r - 如何绑定(bind)行而不丢失带有字符(0)的行？

Question

我有一个类似 L 的列表(来自矢量 split )。

L <- strsplit(c("1 5 9", "", "3 7 11", ""), " ")

# [[1]]
# [1] "1" "5" "9"
# 
# [[2]]
# character(0)
# 
# [[3]]
# [1] "3"  "7"  "11"
# 
# [[4]]
# character(0)

当我做一个普通的 rbind如下，我失去了所有的 character(0)行。

do.call(rbind, L)
#      [,1] [,2] [,3]
# [1,] "1"  "5"  "9" 
# [2,] "3"  "7"  "11"

我总是要做一个 lapply像下面这样还是我错过了什么？

do.call(rbind, lapply(L, function(x) 
    if (length(x) == 0)  rep("", 3) else x))
#      [,1] [,2] [,3]
# [1,] "1"  "5"  "9" 
# [2,] ""   ""   ""  
# [3,] "3"  "7"  "11"
# [4,] ""   ""   ""  

首选基本 R 答案。

Answer 1

这是此类场景的定义行为。如 ?rbind 中所写:

For cbind (rbind), vectors of zero length (including NULL) are ignored unless the result would have zero rows (columns), for S compatibility. (Zero-extent matrices do not occur in S3 and are not ignored in R.)

当您检查您的元素时，您会发现这是真的:

length(L[[1]])

[1] 3

length(L[[2]])

[1] 0

但是，如您所见，可能有多种解决方法。