减少元组的可变数量在Scala中映射[Key,List [Value]]的序列

Question

我有两个序列:

Seq("a" -> 1, "b" -> 2)
Seq("a" -> 3, "b" -> 4)

我想要的是一个结果Map看起来像这样:

Map(a -> List(3, 1), b -> List(4, 2))

Answer 1

val s1 = Seq("a" -> 1, "b" -> 2)
val s2 = Seq("a" -> 3, "b" -> 4)

val ss = s1 ++ s2

val toMap = ss.groupBy(x => x._1).map { case (k,v) => (k,  v.map(_._2))}

res0: scala.collection.immutable.Map[String,Seq[Int]] = Map(b -> List(2, 4), a -> List(1, 3))

你可以对这个或你想要的东西进行排序.