R转换列表:值转换为名称，名称转换为值

Question

有没有一种方法可以在R中没有循环的情况下将list1转换成list2？

list1 <- list(s1=letters[1:4],
              s2=letters[c(1,2,4)],
              s3=letters[c(1,2,5)],
              s4=letters[1:6])
list2 <- list(a=c('s1', 's2', 's3', 's4'),
                b=c('s1', 's2', 's3', 's4'),
                c=c('s1', 's4'),
                d=c('s1', 's2', 's4'),
                e=c('s3', 's4'),
                f=c('s4'))
> list1
$s1
[1] "a" "b" "c" "d"
$s2
[1] "a" "b" "d"
$s3
[1] "a" "b" "e"
$s4
[1] "a" "b" "c" "d" "e" "f"

> list2
$a
[1] "s1" "s2" "s3" "s4"
$b
[1] "s1" "s2" "s3" "s4"
$c
[1] "s1" "s4"
$d
[1] "s1" "s2" "s4"
$e
[1] "s3" "s4"
$f
[1] "s4"

Answer 1

您可以通过使用[2:1]更改列顺序来使用stack和unstack

unstack(stack(list1)[2:1])
#$a
#[1] "s1" "s2" "s3" "s4"
#
#$b
#[1] "s1" "s2" "s3" "s4"
#
#$c
#[1] "s1" "s4"
#
#$d
#[1] "s1" "s2" "s4"
#
#$e
#[1] "s3" "s4"
#
#$f
#[1] "s4"

Answer 2

使用向量和split()

#Code
v1 <- names(sort(unlist(list1)))
v1 <- substring(v1,1,nchar(v1)-1)
v2 <- split(v1,sort(unlist(list1)))

输出：

v2
$a
[1] "s1" "s2" "s3" "s4"

$b
[1] "s1" "s2" "s3" "s4"

$c
[1] "s1" "s4"

$d
[1] "s1" "s2" "s4"

$e
[1] "s3" "s4"

$f
[1] "s4"

Answer 3

这可以通过嵌套两个列表操作来完成，即lapply()或purrr::map()。purrr::map_lgl()在这里特别有用，可以使代码更简单一些。

library(purrr)
res <-
  map(letters[1:6], function(x)
      names(list1)[map_lgl(list1, function(y)
         any(y %in% x))])
names(res) <- letters[1:6]
res
#> $a
#> [1] "s1" "s2" "s3" "s4"
#> 
#> $b
#> [1] "s1" "s2" "s3" "s4"
#> 
#> $c
#> [1] "s1" "s4"
#> 
#> $d
#> [1] "s1" "s2" "s4"
#> 
#> $e
#> [1] "s3" "s4"
#> 
#> $f
#> [1] "s4"