Simplest way to get the top n elements of a Scala Iterable - algorithm

Question

Is there a simple and efficient solution to determine the top n elements of a Scala Iterable? I mean something like
iter.toList.sortBy(_.myAttr).take(2)
but without having to sort all elements when only the top 2 are of interest. Ideally I'm looking for something like
iter.top(2, _.myAttr)
see also: Solution for the top element using an Ordering: In Scala, how to use Ordering[T] with List.min or List.max and keep code readable
Update:
Thank you all for your solutions. Finally, I took the original solution of user unknown and adopted it to use Iterable and the pimp-my-library pattern:
implicit def iterExt[A](iter: Iterable[A]) = new {
def top[B](n: Int, f: A => B)(implicit ord: Ordering[B]): List[A] = {
def updateSofar (sofar: List [A], el: A): List [A] = {
//println (el + " - " + sofar)
if (ord.compare(f(el), f(sofar.head)) > 0)
(el :: sofar.tail).sortBy (f)
else sofar
}
val (sofar, rest) = iter.splitAt(n)
(sofar.toList.sortBy (f) /: rest) (updateSofar (_, _)).reverse
}
}
case class A(s: String, i: Int)
val li = List (4, 3, 6, 7, 1, 2, 9, 5).map(i => A(i.toString(), i))
println(li.top(3, _.i))

Answer 1

My solution (bound to Int, but should be easily changed to Ordered (a few minutes please):
def top (n: Int, li: List [Int]) : List[Int] = {
def updateSofar (sofar: List [Int], el: Int) : List [Int] = {
// println (el + " - " + sofar)
if (el < sofar.head)
(el :: sofar.tail).sortWith (_ > _)
else sofar
}
/* better readable:
val sofar = li.take (n).sortWith (_ > _)
val rest = li.drop (n)
(sofar /: rest) (updateSofar (_, _)) */
(li.take (n). sortWith (_ > _) /: li.drop (n)) (updateSofar (_, _))
}
usage:
val li = List (4, 3, 6, 7, 1, 2, 9, 5)
top (2, li)
For above list, take the first 2 (4, 3) as starting TopTen (TopTwo).
Sort them, such that the first element is the bigger one (if any).
repeatedly iterate through the rest of the list (li.drop(n)), and compare the current element with the maximum of the list of minimums; replace, if neccessary, and resort again.
Improvements:
Throw away Int, and use ordered.
Throw away (_ > _) and use a user-Ordering to allow BottomTen. (Harder: pick the middle 10 :) )
Throw away List, and use Iterable instead
update (abstraction):
def extremeN [T](n: Int, li: List [T])
(comp1: ((T, T) => Boolean), comp2: ((T, T) => Boolean)):
List[T] = {
def updateSofar (sofar: List [T], el: T) : List [T] =
if (comp1 (el, sofar.head))
(el :: sofar.tail).sortWith (comp2 (_, _))
else sofar
(li.take (n) .sortWith (comp2 (_, _)) /: li.drop (n)) (updateSofar (_, _))
}
/* still bound to Int:
def top (n: Int, li: List [Int]) : List[Int] = {
extremeN (n, li) ((_ < _), (_ > _))
}
def bottom (n: Int, li: List [Int]) : List[Int] = {
extremeN (n, li) ((_ > _), (_ < _))
}
*/
def top [T] (n: Int, li: List [T])
(implicit ord: Ordering[T]): Iterable[T] = {
extremeN (n, li) (ord.lt (_, _), ord.gt (_, _))
}
def bottom [T] (n: Int, li: List [T])
(implicit ord: Ordering[T]): Iterable[T] = {
extremeN (n, li) (ord.gt (_, _), ord.lt (_, _))
}
top (3, li)
bottom (3, li)
val sl = List ("Haus", "Garten", "Boot", "Sumpf", "X", "y", "xkcd", "x11")
bottom (2, sl)
To replace List with Iterable seems to be a bit harder.
As Daniel C. Sobral pointed out in the comments, a high n in topN can lead to much sorting work, so that it could be useful, to do a manual insertion sort instead of repeatedly sorting the whole list of top-n elements:
def extremeN [T](n: Int, li: List [T])
(comp1: ((T, T) => Boolean), comp2: ((T, T) => Boolean)):
List[T] = {
def sortedIns (el: T, list: List[T]): List[T] =
if (list.isEmpty) List (el) else
if (comp2 (el, list.head)) el :: list else
list.head :: sortedIns (el, list.tail)
def updateSofar (sofar: List [T], el: T) : List [T] =
if (comp1 (el, sofar.head))
sortedIns (el, sofar.tail)
else sofar
(li.take (n) .sortWith (comp2 (_, _)) /: li.drop (n)) (updateSofar (_, _))
}
top/bottom method and usage as above. For small groups of top/bottom Elements, the sorting is rarely called, a few times in the beginning, and then less and less often over time. For example, 70 times with top (10) of 10 000, and 90 times with top (10) of 100 000.

