What is the most efficient way to iterate a collection in parallel in Scala (TOP N pattern) - scala

Question

I am new to Scala and would like to build a real-time application to match some people. For a given Person, I would like to get the TOP 50 people with the highest matching score.
The idiom is as follows :
val persons = new mutable.HashSet[Person]() // Collection of people
/* Feed omitted */
val personsPar = persons.par // Make it parall
val person = ... // The given person
res = personsPar
.filter(...) // Some filters
.map{p => (p,computeMatchingScoreAsFloat(person, p))}
.toList
.sortBy(-_._2)
.take(50)
.map(t => t._1 + "=" + t._2).mkString("\n")
In the sample code above, HashSet is used, but it can be any type of collection, as I am pretty sure it is not optimal
The problem is that persons contains over 5M elements, the computeMatchingScoreAsFloat méthods computes a kind a correlation value with 2 vectors of 200 floats. This computation takes ~2s on my computer with 6 cores.
My question is, what is the fastest way of doing this TOPN pattern in Scala ?
Subquestions :
- What implementation of collection (or something else?) should I use ?
- Should I use Futures ?
NOTE: It has to be computed in parallel, the pure computation of computeMatchingScoreAsFloat alone (with no ranking/TOP N) takes more than a second, and < 200 ms if multi-threaded on my computer
EDIT: Thanks to Guillaume, compute time has been decreased from 2s to 700ms
def top[B](n:Int,t: Traversable[B])(implicit ord: Ordering[B]):collection.mutable.PriorityQueue[B] = {
val starter = collection.mutable.PriorityQueue[B]()(ord.reverse) // Need to reverse for us to capture the lowest (of the max) or the greatest (of the min)
t.foldLeft(starter)(
(myQueue,a) => {
if( myQueue.length <= n ){ myQueue.enqueue(a);myQueue}
else if( ord.compare(a,myQueue.head) < 0 ) myQueue
else{
myQueue.dequeue
myQueue.enqueue(a)
myQueue
}
}
)
}
Thanks

Answer 1

I would propose a few changes:
1- I believe that the filter and map steps requires traversing the collection twice. Having a lazy collection would reduce it to one. Either have a lazy collection (like Stream) or converting it to one, for instance for a list:
myList.view
2- the sort step requires sorting all elements. Instead, you can do a FoldLeft with an accumulator storing the top N records. See there for an example of an implementation:
Simplest way to get the top n elements of a Scala Iterable . I would probably test a Priority Queue instead of a list if you want maximum performance (really falling into its wheelhouse). For instance, something like this:
def IntStream(n:Int):Stream[(Int,Int)] = if(n == 0) Stream.empty else (util.Random.nextInt,util.Random.nextInt) #:: IntStream(n-1)
def top[B](n:Int,t: Traversable[B])(implicit ord: Ordering[B]):collection.mutable.PriorityQueue[B] = {
val starter = collection.mutable.PriorityQueue[B]()(ord.reverse) // Need to reverse for us to capture the lowest (of the max) or the greatest (of the min)
t.foldLeft(starter)(
(myQueue,a) => {
if( myQueue.length <= n ){ myQueue.enqueue(a);myQueue}
else if( ord.compare(a,myQueue.head) < 0 ) myQueue
else{
myQueue.dequeue
myQueue.enqueue(a)
myQueue
}
}
)
}
def diff(t2:(Int,Int)) = t2._2
top(10,IntStream(10000))(Ordering.by(diff)) // select top 10
I really think that you problem requires a SINGLE collection traverse so you be able to get your run time down to below 1 sec
Good luck!