[R] reading data
arun
smartpink111 at yahoo.com
Mon Feb 18 20:41:23 CET 2013
Hi,
I am not able to open your graph. I am using linux.
Also, the codes in the function are not reproducible
directT <- direct[grepl("^t", direct)]
directC <- direct[grepl("^c", direct)]
It takes double the time to know what is going on.
dir()
#[1] "a1" "a2" "a3" "b1" "b2" "c1"
direct<- list.files(recursive=TRUE)[grepl("^a|^b",dir())]
direct
#[1] "MSMS_23PepInfo.txt" "MSMS_23PepInfo.txt" "MSMS_23PepInfo.txt"
#[4] "MSMS_23PepInfo.txt" "MSMS_23PepInfo.txt"
directA<- list.files(recursive=TRUE)[grepl("^a",dir())]
directB<- list.files(recursive=TRUE)[grepl("^b",dir())]
lista<- lapply(direct,function(x) read.table(x,header=TRUE,stringsAsFactors=FALSE,sep="\t",fill=TRUE))
listaA<-lapply(directA, function(x) read.table(x,header=TRUE, sep = "\t",fill=TRUE))
listaB<-lapply(directB, function(x) read.table(x,header=TRUE, sep = "\t",fill=TRUE))
#here I am changing the names listaT, z, etc..
count different mm values
cab <- vector()
for (i in 1:length(lista)) {
dc<-lista[[i]][ifelse(lista[[i]]$b<0.01, TRUE, FALSE),]
dc<-table(dc$mm)
cab <- c(cab, names(dc))
}
#Relative freqs to construct the graph
cab <- unique(cab)
d <- matrix(ncol=length(cab))
dci<- d[-1,]
dcf <- d[-1,]
dti <- d[-1,]
dtf <- d[-1,]
########################################
for (i in 1:length(listaA)) {
#Relative freq of all data
dcc<-listaA[[i]]
dcc<-table(factor(dcc$mm, levels=cab))
dci<- rbind(dci, dcc)
rownames(dci)<-rownames(1:(nrow(dci)), do.NULL = FALSE, prefix = "a")
#Relative freq of data with FDR<0.01
dcc1<-listaA[[i]][ifelse(listaA[[i]]$FDR<0.01, TRUE, FALSE),]
dcc1<-table(factor(dcc1$mm, levels=cab))
dcf<- rbind(dcf,dcc1)
rownames(dcf)<-rownames(1:(nrow(dcf)), do.NULL = FALSE, prefix = "a")
}
for (i in 1:length(listaB)) {
#Relative freq of all data
dct<-listaB[[i]]
dct<-table(factor(dct$mm, levels=cab))
dti<- rbind(dti, dct)
rownames(dti)<-rownames(1:(nrow(dti)), do.NULL = FALSE, prefix = "b")
#Relative freq of data with FDR<0.01
dct1<-listaB[[i]][ifelse(listaB[[i]]$FDR<0.01, TRUE, FALSE),]
dct1<-table(factor(dct1$mm, levels=cab))
dtf<- rbind(dtf,dct1)
rownames(dtf)<-rownames(1:(nrow(dtf)), do.NULL = FALSE, prefix = "b")
}
freq.i<-rbind(dci,dti)
freq.f<-rbind(dcf,dtf)
freq.rel.i<-freq.i/apply(freq.i,1,sum)
freq.rel.f<-freq.f/apply(freq.f,1,sum)
freq.i
# 2 3
#a1 4 1
#a2 4 1
#a3 4 1
#b1 4 1
#b2 4 1
#b3 4 1
#b4 4 1
#result from my code.
files<-paste("MSMS_",23,"PepInfo.txt",sep="")
read.data<-function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,sep = "\t",stringsAsFactors=FALSE,fill=TRUE))}
lista<-do.call("c",lapply(list.files(recursive=T)[grep(files,list.files(recursive=T))],read.data))
names(lista)<-paste("group_",gsub("\\d+","",names(lista)),sep="")
res2<-split(lista,names(lista))
res3<- lapply(res2,function(x) {names(x)<-paste(gsub(".*_","",names(x)),1:length(x),sep="");x})
res4<-lapply(seq_along(res3),function(i) do.call(rbind,lapply(res3[[i]], function(x) table(x$mm[x[["b"]]<0.01]))))
names(res4)<- names(res2)
res4
$group_a
# 2 3
#a1 3 1
#a2 3 1
#a3 3 1
#$group_b
# 2 3
#b1 3 1
#b2 3 1
#$group_c
# 2 3
#c1 3 1
There is a difference in output from freq.i and res4. There were only two files under 'group_b`. So, check your codes.
