@Seth 使用 data.table foverlaps 函数提供了解决交叉重叠问题的最快方法。然而，这个解决方案没有考虑到输入的床文件可能有重叠的范围，需要减少到单个区域。 @Martin Morgan 通过使用 GenomicRanges 包的解决方案解决了这个问题，该解决方案同时进行了相交和范围缩小。但是，Martin 的解决方案没有使用 foverlaps 函数。 @Arun 指出，表中不同行中的重叠范围目前无法使用 foverlaps。感谢提供的答案，以及对 stackoverflow 的一些额外研究，我想出了这个混合解决方案。
在每个文件中创建没有重叠区域的示例 BED 文件。
chr <- c(1:22,"X","Y","MT")
#bedA contains 5 million rows
bedA <- data.table(
CHROM = as.vector(sapply(chr, function(x) rep(x,200000))),
START = rep(as.integer(seq(1,200000000,1000)),25),
STOP = rep(as.integer(seq(500,200000000,1000)),25),
key = c("CHROM","START","STOP")
)
#bedB contains 500 thousand rows
bedB <- data.table(
CHROM = as.vector(sapply(chr, function(x) rep(x,20000))),
START = rep(as.integer(seq(200,200000000,10000)),25),
STOP = rep(as.integer(seq(600,200000000,10000)),25),
key = c("CHROM","START","STOP")
)
现在创建一个新的bed文件，其中包含bedA和bedB中的相交区域。
#This solution uses foverlaps
system.time(tmpA <- intersectBedFiles.foverlaps(bedA,bedB))
user system elapsed
1.25 0.02 1.37
#This solution uses GenomicRanges
system.time(tmpB <- intersectBedFiles.GR(bedA,bedB))
user system elapsed
12.95 0.06 13.04
identical(tmpA,tmpB)
[1] TRUE
现在，修改 bedA 和 bedB 使其包含重叠区域：
#Create overlapping ranges
makeOverlaps <- as.integer(c(0,0,600,0,0,0,600,0,0,0))
bedC <- bedA[, STOP := STOP + makeOverlaps, by=CHROM]
bedD <- bedB[, STOP := STOP + makeOverlaps, by=CHROM]
使用 foverlaps 或 GenomicRanges 函数测试具有重叠范围的床文件相交的时间。
#This solution uses foverlaps to find the intersection and then run GenomicRanges on the result
system.time(tmpC <- intersectBedFiles.foverlaps(bedC,bedD))
user system elapsed
1.83 0.05 1.89
#This solution uses GenomicRanges
system.time(tmpD <- intersectBedFiles.GR(bedC,bedD))
user system elapsed
12.95 0.04 12.99
identical(tmpC,tmpD)
[1] TRUE
获胜者：foverlaps！
使用的功能
这是基于 foverlaps 的函数，只有在存在重叠范围（使用 rowShift 函数检查）时才会调用 GenomicRanges 函数 (reduceBed.GenomicRanges)。
intersectBedFiles.foverlaps <- function(bed1,bed2) {
require(data.table)
bedKey <- c("CHROM","START","STOP")
if(nrow(bed1)>nrow(bed2)) {
bed <- foverlaps(bed1, bed2, nomatch = 0)
} else {
bed <- foverlaps(bed2, bed1, nomatch = 0)
}
bed[, START := pmax(START, i.START)]
bed[, STOP := pmin(STOP, i.STOP)]
bed[, `:=`(i.START = NULL, i.STOP = NULL)]
if(!identical(key(bed),bedKey)) setkeyv(bed,bedKey)
if(any(bed[, STOP+1 >= rowShift(START), by=CHROM][,V1], na.rm = T)) {
bed <- reduceBed.GenomicRanges(bed)
}
return(bed)
}
rowShift <- function(x, shiftLen = 1L) {
#Note this function was described in this thread:
#http://stackoverflow.com/questions/14689424/use-a-value-from-the-previous-row-in-an-r-data-table-calculation
r <- (1L + shiftLen):(length(x) + shiftLen)
r[r<1] <- NA
return(x[r])
}
reduceBed.GenomicRanges <- function(bed) {
setnames(bed,colnames(bed),bedKey)
if(!identical(key(bed),bedKey)) setkeyv(bed,bedKey)
grBed <- makeGRangesFromDataFrame(bed,
seqnames.field = "CHROM",start.field="START",end.field="STOP")
grBed <- reduce(grBed)
grBed <- data.table(
CHROM=as.character(seqnames(grBed)),
START=start(grBed),
STOP=end(grBed),
key = c("CHROM","START","STOP"))
return(grBed)
}
此函数严格使用 GenomicRanges 包，产生相同的结果，但比 foverlaps 函数慢约 10 倍。
intersectBedFiles.GR <- function(bed1,bed2) {
require(data.table)
require(GenomicRanges)
bed1 <- makeGRangesFromDataFrame(bed1,
seqnames.field = "CHROM",start.field="START",end.field="STOP")
bed2 <- makeGRangesFromDataFrame(bed2,
seqnames.field = "CHROM",start.field="START",end.field="STOP")
grMerge <- suppressWarnings(intersect(bed1,bed2))
resultTable <- data.table(
CHROM=as.character(seqnames(grMerge)),
START=start(grMerge),
STOP=end(grMerge),
key = c("CHROM","START","STOP"))
return(resultTable)
}
使用 IRanges 的额外比较
我找到了一种使用 IRanges 折叠重叠区域的解决方案，但它比 GenomicRanges 慢 10 倍以上。
reduceBed.IRanges <- function(bed) {
bed.tmp <- bed
bed.tmp[,group := {
ir <- IRanges(START, STOP);
subjectHits(findOverlaps(ir, reduce(ir)))
}, by=CHROM]
bed.tmp <- bed.tmp[, list(CHROM=unique(CHROM),
START=min(START),
STOP=max(STOP)),
by=list(group,CHROM)]
setkeyv(bed.tmp,bedKey)
bed[,group := NULL]
return(bed.tmp[, -(1:2)])
}
system.time(bedC.reduced <- reduceBed.GenomicRanges(bedC))
user system elapsed
10.86 0.01 10.89
system.time(bedD.reduced <- reduceBed.IRanges(bedC))
user system elapsed
137.12 0.14 137.58
identical(bedC.reduced,bedD.reduced)
[1] TRUE