How to merge lists with identical column names to get their union

Question

Say I have N lists that all have the same column names. I want to merge these such that I get a resulting list with same columns, but now containing entries from all N list. Here is a MWE showing what I want:

  ls<-list()
ls[[1]]<-list("a"=1,
              "b"=2)
    ls[[2]]<-list("a"=3,
                  "b"=4)

#how to write a one-liner that produces lsTotal, which is the union of ls[[1]] and ls[[2]]?

lsTotal<-list("a"=c(1,3),
              "b"=c(2,4))

I found this thread, from which I can use Map(c, ls[[1]], ls[[2]]). However, writing it out is tedious if ls is very long. Is there a shortcut?

Note that it's less typing to build your example list in one line: `ls <- list(list(a=1, b=2), list(a=3, b=4))`. — lmo, Nov 08 '17 at 12:24

score 7 · Accepted Answer · answered Nov 08 '17 at 12:05

7

One option is tidyverse

library(purrr)
library(dplyr)
transpose(ls) %>%
        map(unlist)

Or use Map with do.call

do.call(Map, c(f=c, ls))
#$a
#[1] 1 3

#$b
#[1] 2 4

answered Nov 08 '17 at 12:05

akrun

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score 3 · Answer 2 · answered Nov 08 '17 at 13:13

3

Here is a simple two-liner with unlist and split.

# get a named vector
tmp <- unlist(ls)

# split on the names
split(unname(tmp), names(tmp))
$a
[1] 1 3

$b
[1] 2 4

answered Nov 08 '17 at 13:13

lmo

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score 2 · Answer 3 · answered Nov 08 '17 at 14:06

I know this question already has a few answers but another option is to use Reduce with your Map to apply the Map to each of the elements successively in the list:

Reduce(function(x,y) Map(c,x,y), ls)
#$a
#[1] 1 3

#$b
#[1] 2 4

Or for a more complicated example, the results are:

ls <- list(list(a=1, b=2), list(a=3, b=4), list(a=2,b=4), list(a=5,b=2), list(a=3,b=2))

#$a
#[1] 1 3 2 5 3

#$b
#[1] 2 4 4 2 2

How to merge lists with identical column names to get their union

3 Answers3