How to extend `==` behavior to vectors that include NAs?

Question

I've completely failed at searching for other r-help or Stack Overflow discussion of this specific issue. Sorry if it's somewhere obvious. I believe that I'm just looking for the easiest way to get R's == sign to never return NAs.

# Example #

# Say I have two vectors
a <- c( 1 , 2 , 3 )
b <- c( 1 , 2 , 4 )
# And want to test if each element in the first
# is identical to each element in the second:
a == b
# It does what I want perfectly:
# TRUE TRUE FALSE

# But if either vector contains a missing,
# the `==` operator returns an incorrect result:
a <- c( 1 , NA , 3 ) 
b <- c( 1 , NA , 4 )
# Here I'd want   TRUE TRUE FALSE
a == b
# But I get TRUE NA FALSE

a <- c( 1 , NA , 3 ) 
b <- c( 1 , 2 , 4 )
# Here I'd want   TRUE FALSE FALSE
a == b
# But I get TRUE NA FALSE again.

I get the result I want with:

mapply( `%in%` , a , b )

But mapply seems heavy-handed to me.

Is there a more intuitive solution to this?

Answer 1

Another option, but is it better than mapply('%in%', a , b)?:

(!is.na(a) & !is.na(b) & a==b) | (is.na(a) & is.na(b))

Following @AnthonyDamico 's suggestion, creation of the "mutt" operator:

"%==%" <- function(a, b) (!is.na(a) & !is.na(b) & a==b) | (is.na(a) & is.na(b))

Edit: or, slightly different and shorter version by @Frank (which is also more efficient)

"%==%" <- function(a, b) (is.na(a) & is.na(b)) | (!is.na(eq <- a==b) & eq)

With the different examples:

a <- c( 1 , 2 , 3 )
b <- c( 1 , 2 , 4 )
a %==% b
# [1]  TRUE  TRUE FALSE

a <- c( 1 , NA , 3 )
b <- c( 1 , NA , 4 )
a %==% b
# [1]  TRUE  TRUE FALSE

a <- c( 1 , NA , 3 )
b <- c( 1 , 2 , 4 )
a %==% b
#[1]  TRUE FALSE FALSE

a <- c( 1 , NA , 3 )
b <- c( 3 , NA , 1 )
a %==% b
#[1] FALSE  TRUE FALSE

Answer 2

You could try

replace(a, is.na(a), Inf)==replace(b, is.na(b), Inf)

Or a faster variation suggested by @docendo discimus

replace(a, which(is.na(a)), Inf)==replace(b, which(is.na(b)), Inf)

Based on the different scenarios

1.

a <- c( 1 , 2 , 3 )
b <- c( 1 , 2 , 4 )
akrun1()
#[1]  TRUE  TRUE FALSE

2.

 a <- c( 1 , NA , 3 ) 
 b <- c( 1 , NA , 4 )
 akrun1()
 #[1]  TRUE  TRUE FALSE

3.

 a <- c( 1 , NA , 3 ) 
 b <- c( 1 , 2 , 4 )
 akrun1()
#[1]  TRUE FALSE FALSE

Benchmarks

set.seed(24)
a <- sample(c(1:10, NA), 1e6, replace=TRUE)
b <- sample(c(1:20, NA), 1e6, replace=TRUE)
akrun1 <- function() {replace(a, is.na(a), Inf)==replace(b, is.na(b), Inf)}
cathG <- function() {(!is.na(a) & !is.na(b) & a==b) | (is.na(a) & is.na(b))}
anthony <- function() {mapply(`%in%`, a, b)}
webb <- function() {ifelse(is.na(a),is.na(b),a==b)}
docend <- function() {replace(a, which(is.na(a)), Inf)==replace(b,
       which(is.na(b)), Inf)}

library(microbenchmark)
microbenchmark(akrun1(), cathG(), anthony(), webb(),docend(),
  unit='relative', times=20L)
#Unit: relative
#    expr        min         lq       mean     median         uq        max
#  akrun1()   3.050200   3.035625   3.007196   2.963916   2.977490   3.083658
#   cathG()   4.829972   4.893266   4.843585   4.790466   4.816472   4.939316
# anthony() 190.499027 224.389971 215.792965 217.647702 215.503308 212.356051
#    webb()  14.000363  14.366572  15.412527  14.095947  14.671741  19.735746
#  docend()   1.000000   1.000000   1.000000   1.000000   1.000000   1.000000
# neval cld
#    20 a  
#    20 a  
#    20 c
#    20 b 
#    20 a  

Answer 3

Assuming that we don't have a big relative number of NA, The proposed vectorized solution waste some ressources comparing values that have already been settled by a==b.

