Modify/Add column to nested tibble

Question

I have a tibble with variables/columns that include lists of tibbles of differing shapes. I would like to add a variable/column to each (sub-)tibble in one of the variables.

E.g. such data

library("tibble")
aaa <- tibble(qq = c(1,2,3),
              ww = list(tibble(a = c(1,2), s = c("as", "df")),
                        tibble(a = 3, s ="q"),
                        tibble(a = 4, s = "e")),
              ee = list(tibble(a = c(1,2), s = c("as", "df")),
                        tibble(a = c(3,6), s = c("qz", "wz")),
                        tibble(a = 4, s = "ez")))

Which look like:

 > aaa
# A tibble: 3 x 3
     qq ww               ee              
  <dbl> <list>           <list>          
1  1.00 <tibble [2 × 2]> <tibble [2 × 2]>
2  2.00 <tibble [1 × 2]> <tibble [2 × 2]>
3  3.00 <tibble [1 × 2]> <tibble [1 × 2]>

and

> aaa$ee
[[1]]
# A tibble: 2 x 2
      a s    
  <dbl> <chr>
1  1.00 as   
2  2.00 df   

[[2]]
# A tibble: 2 x 2
      a s    
  <dbl> <chr>
1  3.00 qz   
2  6.00 wz   

[[3]]
# A tibble: 1 x 2
      a s    
  <dbl> <chr>
1  4.00 ez   

Let's say I want to create a new variable in each tibble in aaa$ee called lll, which is the first letter of each string in variable s in that table.

I can get these with lapply, e.g.:

lll <- lapply(X = aaa$ee,
       FUN = function(z){substr(z$s, 1,1)})

Resulting in

> lll
[[1]]
[1] "a" "d"

[[2]]
[1] "q" "w"

[[3]]
[1] "e"

But how do I bind each vector in this list as a column in each tibble in aaa$ee?

To produce a result that should look something like

> aaa$ee
 [[1]]
 # A tibble: 2 x 2
      a s    
  <dbl> <chr> <chr>
1  1.00 as    a
2  2.00 df    d

[[2]]
# A tibble: 2 x 2
      a s    
  <dbl> <chr> <chr>
1  3.00 qz    q
2  6.00 wz    w

[[3]]
# A tibble: 1 x 2
      a s    
  <dbl> <chr> <chr>
1  4.00 ez    e

Answer 1

We can loop through the 'ee' column with map and mutate to create the 'new' column

library(purrr)
out <- aaa %>% 
         mutate(ee = map(ee, ~ .x %>% 
                              mutate(new = substr(s, 1, 1))))


out$ee
#[[1]]
## A tibble: 2 x 3
#      a s     new  
#  <dbl> <chr> <chr>
#1     1 as    a    
#2     2 df    d    

#[[2]]
# A tibble: 2 x 3
#      a s     new  
#  <dbl> <chr> <chr>
#1     3 qz    q    
#2     6 wz    w 

#[[3]]
# A tibble: 1 x 3
#      a s     new  
#  <dbl> <chr> <chr>
#1     4 ez    e    

---

If we are using the lapply method (using base R), then with transform create a new column and assign it to the 'ee'

aaa$ee <- lapply(aaa$ee, transform, new = substr(s, 1, 1))