用R data.table foverlaps()或IRanges按预期计数重叠 [英] Counting overlaps as expected with R data.table foverlaps() or IRanges

问题描述

我很难计算间隔的重叠.这是一个R data.table，其间隔由开始到结束定义:

I'm having difficulty counting overlaps of intervals as I would expect. Here is an R data.table with intervals defined by start to end:

> library(data.table)
> dt1 = data.table(start=c(1, 5, 3), end=c(10, 15, 8))
> print(dt1)
   start end
1:     1  10
2:     5  15
3:     3   8

在这里，我将如何考虑这些从0到20的时间间隔的重叠:

Here is how I would consider overlaps for these intervals, from 0 to 20:

[0, 1]: 0 (there are no intervals here)
[1, 3]: 1 (there is only one interval here, from [1, 10])
[3, 5]: 2 (two intervals here, both [1, 10] and [3, 8])
[5, 8]: 3
[8, 10]: 1
[10, 15]: 1
[15, 20]: 0

所以，我想以算法方式输出它.像这样:

So, I would like to algorithmically output this. Something like:

   start end  overlaps
1:     0  1   0
2:     1  3   1
3:     3  5   2
4:     5  8   3      
5:     8  10  2      
6:    10  15  1      
7:    15  20  0   

但是，我无法找到如何使用R data.table中的foverlaps()或IRanges的各种功能来执行此操作.

However, I cannot find out how to do this with foverlaps() in R data.table, or the various functions of IRanges.

> setkey(dt1, start, end)
> foverlaps(dt1, dt1, type="any")
   start end i.start i.end
1:     1  10       1    10
2:     3   8       1    10
3:     5  15       1    10
4:     1  10       3     8
5:     3   8       3     8
6:     5  15       3     8
7:     1  10       5    15
8:     3   8       5    15
9:     5  15       5    15
> foverlaps(dt1, dt1, type="within")
   start end i.start i.end
1:     1  10       1    10
2:     1  10       3     8
3:     3   8       3     8
4:     5  15       5    15

为了计算某个时间间隔内的重叠，这两个元素似乎都不相关.

Neither of these appears to be relevant in order to calculate overlaps over some interval.

查看IRanges并不能完全给出预期的重叠间隔计数:

Looking into IRanges also doesn't quite give the expected overlapping interval counts:

> library(IRanges)
> range1
IRanges object with 3 ranges and 0 metadata columns:
          start       end     width
      <integer> <integer> <integer>
  [1]         1        10        10
  [2]         3         8         6
  [3]         5        15        11
> countOverlaps(range1, range1)
[1] 3 3 3
> countOverlaps(range1, range1, type="within")
[1] 1 2 1

如何计算重叠间隔?

推荐答案

> # Where do the 0 and the 20 come from?
> points <- c(0, sort(c(dt1$start, dt1$end)), 20)
> x <- do.call(IRanges,
+              transpose(Map(c, start=head(points, -1), end=tail(points, -1))))
> x
IRanges object with 7 ranges and 0 metadata columns:
          start       end     width
      <integer> <integer> <integer>
  [1]         0         1         2
  [2]         1         3         3
  [3]         3         5         3
  [4]         5         8         4
  [5]         8        10         3
  [6]        10        15         6
  [7]        15        20         6
> y <- do.call(IRanges, dt1)
> y
IRanges object with 3 ranges and 0 metadata columns:
          start       end     width
      <integer> <integer> <integer>
  [1]         1        10        10
  [2]         3         8         6
  [3]         5        15        11
> countOverlaps(x, y, type="within")
[1] 0 1 2 3 2 1 0

第5个结果略有不同，但是确实存在2个重叠，因为[8，10]与[1，10]和[5，15]重叠.

There is a slight difference in the 5th result, but there are indeed 2 overlaps as [8, 10] overlaps with [1, 10] and [5, 15].

这篇关于用R data.table foverlaps()或IRanges按预期计数重叠的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！