[Spark][Python]groupByKey例子

Posted 2020-10-09

[Spark][Python]sortByKey 例子 的继续:

[Spark][Python]groupByKey例子

In [29]: mydata003.collect()

Out[29]:
[[u‘00001‘, u‘sku933‘],
[u‘00001‘, u‘sku022‘],
[u‘00001‘, u‘sku912‘],
[u‘00001‘, u‘sku331‘],
[u‘00002‘, u‘sku010‘],
[u‘00003‘, u‘sku888‘],
[u‘00004‘, u‘sku411‘]]

In [30]: mydata005=mydata003.groupByKey()

In [32]: mydata005.count()
Out[32]: 4

In [33]: mydata005.collect()
Out[33]:
[(u‘00004‘, <pyspark.resultiterable.ResultIterable at 0x7fcebe436b10>),
(u‘00001‘, <pyspark.resultiterable.ResultIterable at 0x7fcebe436850>),
(u‘00003‘, <pyspark.resultiterable.ResultIterable at 0x7fcebe436050>),
(u‘00002‘, <pyspark.resultiterable.ResultIterable at 0x7fcebe4361d0>)]

那么，对于这种：

(00004,sku411)
(00003,sku888)
(00003,sku022)
(00003,sku010)
(00003,sku594)
(00002,sku912)

理论上变成了这样形式的：

(00002,[sku912,sku331])
(00001,[sku022,sku010,sku933])
(00003,[sku888,sku022,sku010,sku594])
(00004,[sku411])

我们如何把它们都打印输出成如下的格式，我考虑需要用到函数，然后对RDD的每行的Value，看作list，再来遍历。
（等待下次编写）

00002
sku912
sku331

00001
sku022
sku010
sku933

00003
sku088
sku022
sku022
sku010
sku594

00004
sku411