我试过这个,但它不起作用:
val map:Map[String,String] = for {
tuple2 <- someList
} yield tuple2._1 -> tuple2._2
我还能如何将 Tuple2 列表转换为 Map?
最佳答案
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)
:_*
需要给编译器一个使用列表作为可变参数参数的提示。否则它会给你: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))
关于scala - 我将如何在 Scala 中生成一个 immutable.Map?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2763584/