使用 R data.table foverlaps() 或 IRanges 按预期计算重叠

Question

我在计算间隔重叠时遇到了困难，正如我所期望的那样。这是一个 R data.table，间隔由开始到结束定义：

> 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

以下是我将如何考虑这些间隔的重叠，从 0 到 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，我想用算法输出这个。类似于：

   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 的各种函数来执行此操作。

> 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

这些似乎都与计算某个时间间隔内的重叠无关。

查看 IRanges 也没有给出预期的重叠间隔计数：

> 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？

Answer 1

> # 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，但确实有 2 个重叠，因为 [8, 10] 与 [1, 10] 和 [5, 15] 重叠。

使用 R data.table foverlaps() 或 IRanges 按预期计算重叠

Counting overlaps as expected with R data.table foverlaps() or IRanges

r

overlap

intervals

iranges

data.table