Pyspark: repartition vs partitionBy

Question

I'm working through these two concepts right now and would like some clarity. From working through the command line, I've been trying to identify the differences and when a developer would use repartition vs partitionBy.

Here is some sample code:

rdd = sc.parallelize([('a', 1), ('a', 2), ('b', 1), ('b', 3), ('c',1), ('ef',5)])
rdd1 = rdd.repartition(4)
rdd2 = rdd.partitionBy(4)

rdd1.glom().collect()
[[('b', 1), ('ef', 5)], [], [], [('a', 1), ('a', 2), ('b', 3), ('c', 1)]]

rdd2.glom().collect()
[[('a', 1), ('a', 2)], [], [('c', 1)], [('b', 1), ('b', 3), ('ef', 5)]]

I took a look at the implementation of both, and the only difference I've noticed for the most part is that partitionBy can take a partitioning function, or using the portable_hash by default. So in partitionBy, all the same keys should be in the same partition. In repartition, I would expect the values to be distributed more evenly over the partitions, but this isnt the case.

Given this, why would anyone ever use repartition? I suppose the only time I could see it being used is if I'm not working with PairRDD, or I have large data skew?

Is there something that I'm missing, or could someone shed light from a different angle for me?

Answer 1

repartition() already exists in RDDs, and does not handle partitioning by key (or by any other criterion except Ordering). 

repartition() is used for specifying the number of partitions considering the number of cores and the amount of data you have.

partitionBy() is most importantly used for making shuffling functions more efficient, such as reduceByKey(), join(), cogroup() etc.. It is only beneficial in cases where a RDD is used for multiple times, so it is usually followed by persist().

Differences between the two in action:

pairs = sc.parallelize([1, 2, 3, 4, 2, 4, 1, 5, 6, 7, 7, 5, 5, 6, 4]).map(lambda x: (x, x))

pairs.partitionBy(3).glom().collect()

[[(3, 3), (6, 6), (6, 6)],

 [(1, 1), (4, 4), (4, 4), (1, 1), (7, 7), (7, 7), (4, 4)],

 [(2, 2), (2, 2), (5, 5), (5, 5), (5, 5)]]

pairs.repartition(3).glom().collect()

[[(4, 4), (2, 2), (6, 6), (7, 7), (5, 5), (5, 5)],

 [(1, 1), (4, 4), (6, 6), (4, 4)],

 [(2, 2), (3, 3), (1, 1), (5, 5), (7, 7)]]