我试图弄清楚为什么我的groupByKey返回以下内容:
[(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>)]
我有如下所示的flatMapped值:
[(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')]
我只是做一个简单的:
groupRDD = columnRDD.groupByKey()
最佳答案
您得到的是一个对象，该对象使您可以迭代结果。您可以通过在值上调用list()将groupByKey的结果转换为列表，例如
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'])]