Question

所以，我有一个列表，每个元素都是一个向量。我有另一个因素向量，我想用它来合并我的一些列表元素。例如，这里有一些数据：

> l <- list("0"=c("a", "b", "c", "a", "a"), "1"=c("c", "b", "c", "a", "c", "c", "c"), "2"=c("b", "b", "b"), "3"=c("d", "d", "a", "b", "d"))
> l
$`0`
[1] "a" "b" "c" "a" "a"

$`1`
[1] "c" "b" "c" "a" "c" "c" "c"

$`2`
[1] "b" "b" "b"

$`3`
[1] "d" "d" "a" "b" "d"

> f <- factor(c(1,2,2,1))
> f
[1] 1 2 2 1
Levels: 1 2

所以根据这个因素，我想合并元素（1,4）和（2,3），因为它们具有相同的因子标签。我的最终名单应如下所示：

list.final：

$`0`
[1] "a" "b" "c" "a" "a" "d" "d" "a" "b" "d"

$`1`
[1] "c" "b" "c" "a" "c" "c" "c" "b" "b" "b"

因此，只有两个元素，从而合并原始列表中的元素1,4和2,3。有没有办法使用一些应用函数来做到这一点？

Answer 1

  tapply(l, f, FUN=unlist,use.names=F)
  #$`1`
  #[1] "a" "b" "c" "a" "a" "d" "d" "a" "b" "d"

  #$`2`
  #[1] "c" "b" "c" "a" "c" "c" "c" "b" "b" "b"

Answer 2

您可以在因子上拆分列表，然后递归调用c：

> lapply(split(l, f), function(x) unname(c(x, recursive = TRUE)))
$`1`
 [1] "a" "b" "c" "a" "a" "d" "d" "a" "b" "d"

$`2`
 [1] "c" "b" "c" "a" "c" "c" "c" "b" "b" "b"

以类似的方式使用unlist

> lapply(split(l, f), function(x) unname(unlist(x)))
$`1`
 [1] "a" "b" "c" "a" "a" "d" "d" "a" "b" "d"

$`2`
 [1] "c" "b" "c" "a" "c" "c" "c" "b" "b" "b"

Answer 3

由于您只有两个级别，因此可以使用列表索引。

> L <- unname(l)
> list(`0` = unlist(L[f == 1]), `1` = unlist(L[f != 1]))
$`0`
 [1] "a" "b" "c" "a" "a" "d" "d" "a" "b" "d"

$`1`
 [1] "c" "b" "c" "a" "c" "c" "c" "b" "b" "b"

R：根据因子向量合并列表的元素

3 个答案: