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 {
   // <snip>
  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()

Answer 2

Another option, at least from PySpark 1.2.0, is the use of takeOrdered.

In ascending order:

rdd.takeOrdered(10)

In descending order:

rdd.takeOrdered(10, lambda x: -x)

Top k values for k,v pairs:

rdd.takeOrdered(10, lambda (k, v): -v)

Answer 3

If you only need the top 10, use rdd.top(10). It avoids sorting, so it is faster.

rdd.top makes one parallel pass through the data, collecting the top N in each partition in a heap, then merges the heaps. It is an O(rdd.count) operation. Sorting would be O(rdd.count log rdd.count), and incur a lot of data transfer — it does a shuffle, so all of the data would be transmitted over the network.