Why is filter in front of foldLeft slow in Scala? - performance

Question

I wrote an answer to the first Project Euler question:
Add all the natural numbers below one thousand that are multiples of 3 or 5.
The first thing that came to me was:
(1 until 1000).filter(i => (i % 3 == 0 || i % 5 == 0)).foldLeft(0)(_ + _)
but it's slow (it takes 125 ms), so I rewrote it, simply thinking of 'another way' versus 'the faster way'
(1 until 1000).foldLeft(0){
(total, x) =>
x match {
case i if (i % 3 == 0 || i % 5 ==0) => i + total // Add
case _ => total //skip
}
}
This is much faster (only 2 ms). Why? I'm guess the second version uses only the Range generator and doesn't manifest a fully realized collection in any way, doing it all in one pass, both faster and with less memory. Am I right?
Here the code on IdeOne: http://ideone.com/GbKlP

Answer 1

The problem, as others have said, is that filter creates a new collection. The alternative withFilter doesn't, but that doesn't have a foldLeft. Also, using .view, .iterator or .toStream would all avoid creating the new collection in various ways, but they are all slower here than the first method you used, which I thought somewhat strange at first.
But, then... See, 1 until 1000 is a Range, whose size is actually very small, because it doesn't store each element. Also, Range's foreach is extremely optimized, and is even specialized, which is not the case of any of the other collections. Since foldLeft is implemented as a foreach, as long as you stay with a Range you get to enjoy its optimized methods.
(_: Range).foreach:
#inline final override def foreach[#specialized(Unit) U](f: Int => U) {
if (length > 0) {
val last = this.last
var i = start
while (i != last) {
f(i)
i += step
}
f(i)
}
}
(_: Range).view.foreach
def foreach[U](f: A => U): Unit =
iterator.foreach(f)
(_: Range).view.iterator
override def iterator: Iterator[A] = new Elements(0, length)
protected class Elements(start: Int, end: Int) extends BufferedIterator[A] with Serializable {
private var i = start
def hasNext: Boolean = i < end
def next: A =
if (i < end) {
val x = self(i)
i += 1
x
} else Iterator.empty.next
def head =
if (i < end) self(i) else Iterator.empty.next
/** $super
* '''Note:''' `drop` is overridden to enable fast searching in the middle of indexed sequences.
*/
override def drop(n: Int): Iterator[A] =
if (n > 0) new Elements(i + n, end) else this
/** $super
* '''Note:''' `take` is overridden to be symmetric to `drop`.
*/
override def take(n: Int): Iterator[A] =
if (n <= 0) Iterator.empty.buffered
else if (i + n < end) new Elements(i, i + n)
else this
}
(_: Range).view.iterator.foreach
def foreach[U](f: A => U) { while (hasNext) f(next()) }
And that, of course, doesn't even count the filter between view and foldLeft:
override def filter(p: A => Boolean): This = newFiltered(p).asInstanceOf[This]
protected def newFiltered(p: A => Boolean): Transformed[A] = new Filtered { val pred = p }
trait Filtered extends Transformed[A] {
protected[this] val pred: A => Boolean
override def foreach[U](f: A => U) {
for (x <- self)
if (pred(x)) f(x)
}
override def stringPrefix = self.stringPrefix+"F"
}

Answer 2

Try making the collection lazy first, so
(1 until 1000).view.filter...
instead of
(1 until 1000).filter...
That should avoid the cost of building an intermediate collection. You might also get better performance from using sum instead of foldLeft(0)(_ + _), it's always possible that some collection type might have a more efficient way to sum numbers. If not, it's still cleaner and more declarative...

Answer 3

Looking through the code, it looks like filter does build a new Seq on which the foldLeft is called. The second skips that bit. It's not so much the memory, although that can't but help, but that the filtered collection is never built at all. All that work is never done.
Range uses TranversableLike.filter, which looks like this:
def filter(p: A => Boolean): Repr = {
val b = newBuilder
for (x <- this)
if (p(x)) b += x
b.result
}
I think it's the += on line 4 that's the difference. Filtering in foldLeft eliminates it.

Answer 4

filter creates a whole new sequence on which then foldLeft is called. Try:
(1 until 1000).view.filter(i => (i % 3 == 0 || i % 5 == 0)).reduceLeft(_+_)
This will prevent said effect, merely wrapping the original thing. Exchanging foldLeft with reduceLeft is only cosmetic (in this case).

Answer 5

Now the challenge is, can you think of a yet more efficient way? Not that your solution is too slow in this case, but how well does it scale? What if instead of 1000, it was 1000000000? There is a solution that could compute the latter case just as quickly as the former.