Reputation: 49816
I'm trying to randomise the order of elements in an RDD. My current approach is to zip the elements with an RDD of shuffled integers, then later join by those integers.
However, pyspark falls over with only 100000000 integers. I'm using the code below.
My question is: is there a better way to either zip with the random index or otherwise shuffle?
I've tried sorting by a random key, which works, but is slow.
def random_indices(n):
"""
return an iterable of random indices in range(0,n)
"""
indices = range(n)
random.shuffle(indices)
return indices
The following happens in pyspark:
Using Python version 2.7.3 (default, Jun 22 2015 19:33:41)
SparkContext available as sc.
>>> import clean
>>> clean.sc = sc
>>> clean.random_indices(100000000)
Killed
Upvotes: 8
Views: 5805
Reputation: 851
pyspark worked!
from random import randrange
data_rnd = data.sortBy(lambda x: randrange(1000000))
Upvotes: -1
Reputation: 330063
One possible approach is to add random keys using mapParitions
import os
import numpy as np
swap = lambda x: (x[1], x[0])
def add_random_key(it):
# make sure we get a proper random seed
seed = int(os.urandom(4).encode('hex'), 16)
# create separate generator
rs = np.random.RandomState(seed)
# Could be randint if you prefer integers
return ((rs.rand(), swap(x)) for x in it)
rdd_with_keys = (rdd
# It will be used as final key. If you don't accept gaps
# use zipWithIndex but this should be cheaper
.zipWithUniqueId()
.mapPartitions(add_random_key, preservesPartitioning=True))
Next you can repartition, sort each partition and extract values:
n = rdd.getNumPartitions()
(rdd_with_keys
# partition by random key to put data on random partition
.partitionBy(n)
# Sort partition by random value to ensure random order on partition
.mapPartitions(sorted, preservesPartitioning=True)
# Extract (unique_id, value) pairs
.values())
If sorting per partition is still to slow it could be replaced by Fisher–Yates shuffle.
If you simply need a random data then you can use mllib.RandomRDDs
from pyspark.mllib.random import RandomRDDs
RandomRDDs.uniformRDD(sc, n)
Theoretically it could be zipped with input rdd
but it would require matching the number of elements per partition.
Upvotes: 5