In Scala, is there a way to take convert two lists into a Map?

耗尽温柔 提交于 2019-12-02 20:19:35
oxbow_lakes

In 2.8 this is really simple using the CanBuildFrom functionality (as described by Daniel) and using breakOut with a type instruction to the compiler as to what the result type should be:

import scala.collection.breakOut
val m = (listA zip listB)(breakOut): Map[A,B]

The following would also work:

val n: Map[A,B] = (listA zip listB)(breakOut)

And (as EastSun, below, has pointed out) this has been added to the library as toMap

val o = (listA zip listB).toMap

As for reversing the map, you can do:

val r = m.map(_.swap)(breakOut): Map[B, A]

Now that you've got a list of tuples it is easy to make it into a map by writing Map(tuplesOfAB: _*). The : _* notation means to call the varargs overload with the arguments taken from the sequence. This seems like a funny bit of syntax, but it helps to think that varargs are declared like Map[A,B](pairs: (A,B)*) and the : _* is a type annotation to convert to varargs because of the common * part.

To reverse a map m use Map(m.map(_.swap): _*). In scala a map is also a collection of pairs. This transforms those pairs by swapping the elements and passing them to the Map constructor.

There's yet another way to do it, beyond those already shown. Here:

Map() ++ tuplesOfAB
Ashwin
scala> List( "a", "f", "d") zip List(7, 5, 4, 8) toMap
res0: scala.collection.immutable.Map[java.lang.String,Int] = Map(a -> 7, f -> 5, d -> 4)
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