How do I convert a Map[Int, Any] to a SortedMap in Scala? Or a TreeMap?

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

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.

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]
}

test

See also this question describing collection.breakOut: Scala 2.8 breakOut

CHALLENGE

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

"foo".as[Vector]