Ausgabe
val adjList = Map("Logging" -> List("Networking", "Game"))
// val adjList: Map[String, List[String]] = Map(Logging -> List(Networking, Game))
adjList.flatMap { case (v, vs) => vs.map(n => (v, n)) }.toList
// val res7: List[(String, String)] = List((Logging,Game))
adjList.map { case (v, vs) => vs.map(n => (v, n)) }.flatten.toList
// val res8: List[(String, String)] = List((Logging,Networking), (Logging,Game))
Ich bin mir nicht sicher, was hier passiert. Ich habe von beiden das gleiche Ergebnis erwartet.
Lösung
.flatMapist , aber ist . Map_.flatMap.mapIterable.map
Für a Map "Logging" -> "Networking"und "Logging" -> "Game"wird gerade letzteres, "Logging" -> "Game"weil die Schlüssel gleich sind.
val adjList: Map[String, List[String]] = Map("Logging" -> List("Networking", "Game"))
val x0: Map[String, String] = adjList.flatMap { case (v, vs) => vs.map(n => (v, n)) }
//Map(Logging -> Game)
val x: List[(String, String)] = x0.toList
//List((Logging,Game))
val adjList: Map[String, List[String]] = Map("Logging" -> List("Networking", "Game"))
val y0: immutable.Iterable[List[(String, String)]] = adjList.map { case (v, vs) => vs.map(n => (v, n)) }
//List(List((Logging,Networking), (Logging,Game)))
val y1: immutable.Iterable[(String, String)] = y0.flatten
//List((Logging,Networking), (Logging,Game))
val y: List[(String, String)] = y1.toList
//List((Logging,Networking), (Logging,Game))
Auch https://users.scala-lang.org/t/map-flatten-flatmap/4180
Beantwortet von – Dmytro Mitin
Antwort geprüft von – Mary Flores (FixError Volunteer)
