问题
I have this:
In [1]:a = sc.parallelize([a,b,c])
In [2]:a.cartesian(a).collect()
Out[3]: [(a, a), (a, b), (a, c), (b, a), (c, a), (b, b), (b, c), (c, b), (c, c)]
I want the following result:
In [1]:a = sc.parallelize([1,2,3])
In [2]:a.cartesianMoreInteligent(a).collect()
Out[3]: [(a, a), (a, b), (a, c), (b, b), (b, c), (c, c)]
Because my calculus return a symetrical matrix (correlation). What is the best way to achieve this ? (No loop) With a, b and c can be anything, even tuple.
回答1:
Not sure about the python syntax, but in scala you could write:
a.cartesian(a).filter{ case (a,b) => a <= b }.collect()
My guess is in python it would be something like:
a.cartesian(a).filter(lambda a, b: a <= b).collect()
来源：https://stackoverflow.com/questions/37027848/spark-unique-pair-in-cartesian-product