Answer 2

Here's another solution that is simple and has pretty good performance.
def pickTopN[T](k: Int, iterable: Iterable[T])(implicit ord: Ordering[T]): Seq[T] = {
val q = collection.mutable.PriorityQueue[T](iterable.toSeq:_*)
val end = Math.min(k, q.size)
(1 to end).map(_ => q.dequeue())
}
The Big O is O(n + k log n), where k <= n. So the performance is linear for small k and at worst n log n.
The solution can also be optimized to be O(k) for memory but O(n log k) for performance. The idea is to use a MinHeap to track only the top k items at all times. Here's the solution.
def pickTopN[A, B](n: Int, iterable: Iterable[A], f: A => B)(implicit ord: Ordering[B]): Seq[A] = {
val seq = iterable.toSeq
val q = collection.mutable.PriorityQueue[A](seq.take(n):_*)(ord.on(f).reverse) // initialize with first n
// invariant: keep the top k scanned so far
seq.drop(n).foreach(v => {
q += v
q.dequeue()
})
q.dequeueAll.reverse
}

Answer 3

Yet another version:
val big = (1 to 100000)
def maxes[A](n:Int)(l:Traversable[A])(implicit o:Ordering[A]) =
l.foldLeft(collection.immutable.SortedSet.empty[A]) { (xs,y) =>
if (xs.size < n) xs + y
else {
import o._
val first = xs.firstKey
if (first < y) xs - first + y
else xs
}
}
println(maxes(4)(big))
println(maxes(2)(List("a","ab","c","z")))
Using the Set force the list to have unique values:
def maxes2[A](n:Int)(l:Traversable[A])(implicit o:Ordering[A]) =
l.foldLeft(List.empty[A]) { (xs,y) =>
import o._
if (xs.size < n) (y::xs).sort(lt _)
else {
val first = xs.head
if (first < y) (y::(xs - first)).sort(lt _)
else xs
}
}

Answer 4

You don't need to sort the entire collection in order to determine the top N elements. However, I don't believe that this functionality is supplied by the raw library, so you would have to roll you own, possibly using the pimp-my-library pattern.
For example, you can get the nth element of a collection as follows:
class Pimp[A, Repr <% TraversableLike[A, Repr]](self : Repr) {
def nth(n : Int)(implicit ord : Ordering[A]) : A = {
val trav : TraversableLike[A, Repr] = self
var ltp : List[A] = Nil
var etp : List[A] = Nil
var mtp : List[A] = Nil
trav.headOption match {
case None => error("Cannot get " + n + " element of empty collection")
case Some(piv) =>
trav.foreach { a =>
val cf = ord.compare(piv, a)
if (cf == 0) etp ::= a
else if (cf > 0) ltp ::= a
else mtp ::= a
}
if (n < ltp.length)
new Pimp[A, List[A]](ltp.reverse).nth(n)(ord)
else if (n < (ltp.length + etp.length))
piv
else
new Pimp[A, List[A]](mtp.reverse).nth(n - ltp.length - etp.length)(ord)
}
}
}
(This is not very functional; sorry)
It's then trivial to get the top n elements:
def topN(n : Int)(implicit ord : Ordering[A], bf : CanBuildFrom[Repr, A, Repr]) ={
val b = bf()
val elem = new Pimp[A, Repr](self).nth(n)(ord)
import util.control.Breaks._
breakable {
var soFar = 0
self.foreach { tt =>
if (ord.compare(tt, elem) < 0) {
b += tt
soFar += 1
}
}
assert (soFar <= n)
if (soFar < n) {
self.foreach { tt =>
if (ord.compare(tt, elem) == 0) {
b += tt
soFar += 1
}
if (soFar == n) break
}
}
}
b.result()
}
Unfortunately I'm having trouble getting this pimp to be discovered via this implicit:
implicit def t2n[A, Repr <% TraversableLike[A, Repr]](t : Repr) : Pimp[A, Repr]
= new Pimp[A, Repr](t)
I get this:
scala> List(4, 3, 6, 7, 1, 2, 8, 5).topN(4)
<console>:9: error: could not find implicit value for evidence parameter of type (List[Int]) => scala.collection.TraversableLike[A,List[Int]]
List(4, 3, 6, 7, 1, 2, 8, 5).topN(4)
^
However, the code actually works OK:
scala> new Pimp(List(4, 3, 6, 7, 1, 2, 8, 5)).topN(4)
res3: List[Int] = List(3, 1, 2, 4)
And
scala> new Pimp("ioanusdhpisjdmpsdsvfgewqw").topN(6)
res2: java.lang.String = adddfe

Answer 5

If the goal is to not sort the whole list then you could do something like this (of course it could be optimized a tad so that we don't change the list when the number clearly shouldn't be there):
List(1,6,3,7,3,2).foldLeft(List[Int]()){(l, n) => (n :: l).sorted.take(2)}

Answer 6

I implemented such an ranking algorithm recently in the Rank class of Apache Jackrabbit (in Java though). See the take method for the gist of it. The basic idea is to quicksort but terminate prematurely as soon as the top n elements have been found.

