CODEWITHSUNDEEP
r
d.b
d.b

Reputation: 32548

How to split vector into list based on consecutive unique values

I'm trying to see if there is a better way to split vector into list in such a way that all consecutive unique values are put in one group.

Note that the method has to work when x is character too.

#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

Upvotes: 1

Views: 676

Answers (2)

G. Grothendieck
G. Grothendieck

Reputation: 269471

Here are some alternatives:

1) Use rle with rep to form a grouping vector and split on that. No packages are used.

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

giving:

$`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) Using rleid from the data.table package it is even easier:

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

Upvotes: 5

JasonWang
JasonWang

Reputation: 2434

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"

Upvotes: 3

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