The Number of Rows with a Specific Number of Missing Values - r

Question

Imagine a small data set like the one below, composed of three variables:
v1 <- c(0, 1, NA, 1, NA, 0)
v2 <- c(0, 0, NA, 1, NA, NA)
v3 <- c(1, NA, 0, 0, NA, 0)
df <- data.frame(v1, v2, v3)
df
v1 v2 v3
1 0 0 1
2 1 0 NA
3 NA NA 0
4 1 1 0
5 NA NA NA
6 0 NA 0
One can use the is.na command as follows to calculate the number of rows with at least one missing value - and R would return 4:
sum(is.na(df$v1) | is.na(df$v2) | is.na(df$v3))
Or the number of rows with all three values missing - and R would return 1:
sum(is.na(df$v1) & is.na(df$v2) & is.na(df$v3))
Two questions at this point:
(1) How can I calculate the number of rows where "exactly one" or "exactly two" values are missing?
(2) If I am to do the above in a large data set, how can I limit the scope of the calculation to v1, v2 and v3 (that is, without having to create a subset)?
I tried variations of is.na, nrow and df, but could not get any of them to work.
Thanks!

Answer 1

We can use rowSums on the logical matrix (is.na(df)) and check whether the number of NAs are equal to the value of interest.
n1 <- 1
sum(rowSums(is.na(df))==n1)
To make it easier, create a function to do this
f1 <- function(dat, n){
sum(rowSums(is.na(dat)) == n)
}
f1(df, 0)
#[1] 2
f1(df, 1)
#[1] 2
f1(df, 3)
#[1] 1
f1(df, 2)
#[1] 1
NOTE: rowSums is very fast, but if it is a large dataset, then creating a logical matrix can also create problems in memory. So, we can use Reduce after looping through the columns of the dataset (lapply(df, is.na)).
sum(Reduce(`+`, lapply(df, is.na))==1)
#[1] 2
f2 <- function(dat, n){
sum(Reduce(`+`, lapply(dat, is.na))==n)
}
f2(df, 1)

Answer 2

Try this:
num.rows.with.x.NA <- function(df, x, cols=names(df)) {
return(sum(apply(df, 1, function(y) sum(is.na(y[cols])) == x)))
}
df
v1 v2 v3
1 0 0 1
2 1 0 NA
3 NA NA 0
4 1 1 0
5 NA NA NA
6 0 NA 0
num.rows.with.x.NA(df, 0, names(df))
#[1] 2
num.rows.with.x.NA(df, 1, names(df))
#[1] 2
num.rows.with.x.NA(df, 2, names(df))
#[1] 1
num.rows.with.x.NA(df, 3, names(df))
#[1] 1