Returning group maximum and NA using dplyr - r

Question

I need a function I can use that returns both the group maximum and any NA values. Here is toy data:
df <- data.frame(id = rep(1:5,
each = 3),
score = rnorm(15))
df$score[c(3,7,10,14)] <- NA
# id score
# 1 1 -1.4666164
# 2 1 0.4392647
# 3 1 NA
# 4 2 -0.6010311
# 5 2 1.9845774
# 6 2 0.1749082
# 7 3 NA
# 8 3 -0.3089731
# 9 3 0.4427471
# 10 4 NA
# 11 4 1.7156319
# 12 4 -0.2354253
# 13 5 1.1781350
# 14 5 NA
# 15 5 0.0642082
I can use slice_max to get the maximum in each group:
df %>%
group_by(id) %>%
slice_max(score)
# id score
# <int> <dbl>
# 1 1 0.439
# 2 2 1.98
# 3 3 0.443
# 4 4 1.72
# 5 5 1.18
But how do I get the maximum plus any NAs returned?

Answer 1

We can group_by the id column, then use summarize to output the summaries with max. Here, two max are used, with one of them has na.rm = T and the other doesn't. union() is used to combine output that is present in both max.
library(dplyr)
df %>%
group_by(id) %>%
summarize(score = union(max(score, na.rm = T), max(score)))
UPDATE: The above code only works if you have at most one NA per ID. Thanks #KU99 for the reminder.
If you have more than one NA per ID, you need to combine the result of max with the records of NA found by is.na().
df %>%
group_by(id) %>%
summarize(score = c(max(score, na.rm = T), score[is.na(score)]))
Result
# A tibble: 9 × 2
# Groups: id [5]
id score
<int> <dbl>
1 1 0.735
2 1 NA
3 2 0.314
4 3 0.994
5 3 NA
6 4 0.847
7 4 NA
8 5 1.95
9 5 NA
Data
df <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L, 5L), score = c(-1.05089006245306, 0.734652105895187,
NA, -1.31427279695036, -0.250038722057874, 0.314204596436828,
NA, 0.994420599790523, 0.855768431757766, NA, 0.834325037545013,
0.846790152407738, 1.95410525460771, NA, 0.971120269710021)), row.names = c(NA,
-15L), class = "data.frame")

Answer 2

One option would be to use slice and | to create a logical condition with is.na to return the NA rows and the max rows.
library(dplyr)
df %>%
group_by(id) %>%
slice(which(score == max(score, na.rm = T)|is.na(score)))
Another option would be to use slice.max as you did but then to use bind_rows to add the NA values back to the dataframe.
library(dplyr)
df %>%
group_by(id) %>%
slice_max(score) %>%
bind_rows(df %>% filter(is.na(score))) %>%
arrange(id)
Output
id score
<int> <dbl>
1 1 -0.161
2 1 NA
3 2 1.49
4 3 -0.451
5 3 NA
6 4 0.878
7 4 NA
8 5 -0.0652
9 5 NA
Data
df <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L, 5L), score = c(-0.161217942983375, -0.456571996252207,
NA, 0.540071362460494, 1.49325799630099, -0.17985218510166, NA,
-0.451301758592, -0.839100876399644, NA, -0.0432130218441599,
0.87779273806634, -0.339260854059069, NA, -0.065177224102029)), row.names = c(NA,
-15L), class = "data.frame")

Answer 3

Using a custom function you could do:
library(dplyr)
set.seed(123)
slice_max_na <- function(.data, order_by, ..., n, prop, with_ties = TRUE) {
bind_rows(
slice_max(.data, order_by = {{order_by}}, ..., n = n, prop = prop, with_ties = with_ties),
filter(.data, is.na({{order_by}})),
)
}
df %>%
group_by(id) %>%
slice_max_na(score)
#> # A tibble: 9 × 2
#> # Groups: id [5]
#> id score
#> <int> <dbl>
#> 1 1 -0.230
#> 2 2 1.72
#> 3 3 -0.687
#> 4 4 1.22
#> 5 5 0.401
#> 6 1 NA
#> 7 3 NA
#> 8 4 NA
#> 9 5 NA

Answer 4

Here is dplyr version more using rank:
library(dplyr)
df %>%
group_by(id) %>%
mutate(rank = rank(-score, ties.method = "random")) %>%
filter(rank == 1 | is.na(score)) %>%
select(-rank)
id score
<int> <dbl>
1 1 0.505
2 1 NA
3 2 -0.109
4 3 NA
5 3 1.45
6 4 NA
7 4 0.355
8 5 NA
9 5 -0.298