Penny_Q的专栏

# coding:utf-8 from pyspark import SparkContext sc = SparkContext(appName='pr') links = [['A', ['B', 'C']], ['B', ['A', 'C']], ['C', ['A', 'B', 'D']], ['D', ['C']]] links = sc.parallelize(links).partitionBy(4).persist() ranks = links.mapValues(lambda x: 1.