr2 14314039 16048724 -0.35 6 6 Sample2 chr2 37491676 37703009 0 2 7 Sample3 chr2 9896633 14314098 0 3 8 Sample3 chr2 14404467 16031769 0.32 3 9 Sample3 chr2 16036178 37491735 0.45 3 10 Sample3 chr2 37702947 37703009 0 1 The problem that I am facing is for expanding rle function for values and probes. Defintely your code looks simpler, but I would like to read the file by just giving the name of the file as written in my code because my original file contains 150 samples,but how to use lapply or rle function for 150 such samples, if my file contain 150 samples similiar to sample1 and sample2. waiting for your reply, Thanks, Suji On Wed, Sep 28, 2011 at 11:37 AM, jim holtman jholt...@gmail.com wrote: Here one approach: x - read.table(textConnection(Chr start end sample1 sample2 + chr2 9896633 9896683 0 0 + chr2 9896639 9896690 0 0 + chr2 14314039 14314098 0 -0.35 + chr2 14404467 14404502 0 -0.35 + chr2 14421718 14421777 -0.43 -0.35 + chr2 16031710 16031769 -0.43 -0.35 + chr2 16036178 16036237 -0.43 -0.35 + chr2 16048665 16048724 -0.43 -0.35 + chr2 37491676 37491735 0 0 + chr2 37702947 37703009 0 0), header = TRUE, as.is = TRUE) closeAllConnections() result - lapply(c('sample1', 'sample2'), function(.samp){ + # split by breaks in the values + .grps - split(x, cumsum(c(0, diff(x[[.samp]]) != 0))) + + # combine the list of dataframes + .range - do.call(rbind, lapply(.grps, function(.set){ + # create a dataframe of the results + data.frame(Sample = .samp +, Chr = .set$Chr[1L] +, Start = min(.set$start) +, End = max(.set$end) +, Values = .set[[.samp]][1L] +, Probes = nrow(.set) +) + })) + }) # put the list of dataframes together result - do.call(rbind, result) result Sample ChrStart End Values Probes 0 sample1 chr2 9896633 14404502 0.00 4 1 sample1 chr2 14421718 16048724 -0.43 4 2 sample1 chr2 37491676 37703009 0.00 2 01 sample2 chr2 9896633 9896690 0.00 2 11 sample2 chr2 14314039 16048724 -0.35 6 21 sample2 chr2 37491676 37703009 0.00 2 On Mon, Sep 26, 2011 at 10:30 AM, sujitha virith...@gmail.com wrote: Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0 This is the output that I am expecting: Sample Chr Start End Values Probes sample1 chr2 9896633 14404502 0 4 sample1 chr2 14421718 16048724 -0.43 4 sample1 chr2 37491676 37703001 0 2 sample2 chr2 9896633 9896690 0 2 sample2 chr2 14314039 16048724 -0.35 6 sample2 chr2 37491676 37703009 0 2 Here the Chr value is same but can be any other value aswell so unique among the similar values. The Start for the first line would be the least value until values are similiar (4) then the end would be highest value. The values is the unique value among the common values and probes is number of similar values.
