Question

我想将单个值转换为多个“特征”的集合，而不使用可变数据结构来收集值。我喜欢这样的幻想构造，它使用模式匹配，但在第一次匹配后不会停止：

scala> 2 multimatch {
  case i if i > 0 => "Positive"
  case i if i < 0 => "Negative"
  case i if (i % 2 == 0) => "Even"
  //yadda yadda
}
res0: Seq[java.lang.String] = List(Positive, Even)

Answer 1

使用皮条客模式，部分函数和重复参数，这就是我能得到的距离：

class MultiMatcher[A](a: A) {
  def multiMatch[B](pfs: PartialFunction[A, B]*): Seq[B] = {
    pfs.map(_.lift(a)).collect{ case Some(v) => v } 
  }
}

implicit def toMultiMatcher[A](a:A): MultiMatcher[A] = new MultiMatcher(a)

2 multiMatch (
  { case i if i > 0 => "Positive" },
  { case i if i < 0 => "Negative" },
  { case i if i % 2 == 0 => "Even" }
)
// returns Seq[java.lang.String] = ArrayBuffer(Positive, Even)

Answer 2

首先，您将特征定义为从Int到Option [String]

的函数

val pos = (i:Int) => if (i > 0) Some("Positive") else None
val neg = (i:Int) => if (i < 0) Some("Negative") else None
val even = (i:Int) => if (i % 2 == 0) Some("Even") else None

然后列出特征列表。

val characteristics = pos::neg::even::Nil

然后使用flatmap获取适用于某个对象的特征列表。

scala> characteristics.flatMap(f=>f(2))
res6: List[java.lang.String] = List(Positive, Even)

Answer 3

首先，按如下方式定义多匹配函数：

scala> def multimatch[A,B]( value : A,ps: ( A => Boolean, B)*) =
     |           for ( p <- ps
     |                 if (p._1(value))) yield p._2
multimatch: [A,B](value: A,ps: ((A) => Boolean, B)*)Seq[B]

然后，我们走了：

    scala> multimatch(2,
         |           ( ((x :Int) => x > 0) -> "POSITIVE"),
         |           ( ((x :Int) => x < 0) -> "NEGATIVE"),
         |           ( ((x :Int) => x % 2 == 0) -> "EVEN")
         |       )

res4: Seq[java.lang.String] = ArrayBuffer(POSITIVE, EVEN)

或者，不那么混乱：

scala> multimatch(2,
     |           ( (x :Int) => x > 0 , "POSITIVE"),
     |           ( (x :Int) => x < 0, "NEGATIVE"),
     |           ( (x :Int) => x % 2 == 0, "EVEN")
     |       )
res5: Seq[java.lang.String] = ArrayBuffer(POSITIVE, EVEN)

转换A =＆gt;的惯用方法SEQ [B]

3 个答案: