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From Joe L <selme...@yahoo.com>
Subject groupByKey(None) returns partitions according to the keys?
Date Wed, 16 Apr 2014 04:58:37 GMT
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])]



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