A.K.
________________________________
From: Vera Costa <veracosta.rt at gmail.com>
To: arun <smartpink111 at yahoo.com>
Sent: Monday, February 18, 2013 10:27 AM
Subject: Re: reading data
Hi!!!
I'm coming to ask a new question.
I want a function to do my statistics. I start with you had send me:
z.plot <- function(directory,number) {
setwd(directory)
indx<-gsub("[./]","",list.dirs())
indx1<- indx[indx!=""]
print(indx1)
files<-paste("MSMS_",number,"PepInfo.txt",sep="")
read.data<-function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,sep = "\t",stringsAsFactors=FALSE,fill=TRUE))}
lista<-do.call("c",lapply(list.files(recursive=T)[grep(files,list.files(recursive=T))],read.data))
print(lista)
#names(lista)<-paste("group_",gsub("\\d+","",names(lista)),sep="") ve = TRUE)
}
z.plot("C:/Users/Vera Costa/Desktop/dados.lixo",23)
In my lista I can´t merge rows to have the group, because the idea is for each file count frequencies of mm, when b<0.01. after that I want a graph like the graph in attach.
When I had 2 groups and knew the name of the groups, I did the code (but Know I have more groups and, maybe, I don´t know the name of the groups):
z.plot <- function(directory,number) {
#reading data
setwd(directory)
direct<-dir(directory,pattern = paste("MSMS_",number,"PepInfo.txt",sep=""), full.names = FALSE, recursive = TRUE)
directT <- direct[grepl("^t", direct)]
directC <- direct[grepl("^c", direct)]
lista<-lapply(direct, function(x) read.table(x,header=TRUE, sep = "\t"))
listaC<-lapply(directC, function(x) read.table(x,header=TRUE, sep = "\t"))
listaT<-lapply(directT, function(x) read.table(x,header=TRUE, sep = "\t"))
#count different z values
cab <- vector()
for (i in 1:length(lista)) {
dc<-lista[[i]][ifelse(lista[[i]]$FDR<0.01, TRUE, FALSE),]
dc<-table(dc$z)
cab <- c(cab, names(dc))
}
#Relative freqs to construct the graph
cab <- unique(cab)
d <- matrix(ncol=length(cab))
dci<- d[-1,]
dcf <- d[-1,]
dti <- d[-1,]
dtf <- d[-1,]
for (i in 1:length(listaC)) {
#Relative freq of all data
dcc<-listaC[[i]]
dcc<-table(factor(dcc$z, levels=cab))
dci<- rbind(dci, dcc)
rownames(dci)<-rownames(1:(nrow(dci)), do.NULL = FALSE, prefix = "c")
#Relative freq of data with FDR<0.01
dcc1<-listaC[[i]][ifelse(listaC[[i]]$FDR<0.01, TRUE, FALSE),]
dcc1<-table(factor(dcc1$z, levels=cab))
dcf<- rbind(dcf,dcc1)
rownames(dcf)<-rownames(1:(nrow(dcf)), do.NULL = FALSE, prefix = "c")
}
for (i in 1:length(listaT)) {
#Relative freq of all data
dct<-listaT[[i]]
dct<-table(factor(dct$z, levels=cab))
dti<- rbind(dti, dct)
rownames(dti)<-rownames(1:(nrow(dti)), do.NULL = FALSE, prefix = "t")
#Relative freq of data with FDR<0.01
dct1<-listaT[[i]][ifelse(listaT[[i]]$FDR<0.01, TRUE, FALSE),]
dct1<-table(factor(dct1$z, levels=cab))
dtf<- rbind(dtf,dct1)
rownames(dtf)<-rownames(1:(nrow(dtf)), do.NULL = FALSE, prefix = "t")
}
freq.i<-rbind(dci,dti)
freq.f<-rbind(dcf,dtf)
freq.rel.i<-freq.i/apply(freq.i,1,sum)
freq.rel.f<-freq.f/apply(freq.f,1,sum)
#Graph plot
colour<-sample(rainbow(nrow(freq.rel.i)))
par(mfrow=c(1,2))
barplot(freq.rel.i,beside=T,main=("Sample"),xlab="Charge",ylab="Relative Frequencies",col=colour,legend.text = rownames(freq.rel.i))
barplot(freq.rel.f,beside=T,main=("Sample with FDR<0.01"),xlab="Charge",ylab="Relative Frequencies",col=colour,legend.text = rownames(freq.rel.f))
#average of the group (except c1&t1)
freqs<-rbind(dcf[-1,], dtf[-1,])
average<-apply(freqs,2,mean)
#chisquare test function
chisq.test<-function(x,y){
somax<-sum(x)
somay<-sum(y)
nj.<-x+y
nj<-sum(nj.)
