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比较并找到 R 中的重叠范围

[英]comparing and finding overlap range in R

我有两个表,每个表都包含数字范围。 一张桌子是另一张桌子的细分。 我想在第一个表中创建二进制列,显示它们在哪个范围内重叠。

例如:

df1:
start1   end1
 1       6
 6       8
 9       12
 13      15
 15      19
 19      20

df2:
start2   end2
 2        4
 9        11
 14       18

结果:结果是第一个带有显示重叠是否存在的列的表。

  start1   end1   overlap
     1       6       1
     6       8       0
     9       12      1
     13      15      1
     15      19      1
     19      20      0

谢谢。

你也可以尝试foverlapsdata.table

library(data.table)
setkey(setDT(df1), start1, end1)
setkey(setDT(df2), start2, end2)
df1[,overlap:=foverlaps(df1, df2, which=TRUE)[, !is.na(yid),]+0]
df1
#   start1 end1 overlap
#1:      1    6       1
#2:      6    8       0
#3:      9   12       1
#4:     13   15       1
#5:     15   19       1
#6:     19   20       0

有了IRanges

library(IRanges)
ir1 = with(df1, IRanges(start1, end1))
ir2 = with(df2, IRanges(start2, end2))
df1$overlap = countOverlaps(ir1, ir2) != 0

如果这是基因组数据,那么GenomicRanges包是合适的。

这是一种基于生成序列的方法:

nums <- unlist(apply(df2, 1, Reduce, f = seq))

df1$overlap <- as.integer(apply(df1, 1, function(x) any(seq(x[1], x[2]) %in% nums)))
#   start1 end1 overlap
# 1      1    6       1
# 2      6    8       0
# 3      9   12       1
# 4     13   15       1
# 5     15   19       1
# 6     19   20       0

您可以使用 ivs package ,它是 package,专门用于区间向量。 iv_overlaps()返回一个逻辑向量,该向量指定df1中列的每个间隔是否与df2中的任何间隔重叠。

library(dplyr)
library(ivs)

df1 <- tribble(
  ~start1, ~end1,
  1,       6,
  6,       8,
  9,       12,
  13,      15,
  15,      19,
  19,      20
)

df2 <- tribble(
  ~start2,   ~end2,
  2,        4,
  9,        11,
  14,       18
)

df1 <- mutate(df1, range1 = iv(start1, end1), .keep = "unused")
df2 <- mutate(df2, range2 = iv(start2, end2), .keep = "unused")

df1 %>%
  mutate(any_overlap = iv_overlaps(range1, df2$range2))
#> # A tibble: 6 × 2
#>      range1 any_overlap
#>   <iv<dbl>> <lgl>      
#> 1    [1, 6) TRUE       
#> 2    [6, 8) FALSE      
#> 3   [9, 12) TRUE       
#> 4  [13, 15) TRUE       
#> 5  [15, 19) TRUE       
#> 6  [19, 20) FALSE

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