我如何在Scala中生成一个immutable.Map？

Question

我试过这个,但它不起作用:

val map:Map[String,String] = for {
    tuple2 <- someList
  } yield tuple2._1 -> tuple2._2

如何将Tuple2s列表转换为Map？

Answer 1

这简直太简单了:

Map(listOf2Tuples: _*)

apply在Map伴侣对象中使用该方法.

Answer 2

我的第一次尝试是这样的:

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)

但这是最好的解决方案:

scala> val betterConversion = Map(country2capitalList:_*)
betterConversion: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(England -> London, Germany -> Berlin)

该:_*需要给编译器的提示使用该列表作为可变参数的参数.否则会给你:

scala> Map(country2capitalList)
<console>:6: error: type mismatch;
 found   : List[(java.lang.String, java.lang.String)]
 required: (?, ?)
       Map(country2capitalList)
           ^

从Scala 2.8开始,您可以使用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

在2.8中,您可以使用以下toMap方法:

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))

这适用于任何对的集合.请注意,文档中有关于其重复策略的说明:

稍后的键将覆盖重复的键:如果这是一个无序的集合,则生成的映射中的哪个键是未定义的.

Answer 4

在scala 2.8中:

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>