How can I convert immutable.Map to mutable.Map in Scala?

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深忆病人
深忆病人 2020-12-02 21:55

How can I convert immutable.Map to mutable.Map in Scala so I can update the values in Map?

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  • 2020-12-02 22:11

    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.

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  • 2020-12-02 22:11

    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")
    
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  • 2020-12-02 22:12

    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.

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  • 2020-12-02 22:16

    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)
    
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  • 2020-12-02 22:29

    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
    
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  • 2020-12-02 22:34
    val myImmutableMap = collection.immutable.Map(1->"one",2->"two")
    val myMutableMap = collection.mutable.Map() ++ myImmutableMap
    
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