CODEWITHSUNDEEP
scalaapache-spark
Eugene Zhulenev
Eugene Zhulenev

Reputation: 9734

Spark: produce RDD[(X, X)] of all possible combinations from RDD[X]

Is it possible in Spark to implement '.combinations' function from scala collections?

   /** Iterates over combinations.
   *
   *  @return   An Iterator which traverses the possible n-element combinations of this $coll.
   *  @example  `"abbbc".combinations(2) = Iterator(ab, ac, bb, bc)`
   */

For example how can I get from RDD[X] to RDD[List[X]] or RDD[(X,X)] for combinations of size = 2. And lets assume that all values in RDD are unique.

Upvotes: 26

Views: 29633

Answers (4)

Behzad Behzadan
Behzad Behzadan

Reputation: 21

This creates all combinations (n, 2) and works for any RDD without requiring any ordering on the elements of RDD.

val rddWithIndex = rdd.zipWithIndex
rddWithIndex.cartesian(rddWithIndex).filter{case(a, b) => a._2 < b._2}.map{case(a, b) => (a._1, b._1)}

a._2 and b._2 are the indices, while a._1 and b._1 are the elements of the original RDD.

Example:

Note that, no ordering is defined on the maps here.

val m1 = Map('a' -> 1, 'b' -> 2)
val m2 = Map('c' -> 3, 'a' -> 4)
val m3 = Map('e' -> 5, 'c' -> 6, 'b' -> 7)
val rdd = sc.makeRDD(Array(m1, m2, m3))
val rddWithIndex = rdd.zipWithIndex
rddWithIndex.cartesian(rddWithIndex).filter{case(a, b) => a._2 < b._2}.map{case(a, b) => (a._1, b._1)}.collect

Output:

Array((Map(a -> 1, b -> 2),Map(c -> 3, a -> 4)), (Map(a -> 1, b -> 2),Map(e -> 5, c -> 6, b -> 7)), (Map(c -> 3, a -> 4),Map(e -> 5, c -> 6, b -> 7)))

Upvotes: 2

maasg
maasg

Reputation: 37435

As discussed, cartesian will give you n^2 elements of the cartesian product of the RDD with itself. This algorithm computes the combinations (n,2) of an RDD without having to compute the n^2 elements first: (used String as type, generalizing to a type T takes some plumbing with classtags that would obscure the purpose here)

This is probably less time efficient that cartesian + filtering due to the iterative count and take actions that forces the computation of the RDD, but more space efficient as it calculates only the C(n,2) = n!/(2*(n-2))! = (n*(n-1)/2) elements instead of the n^2 of the cartesian product. 

 import org.apache.spark.rdd._

 def combs(rdd:RDD[String]):RDD[(String,String)] = {
    val count = rdd.count
    if (rdd.count < 2) { 
        sc.makeRDD[(String,String)](Seq.empty)
    } else if (rdd.count == 2) {
        val values = rdd.collect
        sc.makeRDD[(String,String)](Seq((values(0), values(1))))
    } else {
        val elem = rdd.take(1)
        val elemRdd = sc.makeRDD(elem)
        val subtracted = rdd.subtract(elemRdd)  
        val comb = subtracted.map(e  => (elem(0),e))
        comb.union(combs(subtracted))
    } 
 }

Upvotes: 3

aaronman
aaronman

Reputation: 18750

Cartesian product and combinations are two different things, the cartesian product will create an RDD of size rdd.size() ^ 2 and combinations will create an RDD of size rdd.size() choose 2 

val rdd = sc.parallelize(1 to 5)
val combinations = rdd.cartesian(rdd).filter{ case (a,b) => a < b }`.
combinations.collect()

Note this will only work if an ordering is defined on the elements of the list, since we use <. This one only works for choosing two but can easily be extended by making sure the relationship a < b for all a and b in the sequence

Upvotes: 32

maasg
maasg

Reputation: 37435

This is supported natively by a Spark RDD with the cartesian transformation.

e.g.:

val rdd = sc.parallelize(1 to 5)
val cartesian = rdd.cartesian(rdd)
cartesian.collect

Array[(Int, Int)] = Array((1,1), (1,2), (1,3), (1,4), (1,5), 
(2,1), (2,2), (2,3), (2,4), (2,5), 
(3,1), (3,2), (3,3), (3,4), (3,5), 
(4,1), (4,2), (4,3), (4,4), (4,5), 
(5,1), (5,2), (5,3), (5,4), (5,5))

Upvotes: 3

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