[R] How to Store the executed values in a dataframe & rle function
jim holtman
jholtman at gmail.com
Wed Sep 28 22:15:31 CEST 2011
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 <viritha.k at 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 <jholtman at 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 <viritha.k at 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 <jholtman at 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 <viritha.k at 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 at 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?
>> >
>> >
>>
>>
>>
>> --
>> Jim Holtman
>> Data Munger Guru
>>
>> What is the problem that you are trying to solve?
>
>
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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