r - 是否有等效于 data.table::rleid 的 dplyr？

Question

data.tablerleid为运行长度编码提供了一个很好的便利功能：

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

我可以在基础R上模仿这个：

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

有谁知道dplyr等效的（？）或者是创建rleid行为的“最佳”方法dplyr是执行以下操作

library(dplyr)

my_rleid = rep(seq_along(rle(df$grp)$values), times = rle(df$grp)$lengths)

df %>%
  mutate(rleid = my_rleid)

Answer 1

您可以这样做（当您同时加载data.table和dplyr时）：

DT <- DT %>% mutate(rlid = rleid(grp))

这给出了：

> 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

当您不想单独加载data.table时，您也可以使用（如评论中@DavidArenburg 所述）：

DT <- DT %>% mutate(rlid = data.table::rleid(grp))

正如@RichardScriven 在他的评论中所说，您可以复制/窃取它：

myrleid <- data.table::rleid

Answer 2

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

Answer 3

您可以lag使用dplyr.

DT <-
    DT %>%
    mutate(rleid = (grp != lag(grp, 1, default = "asdf"))) %>%
    mutate(rleid = cumsum(rleid))

给

> 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

Answer 4

OP 使用的方法的简化（不涉及额外的包）可能是：

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

或者：

DT %>%
 mutate(rleid = rep(seq(ls <- rle(grp)$lengths), ls))