Frequency of two columns counting NAs in one column as zero frequency - r

Question

Part 1 I have the following data table. I want to make a new column which contains the number of occurrences of each id where there is any style value, except for NA. The main problem is that I don't know how to deal with the NA. Currently, when NA is present I get a frequency of 1.
id style
1 A
1 A
2 A
2 B
3 NA
4 A
4 C
5 NA
I tried using the following, but it still counts NA values
dt[, allele_count := .N, by = list(pat_id, style)]
The desired data table would be as follows:
id style count
1 A 2
1 A 2
2 A 2
2 B 2
3 NA 0
4 A 4
4 B 4
4 B 4
4 C 4
5 NA 0
Part2 I would also like to be able to add another column which would have the number of occurrences of each id with a certain style value.
id style count2
1 A 2
1 A 2
2 A 1
2 B 1
3 NA 0
4 A 1
4 B 2
4 B 2
4 C 1
5 NA 0
Bonus Question: Instead of looking at how many times an id occurs with a given style value as in Part2 , How can you calculate the number of different style values for each id, as follows.
id style count3
1 A 1
1 A 1
2 A 2
2 B 2
3 NA 0
4 A 3
4 B 3
4 B 3
4 C 3
5 NA 0

Answer 1

Here's one possibility. Basically we use a row subset to assign the new columns, then replace the NA values in all three new columns with zero at the end.
nna <- !is.na(dt$style) ## so we don't have to call it four times
dt[nna, count := .N, by = id][nna, count2 := .N, by = .(id, style)][
nna, count3 := uniqueN(style), by = id][!nna, names(dt)[3:5] := 0L]
which results in
id style count count2 count3
1: 1 A 2 2 1
2: 1 A 2 2 1
3: 2 A 2 1 2
4: 2 B 2 1 2
5: 3 NA 0 0 0
6: 4 A 2 1 2
7: 4 C 2 1 2
8: 5 NA 0 0 0
Or you can simplify this down to the following, then reorder the columns if necessary.
dt[nna, c("count", "count3") := .(.N, uniqueN(style)), by = id][
nna, count2 := .N, by = .(id, style)][!nna, names(dt)[3:5] := 0L]
Note that this method is very similar to the other posted answer. I am not sure which of the two is the preferred method, row subset or if() statement.

Answer 2

This is a possible solution :
library(data.table)
dt <- data.table(id=c(1,1,2,2,3,4,4,5),
style=c('A','A','A','B',NA,'A','C',NA))
# count = number of ids having ALL styles defined
dt[, count := if(any(is.na(style))) 0L else .N, by = id]
# count2 = number of id-style occurrences (0 if style = NA)
dt[, count2 := if(is.na(style)) 0L else .N, by = .(id, style)]
> dt
id style count count2
1: 1 A 2 2
2: 1 A 2 2
3: 2 A 2 1
4: 2 B 2 1
5: 3 NA 0 0
6: 4 A 2 1
7: 4 C 2 1
8: 5 NA 0 0
BONUS:
dt[, count3 := uniqueN(na.omit(style)), by = id]
> dt
id style count count2 count3
1: 1 A 2 2 1
2: 1 A 2 2 1
3: 2 A 2 1 2
4: 2 B 2 1 2
5: 3 NA 0 0 0
6: 4 A 2 1 2
7: 4 C 2 1 2
8: 5 NA 0 0 0