如何在列表中组合列表向量的成员

Question

I have the following list of list:

nfoo <- list(a = list(x = 1:3, y = 11:13), b = list(x = 1:3, y = 100:102))

It looks like this:

 > nfoo
$a
$a$x
[1] 1 2 3

$a$y
[1] 11 12 13


$b
$b$x
[1] 1 2 3

$b$y
[1] 100 101 102

What I want to do is to combine the list within the list, resulting:

$a
[1] 1 2 3 11 12 13

$b
[1] 1 2 3 100 101 102

How can I achieve that?

Answer 1

We can use lapply with unlist

lapply(nfoo, unlist, use.names = FALSE)

#$a
#[1]  1  2  3 11 12 13

#$b
#[1]  1   2   3 100 101 102

Using purrr , we can do that by using map and flatten.*

library(purrr)
map(nfoo, flatten_int)

Answer 2

If we know that each of the sublists of foo contain vectors x and y (which is true for the example in the question) then:

lapply(foo, with, c(x, y))

giving:

$a
[1]  1  2  3 11 12 13

$b
[1]   1   2   3 100 101 102

Answer 3

Use lapply with docall and the concatenate function (which also works with lists):

output <- lapply(nfoo, function(x) do.call(c, x))
output

$a
x1 x2 x3 y1 y2 y3 
 1  2  3 11 12 13 

$b
 x1  x2  x3  y1  y2  y3 
  1   2   3 100 101 102 

Answer 4

使用lapply遍历列表，然后unlist列表以将每个列表转换为向量：

lapply(nfoo, unlist)

Answer 5

library(tidyverse)
nfoo%>%map(unlist)

# $a
# x1 x2 x3 y1 y2 y3 
#  1  2  3 11 12 13 
# 
# $b
#  x1  x2  x3  y1  y2  y3 
#   1   2   3 100 101 102 

Answer 6

For the sake of completeness, there is another variant which uses Reduce() within lapply()

lapply(nfoo, Reduce, f = c)

 $a [1] 1 2 3 11 12 13 $b [1] 1 2 3 100 101 102 

Answer 7

Using do.call .

Map(do.call, nfoo, w="c")
# $a
# x1 x2 x3 y1 y2 y3 
#  1  2  3 11 12 13 
# 
# $b
# x1  x2  x3  y1  y2  y3 
#  1   2   3 100 101 102