Re: [R] How to Store the executed values in a dataframe rle function
Hi Jim, Thanks for the reply, ok but I dont want to use textConnection and paste each line but want the input to be read from a file like m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric'). So how do I incorporate that in your code. Thanks, Suji On Wed, Sep 28, 2011 at 2:40 PM, jim holtman jholt...@gmail.com wrote: The solution that I sent will handle the 150 different samples; just list the column names in the argument to the top 'lapply'. You don't need the 'rle' in my approach. On Wed, Sep 28, 2011 at 2:13 PM, viritha k virith...@gmail.com wrote: Hi, This is the code that I wrote for 3 samples: code: m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric','numeric')) s-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) names(s)=c(Values,Probes) c-data.frame(Sample=character(s$Probes),Chr=character(s$Probes),Start=numeric(s$Probes),End=numeric(s$Probes),Values=numeric(s$Probes),Probes=numeric(s$Probes),stringsAsFactors=FALSE) G=1 n=4 for(i in 1:length(s$Probes)){ + if(G==1){c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else if((G-1) length(m$Sample1)) { + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:(G+s$Probes[i]-1)]) + c[i,3]-min(m$Start[G:(G+s$Probes[i]-1)]) + c[i,4]-max(m$End[G:(G+s$Probes[i]-1)]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else { + G=1 + n=n+1 + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])}} c Sample ChrStart End Values Probes 1 Sample1 chr2 9896633 14404502 0 4 2 Sample1 chr2 14421718 16048724 -0.43 4 3 Sample1 chr2 37491676 37703009 0 2 4 Sample2 chr2 9896633 9896690 0 2 5 Sample2 chr2 14314039 16048724 -0.35 6 6 Sample2 chr2 37491676 37703009 0 2 7 Sample3 chr2 9896633 14314098 0 3 8 Sample3 chr2 14404467 16031769 0.32 3 9 Sample3 chr2 16036178 37491735 0.45 3 10 Sample3 chr2 37702947 37703009 0 1 The problem that I am facing is for expanding rle function for values and probes. Defintely your code looks simpler, but I would like to read the file by just giving the name of the file as written in my code because my original file contains 150 samples,but how to use lapply or rle function for 150 such samples, if my file contain 150 samples similiar to sample1 and sample2. waiting for your reply, Thanks, Suji On Wed, Sep 28, 2011 at 11:37 AM, jim holtman jholt...@gmail.com wrote: Here one approach: x - read.table(textConnection(Chr start end sample1 sample2 + chr2 9896633 9896683 0 0 + chr2 9896639 9896690 0 0 + chr2 14314039 14314098 0 -0.35 + chr2 14404467 14404502 0 -0.35 + chr2 14421718 14421777 -0.43 -0.35 + chr2 16031710 16031769 -0.43 -0.35 + chr2 16036178 16036237 -0.43 -0.35 + chr2 16048665 16048724 -0.43 -0.35 + chr2 37491676 37491735 0 0 + chr2 37702947 37703009 0 0), header = TRUE, as.is = TRUE) closeAllConnections() result - lapply(c('sample1', 'sample2'), function(.samp){ + # split by breaks in the values + .grps - split(x, cumsum(c(0, diff(x[[.samp]]) != 0))) + + # combine the list of dataframes + .range - do.call(rbind, lapply(.grps, function(.set){ + # create a dataframe of the results + data.frame(Sample = .samp +, Chr = .set$Chr[1L] +, Start = min(.set$start) +, End = max(.set$end) +, Values = .set[[.samp]][1L] +, Probes = nrow(.set) +) + })) + }) # put the list of dataframes together result - do.call(rbind, result) result Sample ChrStart End Values Probes 0 sample1 chr2 9896633 14404502 0.00 4 1 sample1 chr2 14421718 16048724 -0.43 4 2 sample1 chr2 37491676 37703009 0.00 2 01 sample2 chr2 9896633 9896690 0.00 2 11 sample2 chr2 14314039 16048724 -0.35 6 21 sample2 chr2 37491676 37703009 0.00 2 On Mon, Sep 26, 2011 at 10:30 AM, sujitha virith...@gmail.com wrote: Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35
Re: [R] How to Store the executed values in a dataframe rle function
