How to sort RDD

Question

I have scoreTriplets is RDD[ARRAY[String]] which I am sorting by following way.

var ScoreTripletsArray = scoreTriplets.collect()
  if (ScoreTripletsArray.size > 0) {        
    /*Sort the ScoreTripletsArray descending by score field*/        
    scala.util.Sorting.stableSort(ScoreTripletsArray, (e1: Array[String], e2: Array[String]) => e1(3).toInt > e2(3).toInt)
}

But collect() will be heavy If there is elements in lack.

So I need to sort RDD by score without using collect().
scoreTriples is RDD[ARRAY[String]] each line of RDD will store Array of the below variables.
EdgeId sourceID destID score sourceNAme destNAme distance

Please give me any reference or hint.

Solution 1

Sorting will be, due to shuffling, an expensive operation even without collecting but you can use sortBy method:

import scala.util.Random

val data = Seq.fill(10)(Array.fill(3)("") :+ Random.nextInt.toString)
val rdd  = sc.parallelize(data)

val sorted = rdd.sortBy(_.apply(3).toInt)
sorted.take(3)
// Array[Array[String]] = Array(
//   Array("", "", "", -1660860558),
//   Array("", "", "", -1643214719),
//   Array("", "", "", -1206834289))

If you're interested only in the top results then top and takeOrdered are usually preferred.

import scala.math.Ordering

rdd.takeOrdered(2)(Ordering.by[Array[String], Int](_.apply(3).toInt))
// Array[Array[String]] = 
//   Array(Array("", "", "", -1660860558), Array("", "", "", -1643214719))

rdd.top(2)(Ordering.by[Array[String], Int](_.apply(3).toInt))
// Array[Array[String]] = 
//   Array(Array("", "", "", 1920955686), Array("", "", "", 1597012602))

Solution 2

There is sortBy method in RDD (see doc). You can do something like that

scoreTriplets.sortBy( _(3).toInt )