How createCombiner,mergeValue, mergeCombiner works in CombineByKey in Spark ( Using Scala)

Question

I am trying to understand how each step in combineByKeys works.

Can someone please help me understand the same for the below RDD?

val rdd = sc.parallelize(List(
  ("A", 3), ("A", 9), ("A", 12), ("A", 0), ("A", 5),("B", 4),
  ("B", 10), ("B", 11), ("B", 20), ("B", 25),("C", 32), ("C", 91),
   ("C", 122), ("C", 3), ("C", 55)), 2)

rdd.combineByKey(
    (x:Int) => (x, 1),
    (acc:(Int, Int), x) => (acc._1 + x, acc._2 + 1),
    (acc1:(Int, Int), acc2:(Int, Int)) => (acc1._1 + acc2._1, acc1._2 + acc2._2))

Answer 1

Let's break the process down:

First, createCombiner creates the initial value (combiner) for a key's first encounter on a partition if one is not found --> (firstValueEncountered, 1). So, this is merely initializing a tuple with the first value and a key counter of 1.

Then, mergeValue is triggered only if a combiner (tuple in our case) has already been created for the found key on this partition --> (existingTuple._1 + subSequentValue, existingTuple._2 + 1). This adds the existing tuple's value (in the first slot) with the newly encountered value and takes the existing tuple's counter (in the second slot) and increments it. So, we are

Then, mergeCombiner takes the combiners (tuples) created on each partition and merges them together --> (tupleFromPartition._1 + tupleFromPartition2._1, tupleFromPartition1._2 + tupleFromPartition2._2). This is merely adding the values from each tuple together and the counters together into one tuple.

Then, let's break up a subset of your data into partitions and see it in action:

("A", 3), ("A", 9), ("A", 12),("B", 4), ("B", 10), ("B", 11)

Partition 1

A=3 --> createCombiner(3) ==> accum[A] = (3, 1)

A=9 --> mergeValue(accum[A], 9) ==> accum[A] = (3 + 9, 1 + 1)

B=11 --> createCombiner(11) ==> accum[B] = (11, 1)

Partition 2

A=12 --> createCombiner(12) ==> accum[A] = (12, 1)

B=4 --> createCombiner(4) ==> accum[B] = (4, 1)

B=10 --> mergeValue(accum[B], 10) ==> accum[B] = (4 + 10, 1 + 1)

Merge partitions together

A ==> mergeCombiner((12, 2), (12, 1)) ==> (12 + 12, 2 + 1)

B ==> mergeCombiner((11, 1), (14, 2)) ==> (11 + 14, 1 + 2)

Now, you should probably get back an array something like this:

Array((A, (24, 3)), (B, (25, 3)))