Question

我在Ubuntu上使用jupyter。

所以我遇到了下一个问题，这是我的代码：

from pyspark import SparkContext
 sc = SparkContext.getOrCreate()
 ut = sc.textFile("hdfs://localhost:54310/hduser/firstnames")
 rows= ut.map(lambda line: line.split(";"))
 res = rows.filter(lamda row: row[2] >= "2000" and row[2] <= "2004")
 res = res.map(lambda row: ({row[1],row[2]},int(row[3])))

输出：

[({'2001', 'Brussel'}, 9),
 ({'2001', 'Brussel'}, 104),
 ({'2001', 'Vlaanderen'}, 16),
 ({'2002', 'Brussel'}, 12), ...]

我需要输出如下：

[({'2001', 'Brussel'}, 113),
 ({'2001', 'Vlaanderen'}, 16),
 ({'2002', 'Brussel'}, 12)]

我之前和之前尝试过reduceByKey的一些事情已经看到很多关于reduceByKey的问题，但无法弄清楚。提前谢谢。

Answer 1

正如A list as a key for PySpark's reduceByKey zero323中所述，密钥必须实现哈希方法。您可以使用tuples：

>>> from operator import add
... 
... sc.parallelize([
...     (('2001', 'Brussel'), 9), (('2001', 'Brussel'), 104),
...     (('2001', 'Vlaanderen'), 16), (('2002', 'Brussel'), 12)
... ]).reduceByKey(add).take(2)
... 
[(('2002', 'Brussel'), 12), (('2001', 'Brussel'), 113)]

替换：

res.map(lambda row: ({row[1],row[2]},int(row[3])))

带

res.map(lambda row: ((row[1], row[2]), int(row[3])))

或将set替换为frozenset：

>>> sc.parallelize([
...     (frozenset(['2001', 'Brussel']), 9), (frozenset(['2001', 'Brussel']), 104),
...     (frozenset(['2001', 'Vlaanderen']), 16), (frozenset(['2002', 'Brussel']), 12)
... ]).reduceByKey(add).take(2)

[(frozenset({'2002', 'Brussel'}), 12), (frozenset({'2001', 'Brussel'}), 113)]

如何在pyspark中使用reduceByKey来获取多键和单值

1 个答案: