[Rd] Suggestions to speed up median() and has.na()

From: Henrik Bengtsson <hb_at_maths.lth.se>
Date: Mon 10 Apr 2006 - 17:37:54 GMT


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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https://stat.ethz.ch/mailman/listinfo/r-devel Received on Tue Apr 11 04:09:03 2006

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