How to group_by and summarise vectors inside a tibble into a single vector?

July 6, 2022


tibble(x = rep(1:3, 2),
       y = list(1:5, 1:10, 10:20, 20:40, 1:50, 5:10)) -> df

df
#> # A tibble: 6 × 2
#>       x y         
#>   <int> <list>    
#> 1     1 <int [5]> 
#> 2     2 <int [10]>
#> 3     3 <int [11]>
#> 4     1 <int [21]>
#> 5     2 <int [50]>
#> 6     3 <int [6]>

I want to group_by ‘x’ and summmarise the vectors of each group into a single vector. I tried using c(), but it didn’t help.

df %>% 
  group_by(x) %>% 
  summarise(z = c(y))
#> `summarise()` has grouped output by 'x'. You can override using the `.groups`
#> argument.
#> # A tibble: 6 × 2
#> # Groups:   x [3]
#>       x z         
#>   <int> <list>    
#> 1     1 <int [5]> 
#> 2     1 <int [21]>
#> 3     2 <int [10]>
#> 4     2 <int [50]>
#> 5     3 <int [11]>
#> 6     3 <int [6]>

I also want a union of elements in a group or any other similar function applied to these kinds of datasets.

df %>% 
  group_by(x) %>% 
  summarise(z = union(y))
#> Error in `summarise()`:
#> ! Problem while computing `z = union(y)`.
#> ℹ The error occurred in group 1: x = 1.
#> Caused by error in `base::union()`:
#> ! argument "y" is missing, with no default

>Solution :

If you want the data to remain nested, you can do

df %>% 
  group_by(x) %>% 
  summarise(z = list(unlist(y)))

The c() function won’t work because it’ doesn’t unnest-lists. For example, compare

c(list(1:3, 4:5))
unlist(list(1:3, 4:5))

The c function doesn’t return a single vector. But unlist does. This matters because your function will recieve a list of matching row values when you use summarize.

Also note that if you leave off the list(), the values don’t be nested anymore

df %>% 
  group_by(x) %>% 
  summarise(z = unlist(y))
#        x     z
#    <int> <int>
#  1     1     1
#  2     1     2
#  3     1     3
#  4     1     4
#  5     1     5
#  6     1    20
#  7     1    21
#  ...