我有一个类似以下的RDD
dataSource = sc.parallelize( [("user1", (3, "blue")), ("user1", (4, "black")), ("user2", (5, "white"), ("user2", (3, "black")), ("user2", (6, "red")), ("user1", (1, "red"))] )
我想使用reduceByKey
为每个用户找到前2种颜色,因此输出将是RDD,如:
sc.parallelize([("user1", ["black", "blue"]), ("user2", ["red", "white"])])
因此我需要按键减少,然后对每个键的值进行排序,即(数字,颜色)数字,并返回前n个颜色。
我不想使用groupBy
。如果除了reduceByKey
之外还有groupBy
以外的任何内容,那就太好了:)
答案 0 :(得分:1)
例如,您可以使用heap queue。必需的进口:
import heapq
from functools import partial
助手功能:
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")]
数据:
rdd = sc.parallelize([
("user1", (3, "blue")), ("user1", (4, "black")),
("user2", (5, "white")), ("user2", (3, "black")),
("user2", (6, "red")), ("user1", (1, "red"))])
解决方案:
(rdd
.aggregateByKey(zero_value(2), seq_func, partial(merge_func, n=2))
.mapValues(finalize)
.collect())
结果:
[('user2', ['red', 'white']), ('user1', ['black', 'blue'])]