cumsum is.na with rle ignoring consectives NA's

Question

simple problem. Lets say I have the following data:

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

I just want to count missing variables using cumsum(is.na(variable) but ignore consecutive missing ones so my desired output would look like:

   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

I think I need to incorporate rle into my code:

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)))) #?

Maybe map over groups could work. Any suggestions please?

Refs: Identify sets of NA in a vector Count consecutive values in groups with condition with dplyr and rle

Answer 1

df %>%
    group_by(group) %>%
    mutate(new = with(rle(is.na(variable)), rep(cumsum(values), lengths))) %>%
    ungroup()