how to "spread" a list-column?

Question

Consider this simple example

mydf <- data_frame(regular_col = c(1,2),
                   normal_col = c('a','b'),
                   weird_col = list(list('hakuna', 'matata'),
                                 list('squash', 'banana')))

> mydf
# A tibble: 2 x 3
  regular_col normal_col weird_col 
        <dbl> <chr>      <list>    
1           1 a          <list [2]>
2           2 b          <list [2]>

I would like to extract the elements of weird_col (programmatically, the number of elements may change) so that each element is placed on a different column. That is, I expect the following output

> data_frame(regular_col = c(1,2),
+           normal_col = c('a','b'),
+           weirdo_one = c('hakuna', 'squash'),
+           weirdo_two = c('matata', 'banana'))
# A tibble: 2 x 4
  regular_col normal_col weirdo_one weirdo_two
        <dbl> <chr>      <chr>      <chr>     
1           1 a          hakuna     matata
2           2 b          squash     banana    

However, I am unable to do so in simple terms. For instance, using the classic unnest fails here, as it expands the dataframe instead of placing each element of the list in a different column.

> mydf %>% unnest(weird_col)
# A tibble: 4 x 3
  regular_col normal_col weird_col
        <dbl> <chr>      <list>   
1           1 a          <chr [1]>
2           1 a          <chr [1]>
3           2 b          <chr [1]>
4           2 b          <chr [1]>

Is there any solution in the tidyverse for that?

Answer 1

unnest develops lists and vectors vertically, and one row data frames horizontally. So what we can do is change your lists into data frames (with adequate column names) and unnest afterwards.

mydf %>% mutate(weird_col = map(weird_col,~ as_data_frame(
  setNames(.,paste0("weirdo_",1:length(.)))
  ))) %>% 
  unnest

# # A tibble: 2 x 4
#   regular_col normal_col weirdo_1 weirdo_2
#         <dbl>      <chr>    <chr>    <chr>
# 1           1          a   hakuna   matata
# 2           2          b   squash   banana