How's the following function to update an immutable Map's values?
scala> def updateMap[A](map: Map[A, Set[A]], item: A, toAdd: A) = { map.map(x => if(x._1 == item) (x._1, (x._2 + toAdd)) else (x._1, x._2)) } scala> val map = Map(1 -> Set(2), 3 -> Set(4)) scala> updateMap(map, 1, 5) res1 ...= Map(1 -> Set(2, 5), 3 -> Set(4)) You iterate through the whole Map, which is O(n) and therefore slow as hell. Worse, you even recreate every element and put it into another element, which is even slower.
Why not just do:
scala> val map = Map(1 -> Set(5,3), 2 -> Set(7,3)) map: scala.collection.immutable.Map[Int,scala.collection.immutable.Set[Int]] = Map(1 -> Set(5, 3), 2 -> Set(7, 3)) scala> def updateMap[A, B](map: Map[A, Set[B]], key: A, value: B) = map + ((key, map.getOrElse(key, Set()) + value)) updateMap: [A, B](map: Map[A,Set[B]], key: A, value: B)scala.collection.immutable.Map[A,Set[B]] scala> updateMap(map, 1, 9) res0: scala.collection.immutable.Map[Int,Set[Int]] = Map(1 -> Set(5, 3, 9), 2 -> Set(7, 3)) And you probably want to introduce another type parameter B to the signature of updateMap to allow values of a different type than the one of the key.
if (map.contains (key)) would work without O(n) but it's not a succinct. Nice code above, thanks. \$\endgroup\$ Pattern matching is a more readable and important feature of Scala, which you can use.
And in the else condition, you are using (x._1, x._2); you can use x in place of it.
def updateMap[A](map: Map[A, Set[A]], item: A, toAdd: A) = { map.map(x => (x._1 == item) match{ case true=>(x._1, (x._2 + toAdd)) case false=>x }) } 
