是否可以在收集期间打印值而无需修改返回类型?

时间:2012-05-27 17:13:26

标签: scala scala-collections scala-2.9

我有一个像这样的代码段:

def test() : Seq[Int] = 
  List("A", "B", "C") collect {
    case "A" => 1
    case "B" => 2
    //case _ => println(_)
  }

现在我想在输出上打印特定值(仅用于调试),而不向结果集合添加任何元素。如果我取消注释注释行,Scala会将表达式的值推断为Seq[Any],这是完全可以理解的。

任何人都有任何提示如何做到这一点?提前谢谢!

3 个答案:

答案 0 :(得分:9)

def skipped: Nothing = throw new Exception("Oops, not skipped after all!")

List("A", "B", "C") collect {
  case "A" => 1
  case "B" => 2
  case x if { println(x); false } => skipped
}

object PrintSkip {
  def unapply(a: Any): Option[Any] = {
    println(a)
    None
  }
}

List(Some("A"), None, Some("C")) collect {
  case Some("A") => 1
  case None => 2
  case Some(PrintSkip(_)) => skipped
}

答案 1 :(得分:8)

flatMap

List("A", "B", "C") flatMap {
    case "A" => List(1)
    case "B" => List(2)
    case x => println(x); Nil
}

collect / flatten

List("A", "B", "C").collect {
    case "A" => Some(1)
    case "B" => Some(2)
    case x => println(x); None
}.flatten

答案 2 :(得分:3)

使用collect,无需在Option中包装东西。

List("A", "B", "C").collect(new PartialFunction[String, Int] {
  def apply(s:String):Int = s match {
    case "A" => 1
    case "B" => 2
  }

  def isDefinedAt(s:String) = {
    try {
      apply(s)
      true
    } catch {
      case e:MatchError => { println(s); false }
    }

  }
})