Take the 2-minute tour ×

I would like to convert a Map[Int, Any] to a SortedMap or a TreeMap. Is there an easy way to do it?

share|improve this question
add comment

5 Answers

up vote 16 down vote accepted

Assuming you're using immutable maps

val m = Map(1 -> "one")
val t = scala.collection.immutable.TreeMap(m.toArray:_*)

The TreeMap companion object's apply method takes repeated map entry parameters (which are instances of Tuple2[_, _] of the appropriate parameter types). toArray produces an Array[Tuple2[Int, String]] (in this particular case). The : _* tells the compiler that the array's contents are to be treated as repeated parameters.

share|improve this answer
add comment

Here's a general way to convert between various Scala collections.

import collection.generic.CanBuildFrom
import collection.immutable.TreeMap

object test {
  class TraversableW[A](t: Traversable[A]) {
    def as[CC[X] <: Traversable[X]](implicit cbf: CanBuildFrom[Nothing, A, CC[A]]): CC[A] = t.map(identity)(collection.breakOut)
    def to[Result](implicit cbf: CanBuildFrom[Nothing, A, Result]): Result = t.map(identity)(collection.breakOut)

  implicit def ToTraverseableW[A](t: Traversable[A]): TraversableW[A] = new TraversableW[A](t)

  List(1, 2, 3).as[Vector]
  List(1, 2, 3).to[Vector[Int]]
  List((1, 1), (2, 4), (3, 4)).to[Map[Int, Int]]
  List((1, 1), (2, 4), (3, 4)).to[TreeMap[Int, Int]]
  val tm: TreeMap[Int, Int] = List((1, 1), (2, 4), (3, 4)).to
  ("foo": Seq[Char]).as[Vector]


See also this question describing collection.breakOut: http://stackoverflow.com/questions/1715681/scala-2-8-breakout


Is it possible to adjust the implicits such that this works? Or would this only be possible if as were added to Traversable?

share|improve this answer
You can make the implicit receive a view bound instead of a Traversable, but I can't think of a way to retrieve "A" from that. After all, String doesn't have an "A". –  Daniel C. Sobral Jun 20 '10 at 1:15
I had the same problem. The A should be Char. Similar problem for retrieving (Int, Int) as the element type of a Map. –  retronym Jun 20 '10 at 7:28
@retronym - really appreciate your post on my simple "scala newbie" question. Sent me on a good dive with the link you posted on collection.breakout. –  Vonn Jun 21 '10 at 18:51
how about: paste.pocoo.org/show/193809? –  IttayD Jun 22 '10 at 8:59
Assuming the contest is still open, here is how I would do it: ideone.com/pVZVW –  missingfaktor Jul 23 '11 at 19:31
show 2 more comments

An alternative to using :_* as described by sblundy is to append the existing map to an empty SortedMap

import scala.collection.immutable.SortedMap
val m = Map(1 -> ("one":Any))
val sorted = SortedMap[Int, Any]() ++ m
share|improve this answer
add comment

Since internal data structures in implementations are completely different, you'll have to add elements one-by-one anyway. So, do it explicitly:

val m = Map(1 -> "one")

var t = scala.collection.immutable.TreeMap[Int,String]()
t ++= m
share|improve this answer
add comment

Here's a way you can do it with a Scala implicit class:

implicit class ToSortedMap[A,B](tuples: TraversableOnce[(A, B)])
                               (implicit ordering: Ordering[A]) {
  def toSortedMap =
    SortedMap(tuples.toSeq: _*)

Since Map[A,B] has an implicit path to a TraversableOnce[Tuple2[A, B]], the following works:

scala> Map("b" -> 3, "c" -> 3, "a" -> 5).toSortedMap
res6: scala.collection.immutable.SortedMap[String,Int] = Map(a -> 5, b -> 3, c -> 3)

It will even work on a list of Tuple2s, similar to toMap:

scala> List(("c", 1), ("b", 3),("a", 6)).toSortedMap
res7: scala.collection.immutable.SortedMap[String,Int] = Map(a -> 6, b -> 3, c -> 1)
share|improve this answer
add comment

Your Answer


