# PMean: Comparing two vectors with possible missing values

## 2014-03-06

I want to compare two vectors in R that have the same length and identify where, if anywhere, that the two vectors don’t match. It sounds like an easy thing to do, but missing values muddy things up quite a bit.

Suppose you have two vectors in R,

x <- c(1, 2,99, 4,NA, 6,NA) y <- c(1, 2, 3, 4, 5,NA,NA)

These vectors match at the first, second, fourth, and seventh location and mismatch at the other spots. NA, of course, is the code for missing in R. I want a logical function that is true when the vectors match and false otherwise. You would think that this would be just as simple as (x==y), but when you try this, you get a rude surprise.

> (x==y) [1]  TRUE  TRUE FALSE  TRUE    NA    NA    NA

The problem is that when one of the values is unknown, R tells you quite logically that it doesn’t know if it equals the other value. So the fifth case (where x is unknown), the sixth case (where y is unknown), and the seventh case (where both x and y are unknown) evaluate neither to TRUE or FALSE, but rather to NA.

This is an example of a three valued logic system. It seems complicated and prone to error, but a two value logic system is equally prone to error. In SAS, for example, the missing value code is treated in logical statements as smaller than any other number (effectively negative infinity). It’s a bit more complicated than this, but for the purposes of this discussion, you don’t need to know all the details. The problem with SAS occurs when you want to select based on missing values. In SAS, the statement

if (age < 18)

will evaluate to TRUE not just for the children, but for those whose ages are missing. In most situations, you would actually have to use

if (age < 18) & (age ne .)

In any program, missing values are a pain. See, for example, all the hoops you have to jump through with missing values in SQL!

Anyway, in R, you need to convert the last three NAs to FALSE, FALSE, and TRUE, respectively. So you might try

> (x==y) | (is.na(x) & is.na(y)) [1]  TRUE  TRUE FALSE  TRUE    NA    NA  TRUE

Pop quiz. Why do the seventh value get TRUE even though the first half of the expression evaluates to NA? Read the answer at the bottom of this post.

So we’re halfway there. We need to convert the two NA values to FALSE. There are several ways to do this.

> (x==y) & (!is.na(x) & !is.na(y)) | (is.na(x) & is.na(y)) [1]  TRUE  TRUE FALSE  TRUE FALSE FALSE  TRUE

Let’s split this apart. The first two conditions evaluate to TRUE only if the two values are equal and both non-missing. But the way it is coded, it evaluates to FALSE (not NA) when one or both values are missing.

> (x==y) & (!is.na(x) & !is.na(y)) [1]  TRUE  TRUE FALSE  TRUE FALSE FALSE FALSE

The third condition evaluates to TRUE if both values are missing and FALSE otherwise.

> (is.na(x) & is.na(y)) [1] FALSE FALSE FALSE FALSE FALSE FALSE  TRUE

Again, we have sidestepped the dreaded logical missing. Combine these together with OR to get the final answer.

I’m wondering if there is a more clever way to do this, possibly with the xor function. If you have something that is shorter and/or easier to follow, send me an email.

Answer to pop quiz: when you have a logic statement with OR, and one of the two components is TRUE, then it doesn’t matter whether the second component is TRUE or FALSE. So R evaluates TRUE | NA as TRUE.