[Rd] Suggestions to speed up median() and has.na()
Henrik Bengtsson
hb at maths.lth.se
Mon Apr 10 19:37:54 CEST 2006
Hi,
I've got two suggestions how to speed up median() about 50%. For all
iterative methods calling median() in the loops this has a major
impact. The second suggestion will apply to other methods too.
This is what the functions look like today:
> median
function (x, na.rm = FALSE)
{
if (is.factor(x) || mode(x) != "numeric")
stop("need numeric data")
if (na.rm)
x <- x[!is.na(x)]
else if (any(is.na(x)))
return(NA)
n <- length(x)
if (n == 0)
return(NA)
half <- (n + 1)/2
if (n%%2 == 1) {
sort(x, partial = half)[half]
}
else {
sum(sort(x, partial = c(half, half + 1))[c(half, half +
1)])/2
}
}
<environment: namespace:stats>
Suggestion 1:
Replace the sort() calls with the .Internal(psort(x, partial)). This
will avoid unnecessary overhead, especially an expensive second check
for NAs using any(is.na(x)). Simple benchmarking with
x <- rnorm(10e6)
system.time(median(x))/system.time(median2(x))
where median2() is the function with the above replacements, gives
about 20-25% speed up.
Suggestion 2:
Create a has.na(x) function to replace any(is.na(x)) that returns TRUE
as soon as a NA value is detected. In the best case it returns after
the first index with TRUE, in the worst case it returns after the last
index N with FALSE. The cost for is.na(x) is always O(N), and any()
in the best case O(1) and in the worst case O(N) (if any() is
implemented as I hope). An has.na() function would be very useful
elsewhere too.
An poor mans alternative to (2), is to have a third alternative to
'na.rm', say, NA, which indicates that we know that there are no NAs
in 'x'.
The original median() is approx 50% slower (naive benchmarking) than a
version with the above two improvements, if passing a large 'x' with
no NAs;
median2 <- function (x, na.rm = FALSE) {
if (is.factor(x) || mode(x) != "numeric")
stop("need numeric data")
if (is.na(na.rm)) {
} else if (na.rm)
x <- x[!is.na(x)]
else if (any(is.na(x)))
return(NA)
n <- length(x)
if (n == 0)
return(NA)
half <- (n + 1)/2
if (n%%2 == 1) {
.Internal(psort(x, half))[half]
}
else {
sum(.Internal(psort(x, c(half, half + 1)))[c(half, half + 1)])/2
}
}
x <- rnorm(10e5)
K <- 10
t0 <- system.time({
for (kk in 1:K)
y <- median(x);
})
print(t0) # [1] 1.82 0.14 1.98 NA NA
t1 <- system.time({
for (kk in 1:K)
y <- median2(x, na.rm=NA);
})
print(t1) # [1] 1.25 0.06 1.34 NA NA
print(t0/t1) # [1] 1.456000 2.333333 1.477612 NA NA
BTW, without having checked the source code, it looks like is.na() is
unnecessarily slow; is.na(sum(x)) is much faster than any(is.na(x)) on
a vector without NAs. On the other hand, is.na(sum(x)) becomes
awfully slow if 'x' contains NAs.
/Henrik
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