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Есть ли способ преобразовать list1 в list2 без циклов в R?

 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"
 

Вы можете использовать stack и unstack , изменив порядок столбцов с помощью [2:1] :

 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"
 

Используя векторы и split() :

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

Вывод:

 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"
 

Это можно сделать путем вложения двух операций со списком, т.Е. lapply() Или purrr::map() . purrr::map_lgl() здесь особенно полезно, чтобы упростить код.

 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"
 

Опция с enframe

 library(tibble)
library(dplyr)
library(tidyr)
enframe(list1) %>%
     unnest(c(value)) %>%
      select(2:1) %>% 
      {split(.$name, .$value)}