How does HashPartitioner work?

Question

I read up on the documentation of HashPartitioner. Unfortunately nothing much was explained except for the API calls. I am under the assumption that HashPartitioner

Answer 1

RDD is distributed this means it is split on some number of parts. Each of this partitions is potentially on different machine. Hash partitioner with argument numPartitions chooses on what partition to place pair (key, value) in following way:

1. Creates exactly numPartitions partitions.
2. Places (key, value) in partition with number Hash(key) % numPartitions

Answer 2

The HashPartitioner.getPartition method takes a key as its argument and returns the index of the partition which the key belongs to. The partitioner has to know what the valid indices are, so it returns numbers in the right range. The number of partitions is specified through the numPartitions constructor argument.

The implementation returns roughly key.hashCode() % numPartitions. See Partitioner.scala for more details.

Answer 3

Well, lets make your dataset marginally more interesting:

val rdd = sc.parallelize(for {
    x <- 1 to 3
    y <- 1 to 2
} yield (x, None), 8)

We have six elements:

rdd.count

Long = 6

no partitioner:

rdd.partitioner

Option[org.apache.spark.Partitioner] = None

and eight partitions:

rdd.partitions.length

Int = 8

Now lets define small helper to count number of elements per partition:

import org.apache.spark.rdd.RDD

def countByPartition(rdd: RDD[(Int, None.type)]) = {
    rdd.mapPartitions(iter => Iterator(iter.length))
}

Since we don't have partitioner our dataset is distributed uniformly between partitions (Default Partitioning Scheme in Spark):

countByPartition(rdd).collect()

Array[Int] = Array(0, 1, 1, 1, 0, 1, 1, 1)

Now lets repartition our dataset:

import org.apache.spark.HashPartitioner
val rddOneP = rdd.partitionBy(new HashPartitioner(1))

Since parameter passed to HashPartitioner defines number of partitions we have expect one partition:

rddOneP.partitions.length

Int = 1

Since we have only one partition it contains all elements:

countByPartition(rddOneP).collect

Array[Int] = Array(6)

Note that the order of values after the shuffle is non-deterministic.

Same way if we use HashPartitioner(2)

val rddTwoP = rdd.partitionBy(new HashPartitioner(2))

we'll get 2 partitions:

rddTwoP.partitions.length

Int = 2

Since rdd is partitioned by key data won't be distributed uniformly anymore:

countByPartition(rddTwoP).collect()

Array[Int] = Array(2, 4)

Because with have three keys and only two different values of hashCode mod numPartitions there is nothing unexpected here:

(1 to 3).map((k: Int) => (k, k.hashCode, k.hashCode % 2))

scala.collection.immutable.IndexedSeq[(Int, Int, Int)] = Vector((1,1,1), (2,2,0), (3,3,1))

Just to confirm the above:

rddTwoP.mapPartitions(iter => Iterator(iter.map(_._1).toSet)).collect()

Array[scala.collection.immutable.Set[Int]] = Array(Set(2), Set(1, 3))

Finally with HashPartitioner(7) we get seven partitions, three non-empty with 2 elements each:

val rddSevenP = rdd.partitionBy(new HashPartitioner(7))
rddSevenP.partitions.length

Int = 7

countByPartition(rddTenP).collect()

Array[Int] = Array(0, 2, 2, 2, 0, 0, 0)

Summary and Notes

• HashPartitioner takes a single argument which defines number of partitions

• values are assigned to partitions using hash of keys. hash function may differ depending on the language (Scala RDD may use hashCode, DataSets use MurmurHash 3, PySpark, portable_hash).

  In simple case like this, where key is a small integer, you can assume that hash is an identity (i = hash(i)).

  Scala API uses nonNegativeMod to determine partition based on computed hash,

• if distribution of keys is not uniform you can end up in situations when part of your cluster is idle

• keys have to be hashable. You can check my answer for A list as a key for PySpark's reduceByKey to read about PySpark specific issues. Another possible problem is highlighted by HashPartitioner documentation:

  Java arrays have hashCodes that are based on the arrays' identities rather than their contents, so attempting to partition an RDD[Array[]] or RDD[(Array[], _)] using a HashPartitioner will produce an unexpected or incorrect result.

• In Python 3 you have to make sure that hashing is consistent. See What does Exception: Randomness of hash of string should be disabled via PYTHONHASHSEED mean in pyspark?

• Hash partitioner is neither injective nor surjective. Multiple keys can be assigned to a single partition and some partitions can remain empty.

• Please note that currently hash based methods don't work in Scala when combined with REPL defined case classes (Case class equality in Apache Spark).

• HashPartitioner (or any other Partitioner) shuffles the data. Unless partitioning is reused between multiple operations it doesn't reduce amount of data to be shuffled.