is ifelse ever appropriate in a non-vectorized situation and vice-versa?

Question

(Background info: ifelse evaluates both of the expressions, even though only one will be returned. EDIT: This is an incorrect statement. See Tommy\'s r

Answer 1

First, ifelse does NOT always evaluate both expressions - only if there are both TRUE and FALSE elements in the test vector.

ifelse(TRUE, 'foo', stop('bar')) # "foo"

And in my opinion:

ifelse should not be used in a non-vectorized situation. It is always slower and more error prone to use ifelse over if / else:

# This is fairly common if/else code
if (length(letters) > 0) letters else LETTERS

# But this "equivalent" code will yield a very different result - TRY IT!
ifelse(length(letters) > 0, letters, LETTERS)

In vectorized situations though, ifelse can be a good choice - but beware that the length and attributes of the result might not be what you expect (as above, and I consider ifelse broken in that respect).

Here's an example: tst is of length 5 and has a class. I'd expect the result to be of length 10 and have no class, but that isn't what happens - it gets an incompatible class and length 5!

# a logical vector of class 'mybool'
tst <- structure(1:5 %%2 > 0, class='mybool')

# produces a numeric vector of class 'mybool'!
ifelse(tst, 101:110, 201:210)
#[1] 101 202 103 204 105
#attr(,"class")
#[1] "mybool"

Why would I expect the length to be 10? Because most functions in R "cycle" the shorter vector to match the longer:

1:5 + 1:10 # returns a vector of length 10.

...But ifelse only cycles the yes/no arguments to match the length of the tst argument.

Why would I expect the class (and other attributes) to not be copied from the test object? Because < which returns a logical vector does not copy class and attributes from its (typically numeric) arguments. It doesn't do that because it would typically be very wrong.

1:5 < structure(1:10, class='mynum') # returns a logical vector without class

Finally, can it be more efficient to "do it yourself"? Well, it seems that ifelse is not a primitive like if, and it needs some special code to handle NA. If you don't have NAs, it can be faster to do it yourself.

tst <- 1:1e7 %%2 == 0
a <- rep(1, 1e7)
b <- rep(2, 1e7)
system.time( r1 <- ifelse(tst, a, b) )            # 2.58 sec

# If we know that a and b are of the same length as tst, and that
# tst doesn't have NAs, then we can do like this:
system.time( { r2 <- b; r2[tst] <- a[tst]; r2 } ) # 0.46 secs

identical(r1, r2) # TRUE