R：从“中的向量”创建列表；三角形“；形式

我发现很难表述这个问题，但我想找到一种聪明的方法（不使用循环）来获得以下结果：

> my.vector = letters[1:6]
> print(my.vector)
[1] "a" "b" "c" "d" "e" "f"
> 
> my.list = (rep(list(NA),6))
> for (i in 1:length(my.vector)){
+   x = my.vector[1:i]
+   my.list[[i]] = x
+ }
> print(my.list)
[[1]]
[1] "a"

[[2]]
[1] "a" "b"

[[3]]
[1] "a" "b" "c"

[[4]]
[1] "a" "b" "c" "d"

[[5]]
[1] "a" "b" "c" "d" "e"

[[6]]
[1] "a" "b" "c" "d" "e" "f"

提前感谢,

加布里埃尔。

我们可以使用

v1 <- my.vector[sequence(seq_along(my.vector))]
split(v1, cumsum(v1=='a'))

v1这里有一种方法（比@akrun更详细，但不依赖于原始向量中的实际值）


如果您想要一个
矩阵
而不是
列表
，您可以尝试：

> x <- t(replicate(length(my.vector), my.vector))
> x[upper.tri(x)] <- ""
> x
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,] "a"  ""   ""   ""   ""   ""  
[2,] "a"  "b"  ""   ""   ""   ""  
[3,] "a"  "b"  "c"  ""   ""   ""  
[4,] "a"  "b"  "c"  "d"  ""   ""  
[5,] "a"  "b"  "c"  "d"  "e"  ""  
[6,] "a"  "b"  "c"  "d"  "e"  "f" 

>x x[上三（x）]x
[,1] [,2] [,3] [,4] [,5] [,6]
[1，]“a”“”“”
[2，]a“b”
[3]“a”“b”“c”
[4，]a“b”c“d”
[5，]a“b”c“d”e
[6，]“a”“b”“c”“d”“e”“f”
您可以执行以下操作：

lapply(seq_along(my.vector), head, x = my.vector)

lapply(seq_along(my.vector), head, x = my.vector)