Given a named list of character vectors, output a new list where the unique elements of all character vectors reverse place with names of origin list

Question

This is a simplified example. My real dataset is very large, so the answer should be computationally inexpensive.

start.list <- list(cb_1 = c("VR_1", "VR_2", "VR_3"), cb_2 = c("VR_2", "VR_3", "VR_4"), 
cb_3 = c("VR_3", "VR_4", "VR_5"))

> start.list  
$cb_1  
[1] "VR_1" "VR_2" "VR_3"

$cb_2  
[1] "VR_2" "VR_3" "VR_4"

$cb_3  
[1] "VR_3" "VR_4" "VR_5"

Desired output list:

> output.list  
$VR_1  
[1] "cb_1"

$VR_2  
[1] "cb_1" "cb_2"

$VR_3  
[1] "cb_1" "cb_2" "cb_3"

$VR_4  
[1] "cb_2" "cb_3"

$VR_5  
[1] "cb_3"

You could do `x <- stack(start.list); split(as.character(x$ind), x$value)` — MrFlick, May 23 '21 at 05:52
Maybe some ideas here too: https://stackoverflow.com/questions/15263146/revert-list-structure — MrFlick, May 23 '21 at 05:56

score 0 · Answer 1 · answered May 23 '21 at 06:20

0

Using enframe and unnest -

tibble::enframe(start.list) %>%
  tidyr::unnest(value) %>%
  {split(.$name, .$value)}

#$VR_1
#[1] "cb_1"

#$VR_2
#[1] "cb_1" "cb_2"

#$VR_3
#[1] "cb_1" "cb_2" "cb_3"

#$VR_4
#[1] "cb_2" "cb_3"

#$VR_5
#[1] "cb_3"

answered May 23 '21 at 06:20

Ronak Shah

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Given a named list of character vectors, output a new list where the unique elements of all character vectors reverse place with names of origin list

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