递归绑定多个列表中的项目 [英] cbind items from multiple lists recursively

问题描述

给出三个(或n列表):

one   <- list(a=1:2,b="one")
two   <- list(a=2:3,b="two")
three <- list(a=3:4,b="three")

cbind找到n列表中的每个列表项的更有效方法是什么?

What would be a more efficient way of cbindind each list item across the n lists, to get this result?

mapply(cbind,mapply(cbind,one,two,SIMPLIFY=FALSE),three,SIMPLIFY=FALSE)

$a
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    2    3    4

$b
     [,1]  [,2]  [,3]   
[1,] "one" "two" "three"

当n为2或3时，此方法可以正常工作，但很快就会变得非常复杂. 在这方面有更有效的变化吗?我在S.O上看到了类似的问题.但一直在努力适应它们.

This works okay when n is 2 or 3 but is quickly going to become ludicrously complex. Is there a more efficient variation on this? I have seen similar questions on S.O. but have struggled to adapt them.

推荐答案

还是这样:

mapply(cbind, one, two, three)

或者像这样:

mylist <- list(one, two, three)
do.call(mapply, c(cbind, mylist))

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