Answer 7

Here is asymptotically O(n) solution.
def top[T](data: List[T], n: Int)(implicit ord: Ordering[T]): List[T] = {
require( n < data.size)
def partition_inner(shuffledData: List[T], pivot: T): List[T] =
shuffledData.partition( e => ord.compare(e, pivot) > 0 ) match {
case (left, right) if left.size == n => left
case (left, x :: rest) if left.size < n =>
partition_inner(util.Random.shuffle(data), x)
case (left # y :: rest, right) if left.size > n =>
partition_inner(util.Random.shuffle(data), y)
}
val shuffled = util.Random.shuffle(data)
partition_inner(shuffled, shuffled.head)
}
scala> top(List.range(1,10000000), 5)
Due to recursion, this solution will take longer than some non-linear solutions above and can cause java.lang.OutOfMemoryError: GC overhead limit exceeded.
But slightly more readable IMHO and functional style. Just for job interview ;).
What is more important, that this solution can be easily parallelized.
def top[T](data: List[T], n: Int)(implicit ord: Ordering[T]): List[T] = {
require( n < data.size)
#tailrec
def partition_inner(shuffledData: List[T], pivot: T): List[T] =
shuffledData.par.partition( e => ord.compare(e, pivot) > 0 ) match {
case (left, right) if left.size == n => left.toList
case (left, right) if left.size < n =>
partition_inner(util.Random.shuffle(data), right.head)
case (left, right) if left.size > n =>
partition_inner(util.Random.shuffle(data), left.head)
}
val shuffled = util.Random.shuffle(data)
partition_inner(shuffled, shuffled.head)
}

Answer 8

For small values of n and large lists, getting the top n elements can be implemented by picking out the max element n times:
def top[T](n:Int, iter:Iterable[T])(implicit ord: Ordering[T]): Iterable[T] = {
def partitionMax(acc: Iterable[T], it: Iterable[T]): Iterable[T] = {
val max = it.max(ord)
val (nextElems, rest) = it.partition(ord.gteq(_, max))
val maxElems = acc ++ nextElems
if (maxElems.size >= n || rest.isEmpty) maxElems.take(n)
else partitionMax(maxElems, rest)
}
if (iter.isEmpty) iter.take(0)
else partitionMax(iter.take(0), iter)
}
This does not sort the entire list and takes an Ordering. I believe every method I call in partitionMax is O(list size) and I only expect to call it n times at most, so the overall efficiency for small n will be proportional to the size of the iterator.
scala> top(5, List.range(1,1000000))
res13: Iterable[Int] = List(999999, 999998, 999997, 999996, 999995)
scala> top(5, List.range(1,1000000))(Ordering[Int].on(- _))
res14: Iterable[Int] = List(1, 2, 3, 4, 5)
You could also add a branch for when n gets close to size of the iterable, and switch to iter.toList.sortBy(_.myAttr).take(n).
It does not return the type of collection provided, but you can look at How do I apply the enrich-my-library pattern to Scala collections? if this is a requirement.

Answer 9

An optimised solution using PriorityQueue with Time Complexity of O(nlogk). In the approach given in the update, you are sorting the sofar list every time which is not needed and below it is optimised by using PriorityQueue.
import scala.language.implicitConversions
import scala.language.reflectiveCalls
import collection.mutable.PriorityQueue
implicit def iterExt[A](iter: Iterable[A]) = new {
def top[B](n: Int, f: A => B)(implicit ord: Ordering[B]) : List[A] = {
def updateSofar (sofar: PriorityQueue[A], el: A): PriorityQueue[A] = {
if (ord.compare(f(el), f(sofar.head)) < 0){
sofar.dequeue
sofar.enqueue(el)
}
sofar
}
val (sofar, rest) = iter.splitAt(n)
(PriorityQueue(sofar.toSeq:_*)( Ordering.by( (x :A) => f(x) ) ) /: rest) (updateSofar (_, _)).dequeueAll.toList.reverse
}
}
case class A(s: String, i: Int)
val li = List (4, 3, 6, 7, 1, 2, 9, 5).map(i => A(i.toString(), i))
println(li.top(3, -_.i))