grouped data frame to list

Question

I've got a data frame that contains names that are grouped, like so:

df <- data.frame(group = rep(letters[1:2], each=2),
                 name = LETTERS[1:4])
> df
  group name
1     a    A
2     a    B
3     b    C
4     b    D

I would like to convert this into a list that is keyed on the group names and contains the names. Example output:

df_out <- list(a=c('A', 'B'),
               b=c('C', 'D'))

> df_out
$a
[1] "A" "B"

$b
[1] "C" "D"

This is not a new question but I would like to do this wholly within the tidyverse.

Answer 1

There is no such function yet in the tidyverse as far as I know. Thus, you will have to write your own:

split_tibble <- function(tibble, col = 'col') tibble %>% split(., .[, col])

Then:

dflist <- split_tibble(df, 'group')

results in a list of dataframes:

> dflist
$a
  group name
1     a    A
2     a    B

$b
  group name
3     b    C
4     b    D

> sapply(dflist, class)
           a            b 
"data.frame" "data.frame"

To get the desired output, you'll have to extend the function a bit:

split_tibble <- function(tibble, column = 'col') {
  tibble %>% split(., .[,column]) %>% lapply(., function(x) x[,setdiff(names(x),column)])
}

Now:

split_tibble(df, 'group')

results in:

$a
[1] A B
Levels: A B C D

$b
[1] C D
Levels: A B C D

Considering the alternatives in the comments and both answers, leads to the following conclusion: using the base R alternative split(df$name, df$group) is much wiser.

Answer 2

Use tidyverse

library(tidyr)
library(dplyr)

df$ID <- 1:nrow(df)  # unique variable
lst <- df %>% spread(group, name) %>% select(-ID) %>% as.list()
lapply(lst, function(x) x[!is.na(x)])