ejx<-(nj./nj)*somax
ejy<-(nj./nj)*somay
ETx<-((x-ejx)^2)/ejx
ETy<-((y-ejy)^2)/ejy
ETobs<-sum(ETx)+sum(ETy)
pvalue<-1-pchisq(c(ETobs),df=length(x|y)-1,lower.tail=TRUE)
return(pvalue)
}
#pvalues of the chisquare test between sample and average (H0: two samples has the same distribution)
pvalues<-c()
for (i in 1:(nrow(freqs))){
a<-chisq.test(freqs[i,],average)
pvalues<-c(pvalues,a)
}
#data frame with final p-values
dataframe<-data.frame(c(rownames(freqs)), c(pvalues))
colnames(dataframe)<-c("sample name","pvalue")
print(dataframe)
}
z.plot("C:/Users/Vera/Desktop/data",23)
Thank you again
2013/2/17 arun <smartpink111 at yahoo.com>
HI Vera,
>
>No problem. I am cc:ing to r-help.
>
>A.K.
>
>
>
>
>
>
>________________________________
>From: Vera Costa <veracosta.rt at gmail.com>
>To: arun <smartpink111 at yahoo.com>
>Sent: Sunday, February 17, 2013 5:44 AM
>Subject: Re: reading data
>
>
>
>Hi. Thank you. It works now:-)
>And yes, I use windows.
>Thank you very much.
>No dia 17 de Fev de 2013 00:44, "arun" <smartpink111 at yahoo.com> escreveu:
>
>Hi Vera,
>>
>>Have you tried the suggestion?
>>
>>Are you using Windows?
>>Thanks,
>>Arun
>>
>>
>>
>>
>>
>>
>>________________________________
>>From: Vera Costa <veracosta.rt at gmail.com>
>>To: arun <smartpink111 at yahoo.com>
>>Sent: Saturday, February 16, 2013 7:10 PM
>>Subject: Re: reading data
>>
>>
>>Thank you.
>>In mine, I have an error " 'what' must be a character string or a function".
>>I need to do equivalent in my system.
>>Thank you and sorry one more time.
>>No dia 16 de Fev de 2013 23:53, "arun" <smartpink111 at yahoo.com> escreveu:
>>
>>Hi,
>>>You didn't mention what the error message or whether you are reading file names which are not "mmmmm11kk.txt".
>>>
>>>It is workiing on my system as I run it again.
>>>?c() combine values into a vector or list.
>>>
>>> sessionInfo()
>>>R version 2.15.1 (2012-06-22)
>>>Platform: x86_64-pc-linux-gnu (64-bit)
>>>
>>>locale:
>>> [1] LC_CTYPE=en_CA.UTF-8 LC_NUMERIC=C
>>> [3] LC_TIME=en_CA.UTF-8 LC_COLLATE=en_CA.UTF-8
>>> [5] LC_MONETARY=en_CA.UTF-8 LC_MESSAGES=en_CA.UTF-8
>>> [7] LC_PAPER=C LC_NAME=C
>>> [9] LC_ADDRESS=C LC_TELEPHONE=C
>>>[11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C
>>>
>>>attached base packages:
>>>[1] stats graphics grDevices utils datasets methods base
>>>
>>>other attached packages:
>>>[1] stringr_0.6.2 reshape2_1.2.2
>>>
>>>loaded via a namespace (and not attached):
>>>[1] plyr_1.8
>>>
>>>
>>>#code
>>>
>>>
>>>res<-do.call(c,lapply(list.files(recursive=T)[grep("mmmmm11kk",list.files(recursive=T))],function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,stringsAsFactors=FALSE,fill=TRUE))})) #it seems like one of the rows of your file doesn't have 6 elements, so added fill=TRUE
>>> names(res)<-paste("group_",gsub("\\d+","",names(res)),sep="")
>>>res2<-split(res,names(res))
>>>res3<- lapply(res2,function(x) {names(x)<-paste(gsub(".*_","",names(x)),1:length(x),sep="");x})
>>>#result
>>>
>>>res3
>>>#$group_a
>>>#$group_a$a1
>>> Id M mm x b u k j y p v
>>>1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>
>>>$group_a$a2
>>> Id M mm x b u k j y p v
>>>1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>
>>>$group_a$a3
>>> Id M mm x b u k j y p v
>>>1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>
>>>
>>>$group_b
>>>$group_b$b1
>>> Id M mm x b u k j y p v
>>>1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>
>>>$group_b$b2
>>> Id M mm x b u k j y p v
>>>1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>
>>>
>>>$group_c
>>>$group_c$c1
>>> Id M mm x b u k j y p v
>>>1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>
>>>
>>>A.K.
