Element wise multiplication of iterators and calculating sum

Question

Using two Lists, element wise multiplication of these lists and sum of resultant list can be calculated in following way.

(List1 , List2).zipped.foldLeft(0.0) { case (a, (b, c)) => a + b * c }

How can I preform this operation for two iterators in Scala in an optimal and fast way?

Write a benchmark, otherwise it's just guessing. Iterators might be fast because they could traverse the structure without calculating things twice, but I would avoid any statements about opimality of this or any other solution. Your current one might be close to "optiomal" or at least fast enough. — Mateusz Kubuszok, Sep 07 '20 at 12:53
I am also a little bit confused. Apart from the performance part, both `zip` and `fold` are present on Iterators. Not sure what you found confusing about Iterators. The code would look almost identical. Did you implement that and weren't satisfied by it's performance? — sinanspd, Sep 07 '20 at 12:56
what kind of performance improvement you are looking for memory(are those list are too big) or time complexity? your current implementation look good enough. — Raman Mishra, Sep 07 '20 at 13:38
@sinanspd I used aboove code and provide Iterator instead of List. But that code is not working — Asif, Sep 07 '20 at 14:07

score 2 · Answer 1 · answered Sep 07 '20 at 13:14

2

(iterator1 zip iterator2).foldLeft(0.0) { case (a, (b, c)) => a + b * c }

is okay I suppose. If you want to squeeze the last bit of performance out of it, use arrays and a while loop.

answered Sep 07 '20 at 13:14

Martijn

11,964
12
50
96

Luis Miguel Mejía Suárez · Answer 2 · 2020-09-07T15:07:50.577

You can use this piece of code that should work with any collection and any numeric type.
It tries to be efficient by doing everything in one traversal. However, as @Martijn said, if you need it to be the most efficient solution then just use plain Arrays of a primitive type like Int or Double and a while.

def dotProduct[N : Numeric](l1: IterableOnce[N], l2: IterableOnce[N]): N =
  l1.iterator.zip(l2).map {
    case (x, y) => Numeric[N].times(x, y)
  }.sum

(note: this code is intended for 2.13+, for 2.12- you may use Iterable instead of IterableOnce)

Element wise multiplication of iterators and calculating sum

2 Answers2