Q:

Perform a left outer join of self and other.

x = sc.parallelize([("a", 1), ("b", 4)])
y = sc.parallelize([("a", 2)])
sorted(y.rightOuterJoin(x).collect())
# [('a', (2, 1)), ('b', (None, 4))]
0
x = sc.parallelize([("a", 1), ("b", 4)])
y = sc.parallelize([("a", 2)])
sorted(x.leftOuterJoin(y).collect())
# [('a', (1, 2)), ('b', (4, None))]
0

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