I only used textConnection for the sample data. Just put your file name in the read.table; e.g., x-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric')) as you have in your email. I used 'x' in my code, so I replaced your 'm' with 'x'. Try it and see if it works; no reason it shouldn't. On Wed, Sep 28, 2011 at 3:03 PM, viritha k virith...@gmail.com wrote: Hi Jim, Thanks for the reply, ok but I dont want to use textConnection and paste each line but want the input to be read from a file like m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric'). So how do I incorporate that in your code. Thanks, Suji On Wed, Sep 28, 2011 at 2:40 PM, jim holtman jholt...@gmail.com wrote: The solution that I sent will handle the 150 different samples; just list the column names in the argument to the top 'lapply'. You don't need the 'rle' in my approach. On Wed, Sep 28, 2011 at 2:13 PM, viritha k virith...@gmail.com wrote: Hi, This is the code that I wrote for 3 samples: code: m-read.table(test.txt,sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric','numeric')) s-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) names(s)=c(Values,Probes) c-data.frame(Sample=character(s$Probes),Chr=character(s$Probes),Start=numeric(s$Probes),End=numeric(s$Probes),Values=numeric(s$Probes),Probes=numeric(s$Probes),stringsAsFactors=FALSE) G=1 n=4 for(i in 1:length(s$Probes)){ + if(G==1){c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else if((G-1) length(m$Sample1)) { + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:(G+s$Probes[i]-1)]) + c[i,3]-min(m$Start[G:(G+s$Probes[i]-1)]) + c[i,4]-max(m$End[G:(G+s$Probes[i]-1)]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])} + else { + G=1 + n=n+1 + c[i,1]-names(m[n]) + c[i,2]-unique(m$Chr[G:s$Probes[i]]) + c[i,3]-min(m$Start[G:s$Probes[i]]) + c[i,4]-max(m$End[G:s$Probes[i]]) + c[i,]-cbind(c[i,1],c[i,2],c[i,3],c[i,4],s$Values[i],s$Probes[i]) + G=(G+s$Probes[i])}} c Sample Chr Start End Values Probes 1 Sample1 chr2 9896633 14404502 0 4 2 Sample1 chr2 14421718 16048724 -0.43 4 3 Sample1 chr2 37491676 37703009 0 2 4 Sample2 chr2 9896633 9896690 0 2 5 Sample2 chr2 14314039 16048724 -0.35 6 6 Sample2 chr2 37491676 37703009 0 2 7 Sample3 chr2 9896633 14314098 0 3 8 Sample3 chr2 14404467 16031769 0.32 3 9 Sample3 chr2 16036178 37491735 0.45 3 10 Sample3 chr2 37702947 37703009 0 1 The problem that I am facing is for expanding rle function for values and probes. Defintely your code looks simpler, but I would like to read the file by just giving the name of the file as written in my code because my original file contains 150 samples,but how to use lapply or rle function for 150 such samples, if my file contain 150 samples similiar to sample1 and sample2. waiting for your reply, Thanks, Suji On Wed, Sep 28, 2011 at 11:37 AM, jim holtman jholt...@gmail.com wrote: Here one approach: x - read.table(textConnection(Chr start end sample1 sample2 + chr2 9896633 9896683 0 0 + chr2 9896639 9896690 0 0 + chr2 14314039 14314098 0 -0.35 + chr2 14404467 14404502 0 -0.35 + chr2 14421718 14421777 -0.43 -0.35 + chr2 16031710 16031769 -0.43 -0.35 + chr2 16036178 16036237 -0.43 -0.35 + chr2 16048665 16048724 -0.43 -0.35 + chr2 37491676 37491735 0 0 + chr2 37702947 37703009 0 0), header = TRUE, as.is = TRUE) closeAllConnections() result - lapply(c('sample1', 'sample2'), function(.samp){ + # split by breaks in the values + .grps - split(x, cumsum(c(0, diff(x[[.samp]]) != 0))) + + # combine the list of dataframes + .range - do.call(rbind, lapply(.grps, function(.set){ + # create a dataframe of the results + data.frame(Sample = .samp + , Chr = .set$Chr[1L] + , Start = min(.set$start) + , End = max(.set$end) + , Values = .set[[.samp]][1L] + , Probes = nrow(.set) + ) + })) + }) # put the list of dataframes together result - do.call(rbind, result) result Sample Chr Start End Values Probes 0 sample1 chr2 9896633 14404502 0.00 4 1 sample1 chr2 14421718 16048724 -0.43 4 2 sample1 chr2 37491676 37703009 0.00 2 01 sample2 chr2 9896633 9896690 0.00