Merge maps by key

Question

Say I have two maps:

val a = Map(1 -> \"one\", 2 -> \"two\", 3 -> \"three\")
val b = Map(1 -> \"un\", 2 -> \"deux\", 3 -> \"trois\")


        

Answer 1

scala.collection.immutable.IntMap has an intersectionWith method that does precisely what you want (I believe):

import scala.collection.immutable.IntMap

val a = IntMap(1 -> "one", 2 -> "two", 3 -> "three", 4 -> "four")
val b = IntMap(1 -> "un", 2 -> "deux", 3 -> "trois")

val merged = a.intersectionWith(b, (_, av, bv: String) => Seq(av, bv))

This gives you IntMap(1 -> List(one, un), 2 -> List(two, deux), 3 -> List(three, trois)). Note that it correctly ignores the key that only occurs in a.

As a side note: I've often found myself wanting the unionWith, intersectionWith, etc. functions from Haskell's Data.Map in Scala. I don't think there's any principled reason that they should only be available on IntMap, instead of in the base collection.Map trait.

Answer 2

Starting Scala 2.13, you can use groupMap which (as its name suggests) is an equivalent of a groupBy followed by map on values:

// val map1 = Map(1 -> "one", 2 -> "two",  3 -> "three")
// val map2 = Map(1 -> "un",  2 -> "deux", 3 -> "trois")
(map1.toSeq ++ map2).groupMap(_._1)(_._2)
// Map(1 -> List("one", "un"), 2 -> List("two", "deux"), 3 -> List("three", "trois"))

This:

• Concatenates the two maps as a sequence of tuples (List((1, "one"), (2, "two"), (3, "three"))). For conciseness, map2 is implicitly converted to Seq to align with map1.toSeq's type - but you could choose to make it explicit by using map2.toSeq.

• groups elements based on their first tuple part (_._1) (group part of groupMap)

• maps grouped values to their second tuple part (_._2) (map part of groupMap)

Answer 3

def merge[A,B,C,D](b : Map[A,B], c : Map[A,C])(d : (Option[B],Option[C]) => D): Map[A,D] = {
  (b.keySet ++ c.keySet).map(k => k -> d(b.get(k), c.get(k))).toMap
}

def optionSeqBiFunctionK[A]:(Option[A], Option[A]) => Seq[A] = _.toSeq ++ _.toSeq

merge(a,b)(optionSeqBiFunctionK)

Answer 4

val a = Map(1 -> "one", 2 -> "two", 3 -> "three")
val b = Map(1 -> "un", 2 -> "deux", 3 -> "trois")

val c = a.toList ++ b.toList
val d = c.groupBy(_._1).map{case(k, v) => k -> v.map(_._2).toSeq}
//res0: scala.collection.immutable.Map[Int,Seq[java.lang.String]] =
        //Map((2,List(two, deux)), (1,List(one, un), (3,List(three, trois)))

Answer 5

Here is my first approach before looking for the other solutions:

for (x <- a) yield 
  x._1 -> Seq (a.get (x._1), b.get (x._1)).flatten

To avoid elements which happen to exist only in a or b, a filter is handy:

(for (x <- a) yield 
  x._1 -> Seq (a.get (x._1), b.get (x._1)).flatten).filter (_._2.size == 2)

Flatten is needed, because b.get (x._1) returns an Option. To make flatten work, the first element has to be an option too, so we can't just use x._2 here.

For sequences, it works too:

scala> val b = Map (1 -> Seq(1, 11, 111), 2 -> Seq(2, 22), 3 -> Seq(33, 333), 5 -> Seq(55, 5, 5555))
b: scala.collection.immutable.Map[Int,Seq[Int]] = Map(1 -> List(1, 11, 111), 2 -> List(2, 22), 3 -> List(33, 333), 5 -> List(55, 5, 5555))

scala> val a = Map (1 -> Seq(1, 101), 2 -> Seq(2, 212, 222), 3 -> Seq (3, 3443), 4 -> (44, 4, 41214))
a: scala.collection.immutable.Map[Int,ScalaObject with Equals] = Map(1 -> List(1, 101), 2 -> List(2, 212, 222), 3 -> List(3, 3443), 4 -> (44,4,41214))

scala> (for (x <- a) yield x._1 -> Seq (a.get (x._1), b.get (x._1)).flatten).filter (_._2.size == 2) 
res85: scala.collection.immutable.Map[Int,Seq[ScalaObject with Equals]] = Map(1 -> List(List(1, 101), List(1, 11, 111)), 2 -> List(List(2, 212, 222), List(2, 22)), 3 -> List(List(3, 3443), List(33, 333)))

Answer 6

Scalaz adds a method |+| for any type A for which a Semigroup[A] is available.

If you mapped your Maps so that each value was a single-element sequence, then you could use this quite simply:

scala> a.mapValues(Seq(_)) |+| b.mapValues(Seq(_))
res3: scala.collection.immutable.Map[Int,Seq[java.lang.String]] = Map(1 -> List(one, un), 2 -> List(two, deux), 3 -> List(three, trois))