How to get intersection of two RDD[(String, Iterable[String])] - scala

Question

The data consists of two columns
A B
A C
A D
B A
B C
B D
B E
C A
C B
C D
C E
D A
D B
D C
D E
E B
E C
E D
In the first row, think of it as A is friends with B, etc.
How do I find their common friends?
(A,B) -> (C D)
Meaning A and B have common friends C and D. I came as close as doing a groupByKey with the following result.
(B,CompactBuffer(A, C, D, E))
(A,CompactBuffer(B, C, D))
(C,CompactBuffer(A, B, D, E))
(E,CompactBuffer(B, C, D))
(D,CompactBuffer(A, B, C, E))
The code:
val rdd: RDD[String] = spark.sparkContext.textFile("twocols.txt")
val splitrdd: RDD[(String, String)] = rdd.map { s =>
var str = s.split(" ")
new Tuple2(str(0), str(1))
}
val group: RDD[(String, Iterable[String])] = splitrdd.groupByKey()
group.foreach(println)

Answer 1

First swap the elements:
val swapped = splitRDD.map(_.swap)
Then self-join and swap back:
val shared = swapped.join(swapped).map(_.swap)
Finally filter out duplicates (if needed) and groupByKey:
shared.filter { case ((x, y), _) => x < y }.groupByKey

Answer 2

This is just an ugly attempt:
Suppose you have converted your two columns into Array[Array[String]] (or List[List[String]], it's really the same), say
val pairs=Array(
Array("A","B"),
Array("A","C"),
Array("A","D"),
Array("B","A"),
Array("B","C"),
Array("B","D"),
Array("B","E"),
Array("C","A"),
Array("C","B"),
Array("C","D"),
Array("C","E"),
Array("D","A"),
Array("D","B"),
Array("D","C"),
Array("D","E"),
Array("E","B"),
Array("E","C"),
Array("E","D")
)
Define the group for which you want to find their common friends:
val group=Array("C","D")
The following will find the friends for each member in your group
val friendsByMemberOfGroup=group.map(
i => pairs.filter(x=> x(1) contains i)
.map(x=>x(0))
)
For example, pairs.filter(x=>x(1) contains "C").map(x=>x(0)) returns the friends of "C" where "C" is being taken from the second column and its friends are taken from the first column:
scala> pairs.filter(x=> x(1) contains "C").map(x=>x(0))
res212: Array[String] = Array(A, B, D, E)
And the following loop will find the common friends of all the members in your group
var commonFriendsOfGroup=friendsByMemberOfGroup(0).toSet
for(i <- 1 to friendsByMemberOfGroup.size-1){
commonFriendsOfGroup=
commonFriendsOfGroup.intersect(friendsByMemberOfGroup(i).toSet)
}
So you get
scala> commonFriendsOfGroup.toArray
res228: Array[String] = Array(A, B, E)
If you change your group to val group=Array("A","B","E") and apply the previous lines then you will get
scala> commonFriendsOfGroup.toArray
res230: Array[String] = Array(C, D)

Answer 3

Continuing from where you left off:
val group: RDD[(String, Iterable[String])] = splitrdd.groupByKey()
val group_map = group.collectAsMap
val common_friends = group
.flatMap{case (x, friends) =>
friends.map{y =>
((x,y),group_map.get(y).get.toSet.intersect(friends.toSet))
}
}
scala> common_friends.foreach(println)
((B,A),Set(C, D))
((B,C),Set(A, D, E))
((B,D),Set(A, C, E))
((B,E),Set(C, D))
((D,A),Set(B, C))
((D,B),Set(A, C, E))
((D,C),Set(A, B, E))
((D,E),Set(B, C))
((A,B),Set(C, D))
((A,C),Set(B, D))
((A,D),Set(B, C))
((C,A),Set(B, D))
((C,B),Set(A, D, E))
((C,D),Set(A, B, E))
((C,E),Set(B, D))
((E,B),Set(C, D))
((E,C),Set(B, D))
((E,D),Set(B, C))
Note: this assumes your data has the relationship in both directions like in your example: (A B and B A). If it's not the case you need to add some code to deal with the fact that group_map.get(y) might return None.

Answer 4

So I ended up doing this on the client side. DO NOT DO THIS
val arr: Array[(String, Iterable[String])] = group.collect()
//arr.foreach(println)
var arr2 = scala.collection.mutable.Set[((String, String), List[String])]()
for (i <- arr)
for (j <- arr)
if (i != j) {
val s1 = i._2.toSet
val s2 = j._2.toSet
val s3 = s1.intersect(s2).toList
//println(s3)
val pair = if (i._1 < j._1) (i._1, j._1) else (j._1, i._1)
arr2 += ((pair, s3))
}
arr2.foreach(println)
The result is
((B,E),List(C, D))
((A,C),List(B, D))
((A,B),List(C, D))
((A,D),List(B, C))
((B,D),List(A, C, E))
((C,D),List(A, B, E))
((B,C),List(A, D, E))
((C,E),List(B, D))
((D,E),List(B, C))
((A,E),List(B, C, D))
I am wondering if I can do this using transformations within Spark.