Take top N elements from each group in PySpark RDD (without using groupByKey)

问题

I have an RDD like the following

dataSource = sc.parallelize( [("user1", (3, "blue")), ("user1", (4, "black")), ("user2", (5, "white"), ("user2", (3, "black")), ("user2", (6, "red")), ("user1", (1, "red"))] )

I want to use reduceByKey to find Top 2 colors for each user so the output would be an RDD like:

sc.parallelize([("user1", ["black", "blue"]), ("user2", ["red", "white"])])

so I need to reduce by key and then sort each key's values, i.e. (number, color) on number and return top n colors.

I don't want to use groupBy. If there is anything better than reduceByKey other than groupBy, it would be great :)

回答1:

You can for example use a heap queue. Required imports:

import heapq
from functools import partial

Helper functions:

def zero_value(n):
    """Initialize a queue. If n is large
    it could be more efficient to track a number of the elements
    on heap (cnt, heap) and switch between heappush and heappushpop
    if we exceed n. I leave this as an exercise for the reader."""
    return [(float("-inf"), None) for _ in range(n)]

def seq_func(acc, x):
    heapq.heappushpop(acc, x)
    return acc

def merge_func(acc1, acc2, n):
    return heapq.nlargest(n, heapq.merge(acc1, acc2))

def finalize(kvs):
    return [v for (k, v) in kvs if k != float("-inf")]

Data:

rdd = sc.parallelize([
    ("user1", (3, "blue")), ("user1", (4, "black")),
    ("user2", (5, "white")), ("user2", (3, "black")),
    ("user2", (6, "red")), ("user1", (1, "red"))])

Solution:

(rdd
    .aggregateByKey(zero_value(2), seq_func, partial(merge_func, n=2))
    .mapValues(finalize)
    .collect())

Result:

[('user2', ['red', 'white']), ('user1', ['black', 'blue'])]

来源：https://stackoverflow.com/questions/41903310/take-top-n-elements-from-each-group-in-pyspark-rdd-without-using-groupbykey

标签

apache-spark

pyspark