How can I convert immutable.Map to mutable.Map in Scala?
ScalaDictionaryScala 2.8Scala Problem Overview
How can I convert immutable.Map to mutable.Map in Scala so I can update the values in Map?
Scala Solutions
Solution 1 - Scala
The cleanest way would be to use the mutable.Map varargs factory. Unlike the ++ approach, this uses the CanBuildFrom mechanism, and so has the potential to be more efficient if library code was written to take advantage of this:
val m = collection.immutable.Map(1->"one",2->"Two")
val n = collection.mutable.Map(m.toSeq: _*)
This works because a Map can also be viewed as a sequence of Pairs.
Solution 2 - Scala
val myImmutableMap = collection.immutable.Map(1->"one",2->"two")
val myMutableMap = collection.mutable.Map() ++ myImmutableMap
Solution 3 - Scala
Starting Scala 2.13, via factory builders applied with .to(factory):
Map(1 -> "a", 2 -> "b").to(collection.mutable.Map)
// collection.mutable.Map[Int,String] = HashMap(1 -> "a", 2 -> "b")
Solution 4 - Scala
How about using collection.breakOut?
import collection.{mutable, immutable, breakOut}
val myImmutableMap = immutable.Map(1->"one",2->"two")
val myMutableMap: mutable.Map[Int, String] = myImmutableMap.map(identity)(breakOut)
Solution 5 - Scala
There is a variant to create an empty mutable Map that has default values taken from the immutable Map. You may store a value and override the default at any time:
scala> import collection.immutable.{Map => IMap}
//import collection.immutable.{Map=>IMap}
scala> import collection.mutable.HashMap
//import collection.mutable.HashMap
scala> val iMap = IMap(1 -> "one", 2 -> "two")
//iMap: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two))
scala> val mMap = new HashMap[Int,String] {
| override def default(key: Int): String = iMap(key)
| }
//mMap: scala.collection.mutable.HashMap[Int,String] = Map()
scala> mMap(1)
//res0: String = one
scala> mMap(2)
//res1: String = two
scala> mMap(3)
//java.util.NoSuchElementException: key not found: 3
// at scala.collection.MapLike$class.default(MapLike.scala:223)
// at scala.collection.immutable.Map$Map2.default(Map.scala:110)
// at scala.collection.MapLike$class.apply(MapLike.scala:134)
// at scala.collection.immutable.Map$Map2.apply(Map.scala:110)
// at $anon$1.default(<console>:9)
// at $anon$1.default(<console>:8)
// at scala.collection.MapLike$class.apply(MapLike.scala:134)....
scala> mMap(2) = "three"
scala> mMap(2)
//res4: String = three
Caveat (see the comment by Rex Kerr): You will not be able to remove the elements coming from the immutable map:
scala> mMap.remove(1)
//res5: Option[String] = None
scala> mMap(1)
//res6: String = one
Solution 6 - Scala
With scala 2.13, there are two alternatives: the to method of the source map instance, or the from method of the destination map's companion object.
scala> import scala.collection.mutable
import scala.collection.mutable
scala> val immutable = Map(1 -> 'a', 2 -> 'b');
val immutable: scala.collection.immutable.Map[Int,Char] = Map(1 -> a, 2 -> b)
scala> val mutableMap1 = mutable.Map.from(immutable)
val mutableMap1: scala.collection.mutable.Map[Int,Char] = HashMap(1 -> a, 2 -> b)
scala> val mutableMap2 = immutable.to(mutable.Map)
val mutableMap2: scala.collection.mutable.Map[Int,Char] = HashMap(1 -> a, 2 -> b)
As you can see, the mutable.Map implementation was decided by the library.
If you want to choose a particular implementation, for example mutable.HashMap, replace all occurrences of mutable.Map with mutable.HashMap.