>>>
>>>
>>>
>>>________________________________
>>>From: Vera Costa <veracosta.rt at gmail.com>
>>>To: arun <smartpink111 at yahoo.com>
>>>Sent: Saturday, February 16, 2013 6:32 PM
>>>Subject: Re: reading data
>>>
>>>
>>>Sorry again... In:
>>>res<-do.call(c,lapply(list.files(recursive=T)[grep("...
>>>What is this c? In do.call(c, When I put this row im R, I have an error.
>>>Thank you
>>>No dia 15 de Fev de 2013 18:11, "arun" <smartpink111 at yahoo.com> escreveu:
>>>
>>>Hi,
>>>>No problem.
>>>>
>>>>BTW, these questions are not stupid..
>>>>Arun
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>________________________________
>>>>From: Vera Costa <veracosta.rt at gmail.com>
>>>>To: arun <smartpink111 at yahoo.com>
>>>>Sent: Friday, February 15, 2013 1:08 PM
>>>>Subject: Re: reading data
>>>>
>>>>
>>>>Thank you very much.
>>>>
>>>>I will try to apply and after I tell you if it is ok :-)
>>>>
>>>>Thank you and sorry about this questions (sometimes stupid questions).
>>>>
>>>>
>>>>
>>>>
>>>>2013/2/15 arun <smartpink111 at yahoo.com>
>>>>
>>>>HI,
>>>>>No problem.
>>>>>?c() for concatenate to vector or list().
>>>>>If I use do.call(cbind,..) or do.call(rbind,...)
>>>>>
>>>>>do.call(cbind,lapply(list.files(recursive=T)[grep("mmmmm11kk",list.files(recursive=T))],function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,stringsAsFactors=FALSE,fill=TRUE))}))
>>>>># [,1] [,2] [,3] [,4] [,5] [,6]
>>>>>#a1 List,11 List,11 List,11 List,11 List,11 List,11
>>>>>
>>>>>
>>>>> do.call(rbind,lapply(list.files(recursive=T)[grep("mmmmm11kk",list.files(recursive=T))],function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,stringsAsFactors=FALSE,fill=TRUE))}))
>>>>># a1
>>>>>#[1,] List,11
>>>>>#[2,] List,11
>>>>>#[3,] List,11
>>>>>#[4,] List,11
>>>>>#[5,] List,11
>>>>>#[6,] List,11
>>>>>ie.
>>>>>list within in a list
>>>>>
>>>>> restrial<-lapply(list.files(recursive=T)[grep("mmmmm11kk",list.files(recursive=T))],function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,stringsAsFactors=FALSE,fill=TRUE))})
>>>>> str(restrial)
>>>>>#List of 6
>>>>># $ :List of 1
>>>>> #..$ a1:'data.frame': 6 obs. of 11 variables:
>>>>> .#. ..$ Id: chr [1:6] "aAA" "aAAAA" "aA" "aAA" ...
>>>>> #.. ..$ M : chr [1:6] "1" "1" "2" "1" ...
>>>>> #. ..$ mm: int [1:6] 2 2 1 2 3 2
>>>>> #. ..$ x : int [1:6] 739 2263 1 1965 3660 1972
>>>>> -----------------------------------------------------------------
>>>>>str(res)
>>>>>#List of 6
>>>>># $ a1:'data.frame': 6 obs. of 11 variables:
>>>>> # ..$ Id: chr [1:6] "aAA" "aAAAA" "aA" "aAA" ...
>>>>> #..$ M : chr [1:6] "1" "1" "2" "1" ...
>>>>> # ..$ mm: int [1:6] 2 2 1 2 3 2
>>>>> # ..$ x : int [1:6] 739 2263 1 1965 3660 1972
>>>>>-----------------------------------------------------------------
>>>>>
>>>>>You mentioned about naming this to "group_a","group_b". etc..
>>>>>
>>>>> names(res)<-paste("group_",gsub("\\d+","",names(res)),sep="")
>>>>>res2<-split(res,names(res))
>>>>>
>>>>>res3<- lapply(res2,function(x) {names(x)<-paste(gsub(".*_","",names(x)),1:length(x),sep="");x})
>>>>> res3$group_a
>>>>>$a1
>>>>>
>>>>># Id M mm x b u k j y p v
>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>
>>>>>#$a2
>>>>>
>>>>># Id M mm x b u k j y p v
>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>
>>>>>#$a3
>>>>>
>>>>> # Id M mm x b u k j y p v
>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>A.K.
