Join two list in R by id

Question

I have something like this:

ids <- c("A","B","C")
ls <- list()
ls[[1]] <- c("aa","aaa")
ls[[2]] <- c("bb","bbb")
ls[[3]] <- c("cc","ccc")

and I would like to obtain something like the next:

data.frame(ids = c("A","A","B","B","C","C"), ls = c("aa","aaa","bb","bbb","cc","ccc"))

Would you know how to do it? The length of each element of the ls can vary, but the length of the ids and ls is the same. Dplyr and purrr(tidyverse) is Ok for me

Best

Answer 1

The simplest base R method is to use unlist and rep like this:

data.frame(ids=rep(ids, lengths(ls)), ls=unlist(ls))
  ids  ls
1   A  aa
2   A aaa
3   B  bb
4   B bbb
5   C  cc
6   C ccc

rep repeats the ids in varying lengths with the length of each list element calculated by lengths. Then the nested list is converted to a vector with unlist.

Answer 2

Like this?

library(tidyr)
gather(data.frame(setNames(ls, ids)), ids, ls)
  ids  ls
1   A  aa
2   A aaa
3   B  bb
4   B bbb
5   C  cc
6   C ccc

This works in the following way: first, I assign the names to the list:

named_list <- setNames(ls, ids)
named_list
$A
[1] "aa"  "aaa"
$B
[1] "bb"  "bbb"
$C
[1] "cc"  "ccc"

Then convert it to a dataframe:

test_df <- data.frame(named_list)
test_df
    A   B   C
1  aa  bb  cc
2 aaa bbb ccc

And then I reshape it using either tidyr::gather or reshape2::melt

gather(test_df)
  key value
1   A    aa
2   A   aaa
3   B    bb
4   B   bbb
5   C    cc
6   C   ccc

To supply your column names, just add ids and ls as above.

Answer 3

Here is one option with base R

setNames(stack(setNames(ls, LETTERS[1:3]))[2:1], c("ids", "ls"))
#   ids  ls
#1   A  aa
#2   A aaa
#3   B  bb
#4   B bbb
#5   C  cc
#6   C ccc