Yes. (With exception)
ifelse
calculates both its yes
value and its no
value. Except in the case where the test
condition is either all TRUE
or all FALSE
.
We can see this by generating random numbers and observing how many numbers are actually generated. (by reverting the seed
).
# TEST CONDITION, ALL TRUE
set.seed(1)
dump <- ifelse(rep(TRUE, 200), rnorm(200), rnorm(200))
next.random.number.after.all.true <- rnorm(1)
# TEST CONDITION, ALL FALSE
set.seed(1)
dump <- ifelse(rep(FALSE, 200), rnorm(200), rnorm(200))
next.random.number.after.all.false <- rnorm(1)
# TEST CONDITION, MIXED
set.seed(1)
dump <- ifelse(c(FALSE, rep(TRUE, 199)), rnorm(200), rnorm(200))
next.random.number.after.some.TRUE.some.FALSE <- rnorm(1)
# RESET THE SEED, GENERATE SEVERAL RANDOM NUMBERS TO SEARCH FOR A MATCH
set.seed(1)
r.1000 <- rnorm(1000)
cat("Quantity of random numbers generated during the `ifelse` statement when:",
"
All True ", which(r.1000 == next.random.number.after.all.true) - 1,
"
All False ", which(r.1000 == next.random.number.after.all.false) - 1,
"
Mixed T/F ", which(r.1000 == next.random.number.after.some.TRUE.some.FALSE) - 1
)
Gives the following output:
Quantity of random numbers generated during the `ifelse` statement when:
All True 200
All False 200
Mixed T/F 400 <~~ Notice TWICE AS MANY numbers were
generated when `test` had both
T & F values present
We can also see it in the source code itself:
.
.
if (any(test[!nas]))
ans[test & !nas] <- rep(yes, length.out = length(ans))[test & # <~~~~ This line and the one below
!nas]
if (any(!test[!nas]))
ans[!test & !nas] <- rep(no, length.out = length(ans))[!test & # <~~~~ ... are the cluprits
!nas]
.
.
Notice that yes
and no
are computed only if there
is some non-NA
value of test
that is TRUE
or FALSE
(respectively).
At which point -- and this is the imporant part when it comes to efficiency -- the entirety of each vector is computed.
Ok, but is it slower?
Lets see if we can test it:
library(microbenchmark)
# Create some sample data
N <- 1e4
set.seed(1)
X <- sample(c(seq(100), rep(NA, 100)), N, TRUE)
Y <- ifelse(is.na(X), rnorm(X), NA) # Y has reverse NA/not-NA setup than X
These two statements generate the same results
yesifelse <- quote(sort(ifelse(is.na(X), Y+17, X-17 ) ))
noiflese <- quote(sort(c(Y[is.na(X)]+17, X[is.na(Y)]-17)))
identical(eval(yesifelse), eval(noiflese))
# [1] TRUE
but one is twice as fast as the other
microbenchmark(eval(yesifelse), eval(noiflese), times=50L)
N = 1,000
Unit: milliseconds
expr min lq median uq max neval
eval(yesifelse) 2.286621 2.348590 2.411776 2.537604 10.05973 50
eval(noiflese) 1.088669 1.093864 1.122075 1.149558 61.23110 50
N = 10,000
Unit: milliseconds
expr min lq median uq max neval
eval(yesifelse) 30.32039 36.19569 38.50461 40.84996 98.77294 50
eval(noiflese) 12.70274 13.58295 14.38579 20.03587 21.68665 50