How to sort an RDD in Scala Spark?

Question

Reading Spark method sortByKey :

sortByKey([ascending], [numTasks])   When called on a dataset of (K, V) pairs where K implements Ordered, returns a dataset         


        

Answer 1

Most likely you have already perused the source code:

  class OrderedRDDFunctions {
   // 
  def sortByKey(ascending: Boolean = true, numPartitions: Int = self.partitions.size): RDD[P] = {
    val part = new RangePartitioner(numPartitions, self, ascending)
    val shuffled = new ShuffledRDD[K, V, P](self, part)
    shuffled.mapPartitions(iter => {
      val buf = iter.toArray
      if (ascending) {
        buf.sortWith((x, y) => x._1 < y._1).iterator
      } else {
        buf.sortWith((x, y) => x._1 > y._1).iterator
      }
    }, preservesPartitioning = true)
  }

And, as you say, the entire data must go through the shuffle stage - as seen in the snippet.

However, your concern about subsequently invoking take(K) may not be so accurate. This operation does NOT cycle through all N items:

  /**
   * Take the first num elements of the RDD. It works by first scanning one partition, and use the
   * results from that partition to estimate the number of additional partitions needed to satisfy
   * the limit.
   */
  def take(num: Int): Array[T] = {

So then, it would seem:

O(myRdd.take(K)) << O(myRdd.sortByKey()) ~= O(myRdd.sortByKey.take(k)) (at least for small K) << O(myRdd.sortByKey().collect()