笛卡尔积中的Spark Unique对 [英] Spark Unique pair in cartesian product

问题描述

我有这个:

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)]

我想要以下结果:

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)]

因为我的演算返回了对称矩阵(相关性). 实现此目标的最佳方法是什么? (无循环) 使用a，b和c可以是任何值，甚至可以是元组.

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.

推荐答案

不确定python语法，但是在scala中，您可以编写:

Not sure about the python syntax, but in scala you could write:

a.cartesian(a).filter{ case (a,b) => a <= b }.collect()

我的猜测是在python中，它类似于:

My guess is in python it would be something like:

a.cartesian(a).filter(lambda a, b: a <= b).collect()

这篇关于笛卡尔积中的Spark Unique对的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！