I was wonder if groupByKey returns 2 partitions in the below example?
>>> x = sc.parallelize([("a", 1), ("b", 1), ("a", 1)])
>>> sorted(x.groupByKey().collect())
[('a', [1, 1]), ('b', [1])]-- View this message in context: http://apache-spark-user-list.1001560.n3.nabble.com/groupByKey-None-returns-partitions-according-to-the-keys-tp4318.html Sent from the Apache Spark User List mailing list archive at Nabble.com.
