Scala - reduce/foldLeft

Question

我有一个嵌套的地图m,如:

m = Map("email" -> "a@b.com", "background" -> Map("language" -> "english"))

我有一个阵列 arr = Array("background","language")

如何foldLeft/reduce数组并从地图中找到字符串"english".我试过这个:

arr.foldLeft(m) { (acc,x) => acc.get(x) }

但我得到这个错误:

<console>:10: error: type mismatch;
 found   : Option[java.lang.Object]
 required: scala.collection.immutable.Map[java.lang.String,java.lang.Object]
       arr.foldLeft(m) { (acc,x) => acc.get(x) }

Answer 1

你应该注意类型.在这里,你从m : Map[String, Any]acc 开始.你结合一个字符串x和调用get,返回一个Option[Object].要继续,您必须检查是否有值,检查此值是否为a Map,强制转换(由于类型擦除而未选中,因此很危险).

我认为错误在于你的结构类型Map [String,Any]代表你所拥有的东西相当糟糕.

假设你做了

sealed trait Tree
case class Node(items: Map[String, Tree]) extends Tree
case class Leaf(s: String) extends Tree

您可以添加一些帮助程序来轻松声明树

object Tree {
  implicit def fromString(s: String) = Leaf(s)
  implicit def fromNamedString(nameAndValue: (String, String)) 
    = (nameAndValue._1, Leaf(nameAndValue._2))
}
object Node {
  def apply(items: (String, Tree)*) : Node = Node(Map(items: _*))
}

然后声明树就像第一个版本一样简单,但类型更精确

m = Node("email" -> "a@b.com", "background" -> Node("language" -> "english"))

然后,您可以添加方法,例如 trait Tree

  def get(path: String*) : Option[Tree] = {
    if (path.isEmpty) Some(this)
    else this match {
      case Leaf(_) => None
      case Node(map) => map.get(path.head).flatMap(_.get(path.tail: _*))
    }
  }
  def getLeaf(path: String*): Option[String] 
    = get(path: _*).collect{case Leaf(s) =>s}

或者如果你愿意用折叠来做

  def get(path: String*) = path.foldLeft[Option[Tree]](Some(this)) {
    case (Some(Node(map)), p) => map.get(p)
    case _ => None
  }