We can usually assume that NAs are few so it makes it worth computing a==b first and then deal with the NAs separately, despite the additional steps and temp variables:

`%==%` <- function(a,b){
  x <- a==b
  na_x <- which(is.na(x))
  x[na_x] <- is.na(a[na_x]) & is.na(b[na_x])
  x
}

Check output

a <- c( 1 , 2 , 3 )
b <- c( 1 , 2 , 4 )
a %==% b
# [1]  TRUE  TRUE FALSE

a <- c( 1 , NA , 3 ) 
b <- c( 1 , NA , 4 )
a %==% b
# [1]  TRUE  TRUE FALSE

a <- c( 1 , NA , 3 ) 
b <- c( 1 , 2 , 4 )
a %==% b
# [1]  TRUE FALSE FALSE

Benchmarks

I'm reproducing below @akrun's benchmark with fastest solutions only and n=100.

set.seed(24)
a <- sample(c(1:10, NA), 1e6, replace=TRUE)
b <- sample(c(1:20, NA), 1e6, replace=TRUE)
mm <- function(){
  x <- a==b
  na_x <- which(is.na(x))
  x[na_x] <- is.na(a[na_x]) & is.na(b[na_x])
  x
}
akrun1 <- function() {replace(a, is.na(a), Inf)==replace(b, is.na(b), Inf)}
cathG <- function() {(!is.na(a) & !is.na(b) & a==b) | (is.na(a) & is.na(b))}
docend <- function() {replace(a, which(is.na(a)), Inf)==replace(b, which(is.na(b)), Inf)}

library(microbenchmark)
microbenchmark(mm(),akrun1(),cathG(),docend(),
               unit='relative', times=100L)

# Unit: relative
#     expr      min       lq     mean   median       uq       max neval
#     mm() 1.000000 1.000000 1.000000 1.000000 1.000000 1.0000000   100
# akrun1() 1.667242 1.884185 1.815392 1.642581 1.765238 0.9973017   100
#  cathG() 2.447168 2.449597 2.118306 2.201346 2.358105 1.1421577   100
# docend() 1.683817 1.950970 1.756481 1.745400 2.007889 1.2264461   100

Extending ==

As the original question is really to find :

the easiest way to get R's == sign to never return NAs

Here's a way, where we define a new class na_comparable. Only one of the vector needs to be of this class as the other will be coerced to it.

na_comparable      <- setClass("na_comparable", contains = "numeric")
`==.na_comparable` <- function(a,b){
  x <- unclass(a) == unclass(b) # inefficient but I don't know how to force the default `==`
  na_x <- which(is.na(x))
  x[na_x] <- is.na(a[na_x]) & is.na(b[na_x])
  x
}

`!=.na_comparable` <- Negate(`==.na_comparable`)

a <- na_comparable(a)
a == b
# [1]  TRUE  TRUE FALSE
b == a
# [1]  TRUE  TRUE FALSE
a != b
# [1] FALSE FALSE  TRUE
b != a
# [1] FALSE FALSE  TRUE

In a dplyr chain it could be conveniently used this way :

data.frame(a=c(1,NA,3),b=c(1,NA,4)) %>%
  mutate(a = na_comparable(a),
         c = a==b,
         d= a!=b)

#    a  b     c     d
# 1  1  1  TRUE FALSE
# 2 NA NA  TRUE FALSE
# 3  3  4 FALSE  TRUE

With this approach, in case you need to update code to account for NAs that were absent before, you might be set with a single na_comparable call instead of transforming your initial data or replacing all your == with %==% down the line.