Question

我有一个键值对RDD，其中键是一个actor，而这个actor参与的电影的值是：

["actor 1", "movie 1"]
["actor 1", "movie 2"]
["actor 1", "movie 3"]
...
["actor n", "movie 2"]

我想将其映射到另一个键值对RDD，其中每对由两个参与公共电影的演员组成。

在上面的示例中，这意味着新的RDD将包含["actor 1", "actor n"]对，因为它们都参与"movie 2"。

Answer 1

简单的交换和连接应该可以解决问题。首先让我们创建一些虚拟数据和一个小辅助函数：

actor_movie = sc.parallelize([
    ("actor 1", "movie 1"),
    ("actor 1", "movie 3"),
    ("actor 1", "movie 3"),
    ("actor n", "movie 2")
])

swap = lambda x: (x[1], x[0])

接下来你交换订单：

movie_actor =  (actor_movie.map(swap)
    .partitionBy(actor_movie.getNumPartitions())
    .cache())

加入：

(movie_actor
    .join(movie_actor) # Join by movie
    .values() # Extract values (actors)
    .filter(lambda x: x[0] != x[1]))

Answer 2

这不完全是你要求的，但我认为这已经足够了：

import itertools as iter

movies = sc.parallelize([("P", "SW4"), ("P", "SW5"), ("P", "SW6"),
                         ("A", "SW4"), ("A", "SW5"),
                         ("B", "SW5"), ("B", "SW6"),
                         ("W", "SW4"),
                         ("X", "SW1"), ("X", "SW7"), ("X", "SW2"), ("X", "SW3"),
                         ("Y", "SW1"), ("Y", "SW7"), ("Y", "SW2"), ("Y", "SW3")])

swap_tuple = lambda (k, v): (v, k)

movies = movies.groupByKey().mapValues(list)

all_pairs = movies.flatMap(lambda (movie, actors): map(lambda actors:(movie, actors), iter.combinations(actors, 2)))

print all_pairs.collect()

""" 
    >> [('SW1', ('X', 'Y')),
        ('SW3', ('X', 'Y')),
        ('SW5', ('P', 'A')),
        ('SW5', ('P', 'B')),
        ('SW5', ('A', 'B')),
        ('SW7', ('X', 'Y')),
        ('SW2', ('X', 'Y')),
        ('SW4', ('P', 'A')),
        ('SW4', ('P', 'W')),
        ('SW4', ('A', 'W')),
        ('SW6', ('P', 'B'))]
"""

Here是使用.ipynb

运行的

PySpark - 使用两个共享相同值的键创建对RDD

2 个答案: