Remove duplicates while keeping NA in R - r

Question

I have data that looks like the following:
a<-data.frame(ID=c("A","B","C","C",NA,NA),score=c(1,2,3,3,5,6),stringsAsFactors=FALSE)
print(a)
ID score
A 1
B 2
C 3
C 3
<NA> 5
<NA> 6
I am trying to remove duplicates without R treating <NA> as duplicates to get the following:
b<-data.frame(ID=c("A","B","C",NA,NA),score=c(1,2,3,5,6),stringsAsFactors=FALSE)
print(b)
ID score
A 1
B 2
C 3
<NA> 5
<NA> 6
I have tried the following:
b<-a[!duplicated(a$ID),]
library(dplyr)
b<-distinct(a,ID)
print(b)
But both treat <NA> as a duplicate ID and remove one, but I want to keep all instances of <NA>. Thoughts? Thank you!

Answer 1

A straight forward approach is to break the original data frame down into 2 parts where ID is NA and where it is not. Perform your distinct filter and then combine the data frames back together:
a<-data.frame(ID=c("A","B","C","C",NA,NA),score=c(1,2,3,3,5,6),stringsAsFactors=FALSE)
aprime<-a[!is.na(a$ID),]
aNA<-a[is.na(a$ID),]
b<-aprime[!duplicated(aprime$ID),]
b<-rbind(b, aNA)
With a little work, one can reduce this down to a 1-2 line lines of code.

Answer 2

using dplyr:
b%>%group_by(ID,score)%>%distinct()
# A tibble: 5 x 2
# Groups: ID, score [5]
ID score
<chr> <dbl>
1 A 1
2 B 2
3 C 3
4 <NA> 5
5 <NA> 6

Answer 3

Found a very simple way to do this simply using the base duplicated() function.
b<-a[!duplicated(a$ID, incomparables = NA),]
Setting incomparables = NA makes R read NA duplicates as FALSE, therefore including them in the result dataset.