[英]reduce variable number of tuples Sequences to Map[Key, List[Value]] in Scala
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))
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。
你可以试试
scala> val seq = Seq("a" -> 1, "b" -> 2) ++ Seq("a" -> 3, "b" -> 4)
seq: Seq[(String, Int)] = List((a,1), (b,2), (a,3), (b,4))
scala> seq groupBy(_._1) mapValues(_ map(_._2))
res9: scala.collection.immutable.Map[String,Seq[Int]] = Map(b -> List(2, 4), a -> List(1, 3))
def reduceToMap[K, V](seqs: Seq[(K, V)]*): Map[K, List[V]] = {
seqs.reduce(_ ++ _).foldLeft(Map.empty[K, List[V]])((memo, next) =>
memo.get(next._1) match {
case None => memo.updated(next._1, next._2 :: Nil)
case Some(xs) => memo.updated(next._1, next._2 :: xs)
}
)
}
scala> reduceToMap(Seq("a" -> 1, "b" -> 2), Seq("a" -> 3, "b" -> 4))
res0: Map[String,List[Int]] = Map(a -> List(3, 1), b -> List(4, 2))
scala> reduceToMap(Seq.empty)
res1: Map[Nothing,List[Nothing]] = Map()
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