Question

我正在尝试将Map的内容展平为字符串列表。

所以这个集合：

val d: scala.collection.immutable.Map[String, scala.collection.immutable.Seq[Any]] = 
  Map("b" -> List(Array(1.0, 2.0), 5.333333333333333),
      "d" -> List(Array(3.0, 3.0), 8.0))

应转换为List[String]类型的两个元素：

  b,1,2,5.3
  d 3,3,8.0

要实现此目的，我尝试使用以下方法分解地图并展平其键：

  d.map(m =>
  m match {
    case(k , v) => {
        (k , v.flatten)
    }
    })

但我收到错误：

Multiple markers at this line - No implicit view available from Any => scala.collection.GenTraversableOnce[B]. - 
 not enough arguments for method flatten: (implicit asTraversable: Any => 
 scala.collection.GenTraversableOnce[B])scala.collection.immutable.Seq[B]. Unspecified value parameter 
 asTraversable.

如何将集合scala.collection.immutable.Map[String, scala.collection.immutable.Seq[Any]]转换为List[String]？

更新：

This works : 

val map: Map[String, scala.collection.immutable.Seq[Any]] =
    Map(
      "bsds" -> List(Array(1.0, 2.0), 5.333333333333333),
      "dsdfsd" -> List(Array(3.0, 3.0), 8.0)
    )                                             //> map  : Map[String,scala.collection.immutable.Seq[Any]] = Map(bsds -> List(Ar
                                                  //| ray(1.0, 2.0), 5.333333333333333), dsdfsd -> List(Array(3.0, 3.0), 8.0))

  val flatten = map.map {
    case (k : String , v) =>
      val expanded = v map {
        case arr: Array[_] => Seq(arr: _*)
        case el : Double => Seq(el)
      }
      (k , (expanded.flatten))
  }                                               //> flatten  : scala.collection.immutable.Map[String,scala.collection.immutable.
                                                  //| Seq[Any]] = Map(bsds -> List(1.0, 2.0, 5.333333333333333), dsdfsd -> List(3.
                                                  //| 0, 3.0, 8.0))
  val ll = flatten.map(m => List(m._1) ++ m._2)   //> ll  : scala.collection.immutable.Iterable[List[Any]] = List(List(bsds, 1.0, 
                                                  //| 2.0, 5.333333333333333), List(dsdfsd, 3.0, 3.0, 8.0))
  ll.map(m => m.mkString("|"))                    //> res5: scala.collection.immutable.Iterable[String] = List(bsds|1.0|2.0|5.3333
                                                  //| 33333333333, dsdfsd|3.0|3.0|8.0)

Answer 1

  val map: Map[String, scala.collection.immutable.Seq[Any]] =
    Map(
      "b" -> List(Array(1.0, 2.0), 5.333333333333333),
      "d" -> List(Array(3.0, 3.0), 8.0)
    )

  val flatten = map.map {
    case (k, v) =>
      val expanded = v map {
        case arr: Array[_] => Seq(arr: _*)
        case el => Seq(el)
      }
      (k ++ expanded.flatten).mkString(",")
  }

  flatten.foreach(println)

Answer 2

不像你想要的那样平坦：

scala> import runtime.{ ScalaRunTime => RT }
import runtime.{ScalaRunTime=>RT}

scala> val m = Map("bcde" -> List(Array(1.0, 2.0), 5.333333333333333), "fghij" -> List(Array(3.0, 3.0), 8.0))
m: scala.collection.immutable.Map[String,List[Any]] = Map(bcde -> List(Array(1.0, 2.0), 5.333333333333333), fghij -> List(Array(3.0, 3.0), 8.0))

scala> m map { case (k,v) => s"${RT stringOf k}, ${v map (RT stringOf _) mkString ", "}" }
res7: scala.collection.immutable.Iterable[String] = List(bcde, Array(1.0, 2.0), 5.333333333333333, fghij, Array(3.0, 3.0), 8.0)

但值得做广告。

Answer 3

使用for comprehension，像这样，

for ( (k, List(Array(a,b,_*),v)) <- d) yield Seq(k,a,b,v).mkString(",")

将Map集合转换为字符串列表

3 个答案: