PySpark groupByKey regresando pyspark.resultatiterable.Resultados

Estoy tratando de averiguar por qué mi groupByKey está devolviendo lo siguiente: 

[(0, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a210>), (1, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a4d0>), (2, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a390>), (3, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a290>), (4, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a450>), (5, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a350>), (6, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a1d0>), (7, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a490>), (8, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a050>), (9, <pyspark.resultiterable.ResultIterable object at 0x7fc659e0a650>)]

Tengo valores planos que se ven así: 

[(0, u'D'), (0, u'D'), (0, u'D'), (0, u'D'), (0, u'D'), (0, u'D'), (0, u'D'), (0, u'D'), (0, u'D'), (0, u'D')]

Estoy haciendo un simple: 

groupRDD = columnRDD.groupByKey()
Author: theMadKing, 2015-04-18
Source

4 answers

Lo que está obteniendo es un objeto que le permite iterar sobre los resultados. Puede convertir los resultados de groupByKey en una lista llamando a list () en los valores, por ejemplo

example = sc.parallelize([(0, u'D'), (0, u'D'), (1, u'E'), (2, u'F')])

example.groupByKey().collect()
# Gives [(0, <pyspark.resultiterable.ResultIterable object ......]

example.groupByKey().map(lambda x : (x[0], list(x[1]))).collect()
# Gives [(0, [u'D', u'D']), (1, [u'E']), (2, [u'F'])]
 56
Author: dpeacock,
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2015-04-18 14:52:02

También puedes usar

example.groupByKey().mapValues(list)
 17
Author: Jayaram,
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2016-05-03 19:03:12

En lugar de usar groupByKey(), te sugiero que uses cogroup(). Puede consultar el siguiente ejemplo.

[(x, tuple(map(list, y))) for x, y in sorted(list(x.cogroup(y).collect()))]

Ejemplo: 

>>> x = sc.parallelize([("foo", 1), ("bar", 4)])
>>> y = sc.parallelize([("foo", -1)])
>>> z = [(x, tuple(map(list, y))) for x, y in sorted(list(x.cogroup(y).collect()))]
>>> print(z)

Debe obtener la salida deseada...

 1
Author: Harsha,
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2016-02-17 06:51:10

Ejemplo: 

r1 = sc.parallelize([('a',1),('b',2)])
r2 = sc.parallelize([('b',1),('d',2)])
r1.cogroup(r2).mapValues(lambdax:tuple(reduce(add,__builtin__.map(list,x))))

Resultado: 

[('d', (2,)), ('b', (2, 1)), ('a', (1,))]
 1
Author: bin yan,
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2017-12-07 07:52:58