如何根据连续的唯一值将向量拆分为列表

Question

我正在尝试查看是否有更好的方法将向量拆分为列表，以便将所有连续的唯一值放在一个组中。

请注意，当x 也是字符时，该方法必须有效。

#DATA
x = c(0, 0, 0, 7, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0, 7, 7, 7, 7)
x
#[1] 0 0 0 7 7 7 7 0 0 0 0 0 0 0 7 7 7 7

#DESIRED OUTPUT
L = list(c(0, 0, 0), c(7, 7, 7, 7), c(0, 0, 0, 0, 0, 0, 0), c(7, 7, 7, 7))
L
#[[1]]
#[1] 0 0 0

#[[2]]
#[1] 7 7 7 7

#[[3]]
#[1] 0 0 0 0 0 0 0

#[[4]]
#[1] 7 7 7 7

#CURRENT APPROACH
split_vector = 0
for (i in 2:length(x)){
     split_vector[i] = ifelse(x[i] != x[i-1], max(split_vector) + 1, split_vector[i-1])
}
split(x, split_vector)
#$`0`
#[1] 0 0 0

#$`1`
#[1] 7 7 7 7

#$`2`
#[1] 0 0 0 0 0 0 0

#$`3`
#[1] 7 7 7 7

Answer 1

这里有一些替代方案：

1) 使用rle 和rep 形成一个分组向量并在其上进行拆分。没有使用任何包。

split(x, with(rle(x), rep(seq_along(values), lengths)))

给予：

$`1`
[1] 0 0 0

$`2`
[1] 7 7 7 7

$`3`
[1] 0 0 0 0 0 0 0

$`4`
[1] 7 7 7 7

2) 使用 data.table 包中的rleid 更加容易：

library(data.table)
split(x, rleid(x))

Answer 2

tapply(x, cumsum(c(TRUE, diff(x) != 0)), identity)

$`1`
[1] 0 0 0

$`2`
[1] 7 7 7 7

$`3`
[1] 0 0 0 0 0 0 0

$`4`
[1] 7 7 7 7

# Character example
x <- rep(c("a", "b", "c", "a"), c(4, 3, 2, 4))
x

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

# Character version
tapply(x, cumsum(c(TRUE, x[-1] != x[-length(x)])), identity)

$`1`
[1] "a" "a" "a" "a"

$`2`
[1] "b" "b" "b"

$`3`
[1] "c" "c"

$`4`
[1] "a" "a" "a" "a"