Question

Fake data for illustration:

df <- data.frame(a=c(1,2,3,4,5), b=(c(2,2,2,2,NA)), 
                 c=c(NA,2,3,4,5)))

This would get me the answer I want IF it weren’t for the NA values:

df$count <- with(df, (a==1) + (b==2) + (c==3)) 

Also, would there be an even more elegant way if I was only interested in, e.g. variables==2?

df$count <- with(df, (a==2) + (b==2) + (c==2)) 

Many thanks!

Answer 1

The following works for your specific example, but I have a suspicion that your real use case is more complicated:

df$count <- apply(df,1,function(x){sum(x == 1:3,na.rm = TRUE)})
> df
  a  b  c count
1 1  2 NA     2
2 2  2  2     1
3 3  2  3     2
4 4  2  4     1
5 5 NA  5     0

but this general approach should work. For instance, your second example would be something like this:

df$count <- apply(df,1,function(x){sum(x == 2,na.rm = TRUE)})

or more generally you could allow yourself to pass in a variable for the comparison:

df$count <- apply(df,1,function(x,compare){sum(x == compare,na.rm = TRUE)},compare = 1:3)