Transform dataframe into list by condition in R/dplyr

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I have a dataframe like this:

> df
Person    a    b    c    d
John      1    0    1    1
James     0    1    1    0
Keith     1    0    0    0
Boris     0    1    0    0
...

And I need to transform it into a list of vectors with names of elements being corresponding to column names of dataframe and elements of the list being names of persons that have 1 in a column. For the above example the list should be like this:

> result_list
$a
[1] "John" "Keith"

$b
[1] "James" "Boris"

$c
[1] "John" "James"

$d
[1] "John"

Moving on to what I know, the vector of "switched" names for every column can be obtained like this:

df$Person[which(df$a == 1)]

But I'm not sure how to iterate over it properly, and I think there might be a tidy solution for the task that utilizes dplyr and purrr.

purrr dplyr r
1 Answers

We can reshape to 'long' format and split

library(dplyr)
library(tidyr)
df %>% 
    pivot_longer(cols = -Person) %>% 
    filter(value == 1) %>% 
    {split(.$Person, .$name)}

-output

$a
[1] "John"  "Keith"

$b
[1] "James" "Boris"

$c
[1] "John"  "James"

$d
[1] "John"

data

df <- structure(list(Person = c("John", "James", "Keith", "Boris"), 
    a = c(1L, 0L, 1L, 0L), b = c(0L, 1L, 0L, 1L), c = c(1L, 1L, 
    0L, 0L), d = c(1L, 0L, 0L, 0L)), class = "data.frame", row.names = c(NA, 
-4L))
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