Pyspark:使用具有多重排序条件的 repartitionAndSortWithinPartitions [英] Pyspark: Using repartitionAndSortWithinPartitions with multiple sort Critiria

问题描述

假设我有以下 RDD:

Assuming I am having the following RDD:

rdd = sc.parallelize([('a', (5,1)), ('d', (8,2)), ('2', (6,3)), ('a', (8,2)), ('d', (9,6)), ('b', (3,4)),('c', (8,3))])

我如何使用 repartitionAndSortWithinPartitions 并按 x[0] 和 x[1][0] 之后排序.使用以下我只按键排序(x[0]):

How can I use repartitionAndSortWithinPartitions and sort by x[0] and after x[1][0]. Using the following I sort only by the key(x[0]):

Npartitions = sc.defaultParallelism
rdd2 = rdd.repartitionAndSortWithinPartitions(2, lambda x: hash(x) % Npartitions, 2)

一种方法如下，但我想应该有更简单的方法:

A way to do it is the following but there should something more simple I guess:

Npartitions = sc.defaultParallelism 
partitioned_data = rdd
  .partitionBy(2)
  .map(lambda x:(x[0],x[1][0],x[1][1]))
  .toDF(['letter','number2','number3'])
  .sortWithinPartitions(['letter','number2'],ascending=False)
  .map(lambda x:(x.letter,(x.number2,x.number3)))

>>> partitioned_data.glom().collect()

[[],
[(u'd', (9, 6)), (u'd', (8, 2))],
[(u'c', (8, 3)), (u'c', (6, 3))],
[(u'b', (3, 4))],
[(u'a', (8, 2)), (u'a', (5, 1))]

正如所见，我必须将其转换为 Dataframe 才能使用 sortWithinPartitions.还有其他方法吗?使用 repartitionAndSortWIthinPartitions?

As it can be seen I have to convert it to Dataframe in order to use sortWithinPartitions. Is there another way? Using repartitionAndSortWIthinPartitions?

(数据不是全局排序没有关系.我只关心在分区内排序.)

(It doesnt matter that the data is not globally sorted. I care only to be sorted inside the partitions.)

推荐答案

这是可能的，但您必须在组合键中包含所有必需的信息:

It is possible but you'll have to include all required information in the composite key:

from pyspark.rdd import portable_hash

n = 2

def partitioner(n):
    """Partition by the first item in the key tuple"""
    def partitioner_(x):
        return portable_hash(x[0]) % n
    return partitioner_


(rdd
  .keyBy(lambda kv: (kv[0], kv[1][0]))  # Create temporary composite key
  .repartitionAndSortWithinPartitions(
      numPartitions=n, partitionFunc=partitioner(n), ascending=False)
  .map(lambda x: x[1]))  # Drop key (note: there is no partitioner set anymore)

分步说明:

• keyBy(lambda kv: (kv[0], kv[1][0])) 创建一个由原始键和值的第一个元素组成的替代键.换句话说，它会转换:

• keyBy(lambda kv: (kv[0], kv[1][0])) creates a substitute key which consist of original key and the first element of the value. In other words it transforms:

(0, (5,1))

进入

((0, 5), (0, (5, 1)))

在实践中，简单地将数据重塑为

In practice it can be slightly more efficient to simply reshape data to

((0, 5), 1)

partitioner 根据键的第一个元素的散列定义分区函数，因此:

partitioner defines partitioning function based on a hash of the first element of the key so:

partitioner(7)((0, 5))
## 0

partitioner(7)((0, 6))
## 0

partitioner(7)((0, 99))
## 0

partitioner(7)((3, 99))
## 3

如您所见，它是一致的并且忽略了第二位.

as you can see it is consistent and ignores the second bit.

我们使用默认的 keyfunc 函数，它是身份 (lambda x: x) 并依赖于 Python tuple 上定义的字典顺序:

we use default keyfunc function which is identity (lambda x: x) and depend on lexicographic ordering defined on Python tuple:

(0, 5) < (1, 5)
## True

(0, 5) < (0, 4)
## False

如前所述，您可以改为重塑数据:

As mentioned before you could reshape data instead:

rdd.map(lambda kv: ((kv[0], kv[1][0]), kv[1][1]))

并删除最终的 map 以提高性能.

and drop final map to improve performance.

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