如何在data.table non-equi join中保持join列不变？

Question

我试图删除列中data.frame的值posn不在另一个中给出的范围内的行data.frame,具有data.table非equi连接功能.

以下是我的数据的样子:

library(data.table)
df.cov <-
    structure(list(posn = c(1, 2, 3, 165, 1000), att = c("a", "b",
    "c", "d", "e")), .Names = c("posn", "att"), row.names = c(NA,
    -5L), class = "data.frame")
df.exons <-
    structure(list(start = c(2889, 2161, 277, 164, 1), end = c(3329,
    2826, 662, 662, 168)), .Names = c("start", "end"), row.names = c(NA,
    -5L), class = "data.frame")

setDT(df.cov)
setDT(df.exons)

df.cov
#    posn att
# 1:    1   a
# 2:    2   b
# 3:    3   c
# 4:  165   d
# 5: 1000   e
df.exons # ranges of `posn` to include
#    start  end
# 1:  2889 3329
# 2:  2161 2826
# 3:   277  662
# 4:   164  662
# 5:     1  168

这是我尝试过的:

df.cov[df.exons, on = .(posn >= start, posn <= end), nomatch = 0]
#    posn att posn.1
# 1:  164   d    662
# 2:    1   a    168
# 3:    1   b    168
# 4:    1   c    168
# 5:    1   d    168

您可以看到posn列中的列df.cov也已更改.预期结果如下所示:

#    posn att
# 1:  165   d
# 2:    1   a
# 3:    2   b
# 4:    3   c
# 5   165   d
# the row order doesn't matter. I'll sort by posn latter.
# It is also fine if the duplicated rows are removed, otherwise I'll do this in next step.

如何通过data.table非equi连接获得所需的输出？

Answer 1

你也可以使用%inrange%:

df.cov[posn %inrange% df.exons]

这导致:

   posn att
1:    1   a
2:    2   b
3:    3   c
4:  165   d

如您所见,这使得posn-column 的值保持不变.

另一个,虽然更长,但可能性:

df.exons[df.cov
         , on = .(start <= posn, end >= posn)
         , mult ='first'
         , nomatch = 0
         , .(posn = i.posn, att)][]