R - Transform Data.frame elements into a single column tibble()

Question

I am looking for a way to improve my code and move away from loops in R.

Background:

I have a list of Data.frames. Each Data.frame has 4 elements - many of them are NULL. I need to preserve the NULLs and replace them with an NA as I enlist / flatten the the list to then add it as a column to a 'tibble'. I know if I access the elements using map(list, n) it will give me the start point for the loop. If I flatten() this it removes the NULL objects and I won't be able to add it to another tibble I am using as the order isn't preserved.

Input structure is below:

[[1]]
  item1 item2 item3 item4
1  aaaa  bbbb  cccc  dddd

[[2]]
data frame with 0 columns and 0 rows

[[3]]
data frame with 0 columns and 0 rows

[[4]]
  item1 item2 item3 item4
1  ffff  gggg  hhhh  kkkk

Solution so far:

I have written the following loop:

element_tibble = tibble()

for (i in 1:length(list_of_dfs)){
    element = list_of_dfs[[i]]
    if (is.null(element) == TRUE) {
        element = NA
        }
    else {
        element = element
        }
    row = c(element = element)
    element_tibble = rbind(element_tibble, row)
}

The intended output is a one column tibble that is the length of the list with NA preserved for NULL elements in the original list.

# A tibble: n x 1
  element
  <chr>  
1 item2  
2 NA       
3 NA     
4 item2 
etc 

I know the loop is slow but I can't find another way to access the element in the Data.frame then transform it into a usable(flat) column for addition to a tibble as another element for each observation.

Any advice would be greatly appreciated.

Thanks

James

Answer 1

Base R option using sapply -

df <- data.frame(item1 = 'aaaa', item2 = 'bbbb', item3 = 'ccc', item4 = 'ddd')
list_of_dfs <- list(df, data.frame(), df)

result <- data.frame(element = sapply(list_of_dfs, function(x) 
                               if(nrow(x)) x[[2]] else NA))
result

#  element
#1    bbbb
#2    <NA>
#3    bbbb