python - Pyspark: Using repartitionAndSortWithinPartitions with multiple sort Critiria

Question

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))])

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)

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))]

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

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

Answer 1

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)

Explained step-by-step:

• 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))
  

  into

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

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

  ((0, 5), 1)
  

• 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
  

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

• 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
  

As mentioned before you could reshape data instead:

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

and drop final map to improve performance.