我有一个曲棍球比赛数据集。我的目标是创建一个子集，在这个子集中，球员不会把冰球翻转过来。
数据集包含一个标记为"营业额"的二进制列;如果比赛导致失误，则为1;如果比赛成功，则为0。我的目标是为营业额列中连续为0的每个行子集创建一个唯一的id。
# Example data:
set.seed(33)
x <- runif(20)
y <- runif(20)
df <-as.data.frame(cbind(x,y))
df$turnover = ifelse(x<0.5,1,0)
#desired output:
play_id <- c(1,2,3,4,4,4,5,6,7,8,9,10,11,12,13,13,13,14,15,16)
df$play_id <- play_id
df
虽然有点长，但只有dplyr。(虽然data.table::rleid()也是我最喜欢的功能)
df %>% group_by(turnover, play_id = cumsum(turnover)) %>%
mutate(play_id = ifelse(turnover == 0 & as.numeric(row_number()) == 1, 1, turnover)) %>%
ungroup() %>% mutate(play_id = cumsum(play_id))
# A tibble: 20 x 4
x y turnover play_id
<dbl> <dbl> <dbl> <dbl>
1 0.446 0.566 1 1
2 0.395 0.0427 1 2
3 0.484 0.488 1 3
4 0.919 0.351 0 4
5 0.844 0.970 0 4
6 0.517 0.790 0 4
7 0.437 0.597 1 5
8 0.343 0.357 1 6
9 0.0155 0.759 1 7
10 0.118 0.261 1 8
11 0.691 0.494 0 9
12 0.260 0.793 1 10
13 0.225 0.443 1 11
14 0.342 0.333 1 12
15 0.782 0.626 0 13
16 0.843 0.136 0 13
17 0.775 0.549 0 13
18 0.387 0.922 1 14
19 0.136 0.276 1 15
20 0.900 0.917 0 16
只有baseR
df$play_id <- cumsum(replace(df$turnover, with(rle(df$turnover != 0), 1+cumsum(lengths)[values]), 1))
x y turnover play_id
1 0.44594048 0.56645266 1 1
2 0.39465031 0.04273416 1 2
3 0.48372887 0.48831925 1 3
4 0.91887596 0.35122322 0 4
5 0.84388144 0.96966171 0 4
6 0.51734962 0.78957889 0 4
7 0.43712500 0.59663026 1 5
8 0.34319822 0.35667053 1 6
9 0.01551696 0.75870834 1 7
10 0.11799116 0.26105077 1 8
11 0.69098590 0.49398192 0 9
12 0.26048568 0.79289819 1 10
13 0.22505121 0.44299896 1 11
14 0.34238622 0.33326556 1 12
15 0.78188794 0.62576846 0 13
16 0.84324669 0.13619319 0 13
17 0.77474887 0.54868278 0 13
18 0.38719298 0.92183168 1 14
19 0.13576507 0.27603364 1 15
20 0.90035758 0.91741393 0 16
以下是data.table的一种方法:
library(data.table)
setDT(df)[, tmp := if(first(turnover) == 0)
replace(turnover, 1, 1) else turnover, rleid(turnover)]
df[, result := cumsum(tmp)]
df[, tmp := NULL]
df
# x y turnover play_id result
# 1: 0.44594048 0.56645266 1 1 1
# 2: 0.39465031 0.04273416 1 2 2
# 3: 0.48372887 0.48831925 1 3 3
# 4: 0.91887596 0.35122322 0 4 4
# 5: 0.84388144 0.96966171 0 4 4
# 6: 0.51734962 0.78957889 0 4 4
# 7: 0.43712500 0.59663026 1 5 5
# 8: 0.34319822 0.35667053 1 6 6
# 9: 0.01551696 0.75870834 1 7 7
#10: 0.11799116 0.26105077 1 8 8
#11: 0.69098590 0.49398192 0 9 9
#12: 0.26048568 0.79289819 1 10 10
#13: 0.22505121 0.44299896 1 11 11
#14: 0.34238622 0.33326556 1 12 12
#15: 0.78188794 0.62576846 0 13 13
#16: 0.84324669 0.13619319 0 13 13
#17: 0.77474887 0.54868278 0 13 13
#18: 0.38719298 0.92183168 1 14 14
#19: 0.13576507 0.27603364 1 15 15
#20: 0.90035758 0.91741393 0 16 16
对于turnover值为0的组，我们将该组的第一个值改为1，并取其累加和。