Pick element in sublist and make it a column in a data frame

Question

I have a list of 2000 sublists each with 7 elements each of the same form. The fifth of these elements is a numeric vector with 40 elements. My aim is to get a data frame with 2000 rows and 40 columns, where each column represents one of the 40 elements in the fifth element of the 2000 sublists. Here is a downsized example:

sub_list_1 <- list(a = c(1:5), b = "b")
sub_list_2 <- list(a = c(6:10), b = "b")
sub_list_3 <- list(a = c(11:15), b = "b")
top_list <- list(sub_list_1, sub_list_2, sub_list_3)

Now, suppose I want a data frame with 5 columns and three rows, where each columns represents one of the five elements in a and each row represents one sublist. Thus, the output should be

 output
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    6    7    8    9   10
[3,]   11   12   13   14   15

Answer 1

One base R method is to use sapply to extract the values, which returns a matrix when each list extracted element has the same length, transpose the result which t.

 t(sapply(top_list, "[[", 1))
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    6    7    8    9   10
[3,]   11   12   13   14   15

You can also use as.data.frame to convert the matrix to a data.frame.

as.data.frame(t(sapply(top_list, "[[", 1)))
  V1 V2 V3 V4 V5
1  1  2  3  4  5
2  6  7  8  9 10
3 11 12 13 14 15

Answer 2

Another way is this one.

do.call(rbind, sapply(top_list, `[`, 1))

First, sapply extracts the first vector of each list, then do.call rbinds those vectors into a matrix.

Answer 3

Another approach , more efficient without using a loop , is to unlist the list and create a matrix:

vals <- unlist(top_list)
matrix(as.numeric(vals[vals!='b']),ncol=length(top_list[[1]]$a),byrow = TRUE)

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    6    7    8    9   10
[3,]   11   12   13   14   15