>>>>>
>>>>>________________________________
>>>>>From: Vera Costa <veracosta.rt at gmail.com>
>>>>>To: arun <smartpink111 at yahoo.com>
>>>>>Sent: Friday, February 15, 2013 12:39 PM
>>>>>Subject: Re: reading data
>>>>>
>>>>>
>>>>>
>>>>>Thank you very much and sorry my questions.
>>>>>
>>>>>But this code isn't grouping for letters sure? I mean, a1,a2,a3 is the same group, (the first letter give me the name of the group)
>>>>>
>>>>>Another question, in do.call, you did do.call (c,.....) .What is c?
>>>>>
>>>>>Sorry
>>>>>
>>>>>
>>>>>
>>>>>2013/2/15 arun <smartpink111 at yahoo.com>
>>>>>
>>>>>HI,
>>>>>>
>>>>>>Just to add:
>>>>>>
>>>>>>
>>>>>>res<-do.call(c,lapply(list.files(recursive=T)[grep("mmmmm11kk",list.files(recursive=T))],function(x) {names(x)<-gsub("^(.*)\\/.*","\\1",x); lapply(x,function(y) read.table(y,header=TRUE,stringsAsFactors=FALSE,fill=TRUE))})) #it seems like one of the rows of your file doesn't have 6 elements, so added fill=TRUE
>>>>>>
>>>>>> names(res)<-paste("group_",gsub("\\d+","",names(res)),sep="")
>>>>>>res[grep("group_b",names(res))]
>>>>>>
>>>>>>I am not sure how you want the grouped data to look like. If you want something like this:
>>>>>>res1<-do.call(rbind,res)
>>>>>>res2<-lapply(split(res1,gsub("[.0-9]","",row.names(res1))),function(x) {row.names(x)<-1:nrow(x);x})
>>>>>>res2
>>>>>>#$group_a
>>>>>>
>>>>>> # Id M mm x b u k j y p v
>>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>#7 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#8 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#9 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#10 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#11 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#12 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>#13 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#14 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#15 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#16 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#17 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#18 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>
>>>>>>
>>>>>>#$group_b
>>>>>> # Id M mm x b u k j y p v
>>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>#7 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#8 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#9 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#10 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#11 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#12 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>
>>>>>>#$group_c
>>>>>>
>>>>>> # Id M mm x b u k j y p v
>>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>
>>>>>>
>>>>>>#or if you want it like this:
>>>>>>res2<-split(res,names(res))
>>>>>>
>>>>>>res2[["group_b"]]
>>>>>>
>>>>>>#$group_b
>>>>>># Id M mm x b u k j y p v
>>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>
>>>>>>#$group_b
>>>>>> # Id M mm x b u k j y p v
>>>>>>#1 aAA 1 2 739 0.1257000 2 2 AA 2 8867 8926
>>>>>>#2 aAAAA 1 2 2263 0.0004000 2 2 AR 4 7640 8926
>>>>>>#3 aA 2 1 1 0.0845435 2 AA 2 6790 734,1092 NA
>>>>>>#4 aAA 1 2 1965 0.0007000 4 3 AR 2 11616 8926
>>>>>>#5 aAAA 1 3 3660 0.0008600 18 3 AA 2 20392 496
>>>>>>#6 AA na 2 1972 0.0007000 11 3 AR 25 509 734
>>>>>>
>>>>>>Hope this helps.
>>>>>>
>>>>>>A.K.
>>>>>>
>>>>>>
>>>>>>
>>>>>>----- Original Message -----
>>>>>>From: "veracosta.rt at gmail.com" <veracosta.rt at gmail.com>
>>>>>>To: smartpink111 at yahoo.com
>>>>>>Cc:
>>>>>>Sent: Friday, February 15, 2013 9:15 AM
>>>>>>Subject: reading data
>>>>>>
>>>>>>Hi,
>>>>>>I post yesterday and you helped me. I have little problem.
>>>>>>
>>>>>>At first, I never worked with regular expressions...
>>>>>>
>>>>>>The code that you gave me it's ok, but my files are inside the folders a1,a2,a3. I try to explain better.
>>>>>>
>>>>>>I have one folder named "data". Inside this folder I have some other folders named "a1","a2","b1",b2",...and inside of each one of that I have some files. I want only the file "mmmmmm.txt" (in all folders I have One file with this name).
>>>>>>The name of the folder give me the name of the group,but I need to read the file inside. And after, have "group_a", group_"b"...because I need to work with this data grouped (and know the name of the group).
>>>>>>
>>>>>>Thank you.
>>>>>>
>>>>>
>>>>
>>>
>>
>
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