Merge duplicate GRanges ranges

GRanges with 6 ranges and 0 metadata columns: seqnames ranges strand <Rle> <IRanges> <Rle> [1] Chr1 [2, 24] + [2] Chr1 [2, 25] + [3] Chr1 [2, 25] + [4] Chr1 [4, 25] + [5] Chr1 [5, 25] + [6] Chr1 [6, 21] + --- seqlengths: Chr1 NA

GRanges with 5 ranges and 1 metadata column: seqnames ranges strand | hits <Rle> <IRanges> <Rle> | <numeric> [1] Chr1 [2, 24] + | 1 [2] Chr1 [2, 25] + | 2 [3] Chr1 [4, 25] + | 1 [4] Chr1 [5, 25] + | 1 [5] Chr1 [6, 21] + | 1 --- seqlengths: Chr1 NA

6.4 years ago by

Devon Ryan ♦ 98k

Freiburg, Germany

Devon Ryan ♦ 98k wrote:

You're looking for the countOverlaps() function:

g <- GRanges(seqnames=c(rep("Chr1",6)), strand=rep("+",6), ranges=IRanges(start=c(2,2,2,4,5,6), end=c(24,25,25,25,25,21)))
g2 <- unique(g)
g2$hits<- countOverlaps(g2, g, type="equal")

You could also run countOverlaps(g,g,type="equal") first and then use unique(). You'll get the same result either way.

ADD COMMENT • link modified 6.4 years ago • written 6.4 years ago by Devon Ryan ♦ 98k

I am testing your approach on a sorted 5 million ranges object, it took hours before Rstudio encountered a fetal error. BTW, countOverlaps is really RAM consuming, it ate up all RAM (64G) on my computer.

ADD REPLY • link written 6.4 years ago by ifreecell • 180

Yeah, I imagine that that could prove pretty memory heavy, since I don't think it assumes that things are sorted. You might split() by chromosome and then use lapply(). Perhaps that'll be a bit more reasonable.

ADD REPLY • link written 6.4 years ago by Devon Ryan ♦ 98k

There is only one chromosome in this GRanges, I don't think split() will help. I wrote the GRanges to a bed file, and it took one second for uniq -c to compute. I really appreciate it if the entire workflow can be done in Bioconductor, is there other workaround?

ADD REPLY • link written 6.4 years ago by ifreecell • 180

Is it the unique() step that takes forever or just countOverlaps (or both)? I would guess that it's just the countOverlaps() function. Since you only ever care about exact overlaps/matches, you might be best off writing a custom function to do this. At least if you have to do this more than once or twice.

ADD REPLY • link written 6.4 years ago by Devon Ryan ♦ 98k

You are right, it's just the countOverlaps() function. Thank you anyway, I will try to write a custom function.

ADD REPLY • link written 6.4 years ago by ifreecell • 180

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