I would like to convert a Map[Int, Any] to a SortedMap or a TreeMap. Is there an easy way to do it?
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6 Answers
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.
3 Comments
WestCoastProjects
This is a correct answer. But it shows an ugly side of the scala collections: it is too awkward.
Rag
Is this going to push the entire contents of m onto the stack as method arguments? Or is Scala/Java smarter than that?
reggert
This solution is inefficient, as it first copies the entire copies of the
Map to an array, and then copies the array into a TreeMap.Builder.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]
7 Comments
Daniel C. Sobral
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".
retronym
I had the same problem. The
A should be Char. Similar problem for retrieving (Int, Int) as the element type of a Map.
Vonn
@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.
retronym
IttayD: I suspect that solves a slightly different problem, but need to look at it more closely. Thanks!
missingfaktor
Assuming the contest is still open, here is how I would do it: ideone.com/pVZVW
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Starting Scala 2.13, via factory builders applied with .to(factory):
Map(1 -> "a", 2 -> "b").to(collection.immutable.SortedMap)
// collection.immutable.SortedMap[Int,String] = TreeMap(1 -> "a", 2 -> "b")
Map(1 -> "a", 2 -> "b").to(collection.immutable.TreeMap)
// collection.immutable.TreeMap[Int,String] = TreeMap(1 -> "a", 2 -> "b")
Comments
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)
Comments
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
