How would I yield an immutable.Map in Scala? - scala

Question

I have tried this but it does not work:
val map:Map[String,String] = for {
tuple2 <- someList
} yield tuple2._1 -> tuple2._2
How else would I convert a List of Tuple2s into a Map?

Answer 1

It couldn't be simpler:
Map(listOf2Tuples: _*)
using the apply method in Map companion object.

Answer 2

My First try is this:
scala> val country2capitalList = List("England" -> "London", "Germany" -> "Berlin")
country2capitalList: List[(java.lang.String, java.lang.String)] = List((England,London), (Germany,Berlin))
scala> val country2capitalMap = country2capital.groupBy(e => e._1).map(e => (e._1, e._2(0)._2))
country2capitalMap: scala.collection.Map[java.lang.String,java.lang.String] = Map(England -> London, Germany -> Berlin)
But here is the best solution:
scala> val betterConversion = Map(country2capitalList:_*)
betterConversion: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(England -> London, Germany -> Berlin)
The :_* is needed to give the compiler a hint to use the list as a varargs argument. Otherwise it will give you:
scala> Map(country2capitalList)
<console>:6: error: type mismatch;
found : List[(java.lang.String, java.lang.String)]
required: (?, ?)
Map(country2capitalList)
^
From Scala 2.8 on you can use toMap:
scala> val someList = List((1, "one"), (2, "two"))
someList: List[(Int, java.lang.String)] = List((1,one), (2,two))
scala> someList.toMap
res0: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two))

Answer 3

In 2.8, you can use the toMap method:
scala> val someList = List((1, "one"), (2, "two"))
someList: List[(Int, java.lang.String)] = List((1,one), (2,two))
scala> someList.toMap
res0: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two))
This will work for any collection of pairs. Note that the documentation has this to say about its duplicate policy:
Duplicate keys will be overwritten by
later keys: if this is an unordered
collection, which key is in the
resulting map is undefined.

Answer 4

In scala 2.8:
scala> import scala.collection.breakOut
import scala.collection.breakOut
scala> val ls = List("a","bb","ccc")
ls: List[java.lang.String] = List(a, bb, ccc)
scala> val map: Map[String,Int] = ls.map{ s => (s,s.length) }(breakOut)
map: Map[String,Int] = Map((a,1), (bb,2), (ccc,3))
scala> val map2: Map[String,Int] = ls.map{ s => (s,s.length) }.toMap
map2: Map[String,Int] = Map((a,1), (bb,2), (ccc,3))
scala>