On Jul 2, 2011, at 4:46 PM, Bansal, Vikas wrote:
DEAR ALL, I TRIED THIS CODE AND THIS IS RUNNING PERFECTLY... df=read.table("Case2.pileup",fill=T,sep="\t",colClasses = "character") txt=df[,9] txtvec <- readLines(textConnection(txt))dad=data.frame(A = unlist(sapply(gregexpr("A|a", txtvec), function(x) if ( x[[1]] != -1)length(x) else 0 )),C = unlist(sapply(gregexpr("C|c", txtvec), function(x) if ( x[[1]] ! = -1)length(x) else 0 )),G = unlist(sapply(gregexpr("G|g", txtvec), function(x) if ( x[[1]] ! = -1)length(x) else 0 )),T = unlist(sapply(gregexpr("T|t", txtvec), function(x) if ( x[[1]] ! = -1)length(x) else 0 )),N = unlist(sapply(gregexpr("\\,|\\.", txtvec), function(x) if ( x[[1]] != -1)length(x) else 0 )))
The unlist operation is unnecessary since the sapply operation returns a vector. (It doesn't hurt, but it is unnecessary.)
Thanking you, Warm Regards Vikas Bansal Msc Bioinformatics Kings College London ________________________________________ From: David Winsemius [dwinsem...@comcast.net] Sent: Saturday, July 02, 2011 9:04 PM To: Dennis Murphy Cc: r-help@r-project.org; Bansal, Vikas Subject: Re: [R] For help in R coding On reflection and a bit of testing I think the best approach would be to use gregexpr. For counting the number of commas, this appears quite straightforward.sapply(gregexpr("\\,", txtvec), function(x) if ( x[[1]] != -1)length(x) else 0 ) [1] 3 3 3 4 3 3 2 6 4 6 6 It easily generalizes to period and the `|` (or) operation on letters. ( did need to add the check since the length of gregexpr is always at least one but ihas value -1 when there is no matchsapply(gregexpr("t|T", txtvec), function(x) if ( x[[1]] != -1)length(x) else 0 ) [1] 0 2 0 0 3 0 0 0 1 0 0 On Jul 2, 2011, at 3:22 PM, Dennis Murphy wrote:Hi:There seems to be a problem if the string ends in , or . , which makesit difficult for strsplit() to pick up if it is splitting on thosecharacters. Here is an alternative, splitting on individual charactersand using charmatch() instead: charsum <- function(s, char) { u <- strsplit(s, "") sum(sapply(u, function(x) charmatch(x, char)), na.rm = TRUE) } unname(sapply(txtvec, function(x) charsum(x, ','))) unname(sapply(txtvec, function(x) charsum(x, '.'))) Putting this into a data frame, dfout <- data.frame(periods = unname(sapply(txtvec, function(x) charsum(x, '.'))), commas = unname(sapply(txtvec, function(x) charsum(x, '.'))) ) txtvec HTH, Dennis On Sat, Jul 2, 2011 at 10:19 AM, David Winsemius <dwinsem...@comcast.netwrote: On Jul 2, 2011, at 12:34 PM, Bansal, Vikas wrote:Dear all, I am doing a project on variant calling using R.I am working on pileup file.There are 10 columns in my data frame and I want to count the number of A,C,G and T in each row for column 9.example of column 9 is given below- .a,g,, .t,t,, .,c,c, .,a,,, .,t,t,t .c,,g,^!. .g,ggg.^!, .$,,,,,., a,g,,t, ,,,,,.,^!. ,$,,,,.,. This is a bit confusing for me as these characters are in one column and how can we scan them for each row to print number of A,C,G and T for each row.Seems a bit clunky but this does the job (first the data):txt <- " .a,g,,+ .t,t,, + .,c,c, + .,a,,, + .,t,t,t + .c,,g,^!. + .g,ggg.^!, + .$,,,,,., + a,g,,t, + ,,,,,.,^!. + ,$,,,,.,."txtvec <- readLines(textConnection(txt))Now the clunky solution, Basically subtracts 1 from the counts of "fragments" that result from splitting on each letter in turn. Could be made prettier with a function that did the job.data.frame(A = unlist(lapply( lapply( sapply(txtvec, strsplit,split="a"), length) , "-", 1)), + C = unlist(lapply( lapply( sapply(txtvec, strsplit, split="c"), length) , "-", 1)), + G = unlist(lapply( lapply( sapply(txtvec, strsplit, split="g"), length) , "-", 1)), + T = unlist(lapply( lapply( sapply(txtvec, strsplit, split="t"), length) , "-", 1)) ) A C G T .a,g,, 1 0 1 0 .t,t,, 0 0 0 2 .,c,c, 0 2 0 0 .,a,,, 1 0 0 0 .,t,t,t 0 0 0 2 .c,,g,^!. 0 1 1 0 .g,ggg.^!, 0 0 4 0 .$,,,,,., 0 0 0 0 a,g,,t, 1 0 1 1 ,,,,,.,^!. 0 0 0 0 ,$,,,,.,. 0 0 0 0 Has the advantage that the input data ends up as rownames, which was a surprise. If you wanted to count "A" and "a" as equivalent, then the split argument should be "a|A"AS YOU MENTIONED THAT IF I WANT TO COUNT A AND a I SHOULD SPLIT LIKE THIS.BUT CAN I COUNT . AND , ALSO USING- data.frame(A = unlist(lapply( lapply( sapply(txtvec, strsplit, split=".|,"), length) , "-", 1)), I TRIED IT BUT ITS NOT WORKING.IT IS GIVING THE OUTPUT BUT AT SOME PLACES IT IS SHOWING MORE NUMBER OF . AND , AND SOMEWHERE IT IS NOT EVEN CALCULATING AND JUST SHOWING 0.You need to use valid regex expressions for 'split'. Since "." and "," are special characters they need to be escaped when you wnat the literals to be recognized as such.I haven't figured out why but you need to drop the final operation ofsubtracting 1 from the values when counting commas: data.frame(periods = unlist(lapply( lapply( sapply(txtvec, strsplit, split="\\."), length) , "-", 1)) ,commas = unlist( lapply( sapply(txtvec, strsplit, split="\\,"), length) ) ) periods commas .a,g,, 1 3 .t,t,, 1 3 .,c,c, 1 3 .,a,,, 1 4 .,t,t,t 1 4 .c,,g,^!. 1 4 .g,ggg.^!, 2 2 .$,,,,,., 2 6 a,g,,t, 0 4 ,,,,,.,^!. 1 7 ,$,,,,.,. 1 7 -- David Winsemius, MD West Hartford, CT ______________________________________________ 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.David Winsemius, MD West Hartford, CT
David Winsemius, MD West Hartford, CT ______________________________________________ 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.
