R Filling missing values with NA for a data frame - r

Question

I am currently trying to create a data-frame with the following lists
location <- list("USA","Singapore","UK")
organization <- list("Microsoft","University of London","Boeing","Apple")
person <- list()
date <- list("1989","2001","2018")
Jobs <- list("CEO","Chairman","VP of sales","General Manager","Director")
When I try and create a data-frame I get the (obvious) error that the lengths of the lists are not equal. I want to find a way to either make the lists the same length, or fill the missing data-frame entries with "NA". After doing some searching I have not been able to find a solution

Answer 1

Here are purrr (part of tidyverse) and base R solutions, assuming you just want to fill remaining values in each list with NA. I'm taking the maximum length of any list as len, then for each list doing rep(NA) for the difference between the length of that list and the maximum length of any list.
library(tidyverse)
location <- list("USA","Singapore","UK")
organization <- list("Microsoft","University of London","Boeing","Apple")
person <- list()
date <- list("1989","2001","2018")
Jobs <- list("CEO","Chairman","VP of sales","General Manager","Director")
all_lists <- list(location, organization, person, date, Jobs)
len <- max(lengths(all_lists))
With purrr::map_dfc, you can map over the list of lists, tack on NAs as needed, convert to character vector, then get a data frame of all those vectors cbinded in one piped call:
map_dfc(all_lists, function(l) {
c(l, rep(NA, len - length(l))) %>%
as.character()
})
#> # A tibble: 5 x 5
#> V1 V2 V3 V4 V5
#> <chr> <chr> <chr> <chr> <chr>
#> 1 USA Microsoft NA 1989 CEO
#> 2 Singapore University of London NA 2001 Chairman
#> 3 UK Boeing NA 2018 VP of sales
#> 4 NA Apple NA NA General Manager
#> 5 NA NA NA NA Director
In base R, you can lapply the same function across the list of lists, then use Reduce to cbind the resulting lists and convert it to a data frame. Takes two steps instead of purrr's one:
cols <- lapply(all_lists, function(l) c(l, rep(NA, len - length(l))))
as.data.frame(Reduce(cbind, cols, init = NULL))
#> V1 V2 V3 V4 V5
#> 1 USA Microsoft NA 1989 CEO
#> 2 Singapore University of London NA 2001 Chairman
#> 3 UK Boeing NA 2018 VP of sales
#> 4 NA Apple NA NA General Manager
#> 5 NA NA NA NA Director
For both of these, you can now set the names however you like.

Answer 2

You could do:
data.frame(sapply(dyem_list, "length<-", max(lengths(dyem_list))))
location organization person date Jobs
1 USA Microsoft NULL 1989 CEO
2 Singapore University of London NULL 2001 Chairman
3 UK Boeing NULL 2018 VP of sales
4 NULL Apple NULL NULL General Manager
5 NULL NULL NULL NULL Director
Where dyem_list is the following:
dyem_list <- list(
location = list("USA","Singapore","UK"),
organization = list("Microsoft","University of London","Boeing","Apple"),
person = list(),
date = list("1989","2001","2018"),
Jobs = list("CEO","Chairman","VP of sales","General Manager","Director")
)