R replace last nth value with NA by group - r

Question

I want to replace value(s) with NA by group.
have <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,4,5,6,7))
want1 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,NA,5,6,NA))
want2 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,NA,NA,5,NA,NA))
want1 corresponds to replacing the last obs of value with NA and want2 corresponds to replacing last obs of value & last 2nd value with NA. I'm currently trying to do with with dplyr package but can't seem to get any traction. Any help would be much appreciated. Thanks!

Answer 1

We can use row_number() to test the current row against n() the total rows in the group.
have |>
group_by(id) |>
mutate(
last1 = ifelse(row_number() == n(), NA, value),
last2 = ifelse(row_number() >= n() - 1, NA, value)
)
# # A tibble: 7 × 4
# # Groups: id [2]
# id value last1 last2
# <dbl> <dbl> <dbl> <dbl>
# 1 1 1 1 1
# 2 1 2 2 2
# 3 1 3 3 NA
# 4 1 4 NA NA
# 5 2 5 5 5
# 6 2 6 6 NA
# 7 2 7 NA NA

Answer 2

And a general way to provide variants as different data frames.
lapply(
1:2,
function(k) {
have %>%
group_by(id) %>%
mutate(value=ifelse(row_number() <= (n() - k), value, NA))
}
)
[[1]]
# A tibble: 7 × 2
# Groups: id [2]
id value
<dbl> <dbl>
1 1 1
2 1 2
3 1 3
4 1 NA
5 2 5
6 2 6
7 2 NA
[[2]]
# A tibble: 7 × 2
# Groups: id [2]
id value
<dbl> <dbl>
1 1 1
2 1 2
3 1 NA
4 1 NA
5 2 5
6 2 NA
7 2 NA

Answer 3

Here is a base R way.
have <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,4,5,6,7))
want1 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,3,NA,5,6,NA))
want2 <- data.frame(id = c(1,1,1,1,2,2,2),
value = c(1,2,NA,NA,5,NA,NA))
with(have, ave(value, id, FUN = \(x){
x[length(x)] <- NA
x
}))
#> [1] 1 2 3 NA 5 6 NA
with(have, ave(value, id, FUN = \(x){
x[length(x)] <- NA
if(length(x) > 1)
x[length(x) - 1L] <- NA
x
}))
#> [1] 1 2 NA NA 5 NA NA
Created on 2022-06-09 by the reprex package (v2.0.1)
Then reassign these results to column value.