R中的交错列表 [英] Interleave lists in R

问题描述

假设我在 R 中有两个列表，长度不一定相等，例如:

Let's say that I have two lists in R, not necessarily of equal length, like:

 a <- list('a.1','a.2', 'a.3')
 b <- list('b.1','b.2', 'b.3', 'b.4')

构造交错元素列表的最佳方法是什么，一旦添加了较短列表的元素，较长列表的其余元素将追加到末尾?，​​例如:

What is the best way to construct a list of interleaved elements where, once the element of the shorter list had been added, the remaining elements of the longer list are append at the end?, like:

interleaved <- list('a.1','b.1','a.2', 'b.2', 'a.3', 'b.3','b.4')

不使用循环.我知道 mapply 适用于两个列表长度相等的情况.

without using a loop. I know that mapply works for the case where both lists have equal length.

推荐答案

这是一种方法:

idx <- order(c(seq_along(a), seq_along(b)))
unlist(c(a,b))[idx]

# [1] "a.1" "b.1" "a.2" "b.2" "a.3" "b.3" "b.4"

正如@James 指出的，既然你需要一个列表，你应该这样做:

As @James points out, since you need a list back, you should do:

(c(a,b))[idx]

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