简单的问题。可以说我有以下数据:
library(tidyverse)
df <- data.frame(group = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2),
variable = c(NA, "a", NA, "b", "c", NA, NA, NA, NA, "a", NA, "c", NA, NA, "d", NA, NA, "a"))
df
group variable
1 1 <NA>
2 1 a
3 1 <NA>
4 1 b
5 1 c
6 1 <NA>
7 1 <NA>
8 1 <NA>
9 1 <NA>
10 1 a
11 1 <NA>
12 1 c
13 1 <NA>
14 1 <NA>
15 1 d
16 2 <NA>
17 2 <NA>
18 2 a
我只想使用
cumsum(is.na(variable)
来计算丢失的变量,但是忽略连续丢失的变量,因此我想要的输出如下所示: group variable newvariable
1 1 <NA> 1
2 1 a 1
3 1 <NA> 2
4 1 b 2
5 1 c 2
6 1 <NA> 3
7 1 <NA> 3
8 1 <NA> 3
9 1 <NA> 3
10 1 a 3
11 1 <NA> 4
12 1 c 4
13 1 <NA> 5
14 1 <NA> 5
15 1 d 5
16 2 <NA> 1
17 2 <NA> 1
18 2 a 1
我想我需要将
rle
合并到我的代码中:df %>%
group_by(group, na_group = {na_group = rle(variable); rep(seq_along(na_group$lengths), na_group$lengths)}) %>%
mutate(newvariable = cumsum((is.na(variable)))) #?
也许组上的
map
可以工作。有什么建议吗?参考:
Identify sets of NA in a vector
Count consecutive values in groups with condition with dplyr and rle
最佳答案
df %>%
group_by(group) %>%
mutate(new = with(rle(is.na(variable)), rep(cumsum(values), lengths))) %>%
ungroup()