R按组用NA替换最后n个值

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 .