Vectorized (non-loop) solution returns wrong result (solution with for-loop returns correct result)

Question

I have theoretically identical solutions, one is vectorized solution and another is with for-loop. But vectorized solution returns wrong result and I want to understand why. Solution's logic is simple: need to replace NA with previous non-NA value in the vector.

# vectorized
f1 <- function(x) {
    idx <- which(is.na(x))
    x[idx] <- x[ifelse(idx > 1, idx - 1, 1)]
    x
}

# non-vectorized
f2 <- function(x) {
    for (i in 2:length(x)) {
        if (is.na(x[i]) && !is.na(x[i - 1])) {
            x[i] <- x[i - 1]
        }
    }
    x
}

v <- c(NA,NA,1,2,3,NA,NA,6,7)
f1(v)
# [1] NA NA  1  2  3  3 NA  6  7
f2(v)
# [1] NA NA  1  2  3  3  3  6  7

score 4 · Accepted Answer · edited May 23 '17 at 12:21

4

The two pieces of code are different.

The first one replace NA with the previous element if this one is not NA.
The second one replace NA with the previous element if this one is not NA, but the previous element can be the result of a previous NA substitution.

Which one is correct really depends on you. The second behaviour is more difficult to vectorize, but there are some already implemented functions like zoo::na.locf.

Or, if you only want to use base packages, you could have a look at this answer.

edited May 23 '17 at 12:21

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answered May 13 '14 at 16:10

digEmAll

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It's possible to vectorize it without using any third-party functions such as na.locf? – Eldar Agalarov May 19 '14 at 17:08
@EldarAgalarov: I've added a link in my answer ;) – digEmAll May 19 '14 at 20:25

score 3 · Answer 2 · answered May 13 '14 at 16:09

These two solutions are not equivalent. The first function is rather like:

f2_as_f1 <- function(x) {
    y <- x # a copy of x
    for (i in 2:length(x)) {
        if (is.na(y[i])) {
            x[i] <- y[i - 1]
        }
    }
    x
}

Note the usage of the y vector.

Vectorized (non-loop) solution returns wrong result (solution with for-loop returns correct result)

2 Answers2

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