On Jul 3, 2011, at 1:07 PM, Bansal, Vikas wrote:
Yes you are right. unlist operation is unnecessary and I have tried it yesterday and it is working without that operation also.But I have one more problem on which I have worked whole day but did not get any solution.As I told you I am new to R,I want to ask that how I can use the (if condition) in the following codedf=read.table("Case2.pileup",fill=T,sep="\t",colClasses = "character") txtvec <- readLines(textConnection(df[,9]))dad=data.frame(A = (sapply(gregexpr("A|a", (df[,9])), function(x) if ( x[[1]] != -1)length(x) else 0 )), C = (sapply(gregexpr("C|c", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 )), G = (sapply(gregexpr("G|g", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 )), T = (sapply(gregexpr("T|t", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 )),N = (sapply(gregexpr("\\,|\\.", (df[,9])), function(x) if ( x[[1]] ! = -1)length(x) else 0 )))Now my problem is in my data frame I have alphabets A,C,G and T in 3rd column also.Now these commas (,)and dots(.) in column 9 are for these alphabets which are in column 3.I want to use if condition like thisif in my dataframe column 3 have A then A = (sapply(gregexpr("\\,|\ \.", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 ))) else (A = (sapply(gregexpr("A|a", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 )),if in my dataframe column 3 haveCA then C = (sapply(gregexpr("\\,|\\.", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 ))) else C = (sapply(gregexpr("C|c", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 )), if in my dataframe column 3 have G then G = (sapply(gregexpr("\\,|\\.", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 ))) else G = (sapply(gregexpr("G|g", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 )) if in my dataframe column 3 have T then T = (sapply(gregexpr("\\,|\\.", (df[,9])), function(x) if ( x[[1]] != -1) length(x) else 0 ))) else T = (sapply(gregexpr("T|t", (df[,9])), function(x) if ( x[[1]] != -1)length(x) else 0 )),
I finally figured out that you wanted this:> dat$newcol <- apply(dat, 1, function(x) gsub("\\,|\\.", x[3], x[9]) )
# So that replaces any instance of "," or "." in col9 with the letter in col3
# Then the same old routine as yesterday> dat$A <- sapply(gregexpr("A|a", (dat[,"newcol"])), function(x) if ( x[[1]] != -1) length(x) else 0 ) > dat$C <- sapply(gregexpr("C|c", (dat[,"newcol"])), function(x) if ( x[[1]] != -1) length(x) else 0 ) > dat$G <- sapply(gregexpr("G|g", (dat[,"newcol"])), function(x) if ( x[[1]] != -1) length(x) else 0 ) > dat$T <- sapply(gregexpr("T|t", (dat[,"newcol"])), function(x) if ( x[[1]] != -1) length(x) else 0 )
> dat[, c("A","C", "G", "T")]
A C G T
1 1 0 1 4
2 4 0 0 2
3 4 2 0 0
4 1 5 0 0
5 0 0 4 3
6 5 1 1 0
7 4 0 4 0
8 8 0 0 0
9 1 4 1 1
10 0 0 0 8
11 0 0 0 8
So I want to code so that it will give the output like this- DATA FRAME (Input) col3 col 9 T .a,g,, A .t,t,, A .,c,c, C .,a,,, G .,t,t,t A .c,,g,^!. A .g,ggg.^!, A .$,,,,,., C a,g,,t, T ,,,,,.,^!. T ,$,,,,.,." output A C G T 1 0 1 4 4 0 0 2 4 2 0 0 1 5 0 0 0 0 4 3 This is the output for first five rows.Can you please help me how to use this if condition in your coding or we can also do it by using some other condition rather than if condition?________________________________________ From: David Winsemius [dwinsem...@comcast.net] Sent: Sunday, July 03, 2011 3:57 AM To: Bansal, Vikas Cc: Dennis Murphy; r-help@r-project.org Subject: Re: [R] For help in R coding 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 beto use gregexpr. For counting the number of commas, this appears quitestraightforward.sapply(gregexpr("\\,", txtvec), function(x) if ( x[[1]] != -1)length(x) else 0 ) [1] 3 3 3 4 3 3 2 6 4 6 6It 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 makes it difficult for strsplit() to pick up if it is splitting on those characters. Here is an alternative, splitting on individual characters and 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 ,$,,,,.,. 0 0 0 0 Has the advantage that the input data ends up as rownames, which was a surprise. If you wanted to count "A" and "a" as equivalent, then the split argument should be "a|A"AS YOU MENTIONED THAT IF I WANT TO COUNT A AND a I SHOULD SPLIT LIKE THIS.BUT CAN I COUNT . AND , ALSO USING- data.frame(A = unlist(lapply( lapply( sapply(txtvec, strsplit, split=".|,"), length) , "-", 1)), I TRIED IT BUT ITS NOT WORKING.IT IS GIVING THE OUTPUT BUT AT SOME PLACES IT IS SHOWING MORE NUMBER OF . AND , AND SOMEWHERE IT IS NOT EVEN CALCULATING AND JUST SHOWING 0.You need to use valid regex expressions for 'split'. Since "." and "," are special characters they need to be escaped when you wnat the literals to be recognized as such. I haven't figured out why but you need to drop the final operation of subtracting 1 from the values when counting commas:data.frame(periods = unlist(lapply( lapply( sapply(txtvec, strsplit,split="\\."), length) , "-", 1)) ,commas = unlist( lapply( sapply(txtvec, strsplit, split="\\,"), length) ) ) periods commas .a,g,, 1 3 .t,t,, 1 3 .,c,c, 1 3 .,a,,, 1 4 .,t,t,t 1 4 .c,,g,^!. 1 4 .g,ggg.^!, 2 2 .$,,,,,., 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, CTDavid 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.
