'Is there a dplyr equivalent to data.table::rleid?
data.table offers a nice convenience function, rleid for run-length encoding:
library(data.table)
DT = data.table(grp=rep(c("A", "B", "C", "A", "B"), c(2, 2, 3, 1, 2)), value=1:10)
rleid(DT$grp)
# [1] 1 1 2 2 3 3 3 4 5 5
I can mimic this in base R with:
df <- data.frame(DT)
rep(seq_along(rle(df$grp)$values), times = rle(df$grp)$lengths)
# [1] 1 1 2 2 3 3 3 4 5 5
Does anyone know of a dplyr equivalent (?) or is the "best" way to create the rleid behavior with dplyr is to do something like the following
library(dplyr)
my_rleid = rep(seq_along(rle(df$grp)$values), times = rle(df$grp)$lengths)
df %>%
mutate(rleid = my_rleid)
Solution 1:[1]
If you want to use just base R and dplyr, the better way is to wrap up your own one or two line version of rleid() as a function and then apply that whenever you need it.
library(dplyr)
myrleid <- function(x) {
x <- rle(x)$lengths
rep(seq_along(x), times=x)
}
## Try it out
DT <- DT %>% mutate(rlid = myrleid(grp))
DT
# grp value rlid
# 1: A 1 1
# 2: A 2 1
# 3: B 3 2
# 4: B 4 2
# 5: C 5 3
# 6: C 6 3
# 7: C 7 3
# 8: A 8 4
# 9: B 9 5
#10: B 10 5
Solution 2:[2]
You can do it using the lag function from dplyr.
DT <-
DT %>%
mutate(rleid = (grp != lag(grp, 1, default = "asdf"))) %>%
mutate(rleid = cumsum(rleid))
gives
> DT
grp value rleid
1: A 1 1
2: A 2 1
3: B 3 2
4: B 4 2
5: C 5 3
6: C 6 3
7: C 7 3
8: A 8 4
9: B 9 5
10: B 10 5
Solution 3:[3]
A simplification (involving no additional package) of the approach used by the OP could be:
DT %>%
mutate(rleid = with(rle(grp), rep(seq_along(lengths), lengths)))
grp value rleid
1 A 1 1
2 A 2 1
3 B 3 2
4 B 4 2
5 C 5 3
6 C 6 3
7 C 7 3
8 A 8 4
9 B 9 5
10 B 10 5
Or:
DT %>%
mutate(rleid = rep(seq(ls <- rle(grp)$lengths), ls))
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Alex |
| Solution 3 |
