CODEWITHSUNDEEP
pythonapache-sparkmapreducepyspark
tormond
tormond

Reputation: 452

Creating combination of value list with existing key - Pyspark

So my rdd consists of data looking like: 

(k, [v1,v2,v3...])

I want to create a combination of all sets of two for the value part.

So the end map should look like:

(k1, (v1,v2))
(k1, (v1,v3))
(k1, (v2,v3))

I know to get the value part, I would use something like 

rdd.cartesian(rdd).filter(case (a,b) => a < b)

However, that requires the entire rdd to be passed (right?) not just the value part. I am unsure how to arrive at my desired end, I suspect its a groupby.

Also, ultimately, I want to get to the k,v looking like 

((k1,v1,v2),1)

I know how to get from what I am looking for to that, but maybe its easier to go straight there?

Thanks.

Upvotes: 1

Views: 3757

Answers (3)

Andy Quiroz
Andy Quiroz

Reputation: 901

I have made this algorithm, but with higher numbers looks like that doesn't work or its very slow, it will run in a cluster of big data(cloudera), so i think that i have to put the function into pyspark, please give a hand if you can.

import pandas as pd import itertools as itts

number_list = [10953, 10423, 10053]

def reducer(nums): def ranges(n): print(n) return range(n, -1, -1)

num_list = list(map(ranges, nums)) return list(itts.product(*num_list))

data=pd.DataFrame(reducer(number_list)) print(data)

Upvotes: 0

Israel Unterman
Israel Unterman

Reputation: 13510

Use itertools to create the combinations. Here is a demo:

import itertools

k, v1, v2, v3 = 'k1 v1 v2 v3'.split()

a = (k, [v1,v2,v3])

b = itertools.combinations(a[1], 2)
data = [(k, pair) for pair in b]

data will be:

[('k1', ('v1', 'v2')), ('k1', ('v1', 'v3')), ('k1', ('v2', 'v3'))]

Upvotes: 1

Alberto Bonsanto
Alberto Bonsanto

Reputation: 18022

I think Israel's answer is a incomplete, so I go a step further. 

import itertools

a = sc.parallelize([
    (1, [1,2,3,4]),
    (2, [3,4,5,6]),
    (3, [-1,2,3,4])
  ])

def combinations(row):
  l = row[1]
  k = row[0]
  return [(k, v) for v in itertools.combinations(l, 2)]

a.map(combinations).flatMap(lambda x: x).take(3)
# [(1, (1, 2)), (1, (1, 3)), (1, (1, 4))]

Upvotes: 2

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