我想应用一个函数,它返回一个矩阵到一个大的data.table对象的每一行(原始文件大约30 GB,我有80 GB ram),并获取一个data.table对象。我想有效地做到这一点。我目前的做法如下:
my.function <- function(x){
alnRanges<-cigarToIRanges(x[6]);
alnStarts<-start(alnRanges)+as.numeric(x[4])-1;
alnEnds<-end(alnRanges)+as.numeric(x[4])-1;
y<-x[-4];
ys<-matrix(rep(y,length(alnRanges)),nrow=length(alnRanges),ncol=length(y),byrow=TRUE);
ys<-cbind(ys,alnStarts,alnEnds);
return(ys); # ys is a matrix
}
my.dt<-fread(my.file.name);
my.list.of.matrices<-apply(my.dt,1,my.function);
new.df<-do.call(rbind.data.frame,my.list.of.matrices);
colnames(new.df)[1:14]<-colnames(my.dt)[-4];
new.dt<-as.data.table(new.df);
注1:我指定my.function只是为了表明它返回一个矩阵,因此我的apply行是一个矩阵列表。
注意2:我不确定我正在做的操作有多慢,但似乎我可以减少行数。例如,将数据帧转换为大型对象的数据表是否很慢?
可重复的例子:
请注意,Arun和Roland让我对这个问题更加思考,所以我还在努力......可能我不需要这些线......
我想获取一个sam文件,然后创建一个新的坐标文件,其中每个读取根据其CIGAR字段进行拆分。
My sam file:
qname rname pos cigar
2218 chr1 24613476 42M2S
2067 chr1 87221030 44M
2129 chr1 79702717 44M
2165 chr1 43113438 44M
2086 chr1 52155089 4M921N40M
code:
library("data.table");
library("GenomicRanges");
sam2bed <- function(x){
alnRanges<-cigarToIRanges(x[4]);
alnStarts<-start(alnRanges)+as.numeric(x[3])-1;
alnEnds<-end(alnRanges)+as.numeric(x[3])-1;
#y<-as.data.frame(x[,pos:=NULL]);
#ys<-y[rep(seq_len(nrow(y)),length(alnRanges)),];
y<-x[-3];
ys<-matrix(rep(y,length(alnRanges)),nrow=length(alnRanges),ncol=length(y),byrow=TRUE);
ys<-cbind(ys,alnStarts,alnEnds);
return(ys);
}
sam.chr.dt<-fread(sam.parent.chr.file);
setnames(sam.chr.dt,old=c("V1","V2","V3","V4"),new=c("qname","rname","pos","cigar"));
bed.chr.lom<-apply(sam.chr.dt,1,sam2bed);
> bed.chr.lom
[[1]]
alnStarts alnEnds
[1,] "2218" "chr1" "42M2S" "24613476" "24613517"
[[2]]
alnStarts alnEnds
[1,] "2067" "chr1" "44M" "87221030" "87221073"
[[3]]
alnStarts alnEnds
[1,] "2129" "chr1" "44M" "79702717" "79702760"
[[4]]
alnStarts alnEnds
[1,] "2165" "chr1" "44M" "43113438" "43113481"
[[5]]
alnStarts alnEnds
[1,] "2086" "chr1" "4M921N40M" "52155089" "52155092"
[2,] "2086" "chr1" "4M921N40M" "52156014" "52156053"
bed.chr.df<-do.call(rbind.data.frame,bed.chr.lom);
> bed.chr.df
V1 V2 V3 alnStarts alnEnds
1 2218 chr1 42M2S 24613476 24613517
2 2067 chr1 44M 87221030 87221073
3 2129 chr1 44M 79702717 79702760
4 2165 chr1 44M 43113438 43113481
5 2086 chr1 4M921N40M 52155089 52155092
6 2086 chr1 4M921N40M 52156014 52156053
bed.chr.dt<-as.data.table(bed.chr.df);
> bed.chr.dt
V1 V2 V3 alnStarts alnEnds
1: 2218 chr1 42M2S 24613476 24613517
2: 2067 chr1 44M 87221030 87221073
3: 2129 chr1 44M 79702717 79702760
4: 2165 chr1 44M 43113438 43113481
5: 2086 chr1 4M921N40M 52155089 52155092
6: 2086 chr1 4M921N40M 52156014 52156053
答案 0 :(得分:3)
假设ff
是data.table
,那么这个怎么样?
splits <- cigarToIRangesListByAlignment(ff$cigar, ff$pos, reduce.ranges = TRUE)
widths <- width(attr(splits, 'partitioning'))
cbind(data.table(qname=rep.int(ff$qname, widths),
rname=rep.int(ff$rname, widths)), as.data.frame(splits))
qname rname space start end width
1: 2218 chr1 1 24613476 24613517 42
2: 2067 chr1 2 87221030 87221073 44
3: 2129 chr1 3 79702717 79702760 44
4: 2165 chr1 4 43113438 43113481 44
5: 2086 chr1 5 52155089 52155092 4
6: 2086 chr1 5 52156014 52156053 40