How can I make a list-column that is a subset of another list-column in R?

Question

I have a data frame with one column as a list-column, ie it is a column that has, for each row, two vectors contained in that column. I would like to be able to make another column in my data frame that is also a list-column, but that only contains a single sub-list (rather than two), and I would like that list to be the first three elements of one of the sub-lists of the column with two sub-lists.

A simple reproducible example is provided below:

df <- data.frame(state = c(rep("Alabama", 5), rep("Alaska", 5), rep("Arizona", 5), rep("Arkansas", 5), rep("California", 5)),
           letter = c("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y"),
           freq = c(8, 7, 4, 3, 1, 19, 15, 7, 4, 2, 10, 6, 3, 2, 2, 11, 10, 10, 5, 4, 50, 33, 22, 11, 1))
df <- nest(df, letter_list = c(letter, freq))

In the context of this reprex, I would like to have a third column in df that has, for each state, a list of the first three elements of letter (which is contained in letter_list ).

I have attempted to use purrr functions, such as map() , in conjunction with the head() function to mutate a new variable, but this has been unsuccessful; my new column is populated with lists of length 0.

If possible, a solution using the tidyverse would be ideal.

Any help would be greatly appreciated!

Answer 1

Use map to loop over the list column, select the 'letter', get the first 3 with either Extract ( [ ) or use slice_head

library(dplyr)
library(purrr)
df %>%
    mutate(letter_new = map(letter_list, ~
         .x %>%
         select(letter) %>% 
         slice_head(n = 3) %>% 
         pull(letter)))

-output

# A tibble: 5 × 3
  state      letter_list      letter_new
  <chr>      <list>           <list>    
1 Alabama    <tibble [5 × 2]> <chr [3]> 
2 Alaska     <tibble [5 × 2]> <chr [3]> 
3 Arizona    <tibble [5 × 2]> <chr [3]> 
4 Arkansas   <tibble [5 × 2]> <chr [3]> 
5 California <tibble [5 × 2]> <chr [3]> 

NOTE: if it needs to be kept as tibble , we don't need the last pull step

---

Or using base R

df$letter_new <- lapply(df$letter_list, \(x) head(x$letter, 3))