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 = min(.set
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
