Returning values after last NA in a vector

Question

Returning values after last NA in a vector

I can remove all NA values from a vector

v1 <- c(1,2,3,NA,5,6,NA,7,8,9,10,11,12)
v2 <- na.omit(v1)
v2

but how do I return a vector with values only after the last NA

c( 7,8,9,10,11,12)

Thank you for your help.

Answer 1

You could detect the last NA with which and add 1 to get the index past the last NA and index until the length(v1) :

 v1[(max(which(is.na(v1)))+1):length(v1)]
[1]  7  8  9 10 11 12

Answer 2

You could detect the last NA with which

v1[(tail(which(is.na(v1)), 1) + 1):length(v1)]
# [1]  7  8  9 10 11 12

However, the most general - as @MrFlick pointed out - seems to be this:

tail(v1, -tail(which(is.na(v1)), 1))
# [1]  7  8  9 10 11 12

which also handles the following case correctly:

v1[13] <- NA
tail(v1, -tail(which(is.na(v1)), 1))
# numeric(0)

To get the null NA case, too,

v1 <- 1:13

we can do

if (any(is.na(v1))) tail(v1, -tail(which(is.na(v1)), 1)) else v1
# [1]  1  2  3  4  5  6  7  8  9 10 11 12 13

Data

v1 <- c(1, 2, 3, NA, 5, 6, NA, 7, 8, 9, 10, 11, 12)

Answer 3

v1 <- c(1,2,3,NA,5,6,NA,7,8,9,10,11,12)
v1[seq_along(v1) > max(0, tail(which(is.na(v1)), 1))]
#[1]  7  8  9 10 11 12

v1 = 1:5
v1[seq_along(v1) > max(0, tail(which(is.na(v1)), 1))]
#[1] 1 2 3 4 5

v1 = c(1:5, NA)
v1[seq_along(v1) > max(0, tail(which(is.na(v1)), 1))]
#integer(0)

Answer 4

The following will do what you want.

i <- which(is.na(v1))
if(i[length(i)] < length(v1)){
  v1[(i[length(i)] + 1):length(v1)]
}else{
  NULL
}
#[1]  7  8  9 10 11 12

Answer 5

Here's an alternative solution that does not use indices and only vectorised operations:

after_last_na = as.logical(rev(cumprod(rev(! is.na(v1)))))
v1[after_last_na]

The idea is to use cumprod to fill the non- NA fields from the last to the end. It's not a terribly useful solution in its own right (I urge you to use the more obvious, index range based solution from other answers) but it shows some interesting techniques.