撰写部分功能

Question

我有两个PartialFunctions f和g.它们没有副作用,可以快速执行.将它们组成另一个部分函数的最佳方法是什么h, h.isDefinedAt(x)iff f.isDefinedAt(x) && g.isDefinedAt(f(x))？

如果h函数返回Option而不是部分函数,​​也可以.

我很失望,f andThen g没有做我想做的事:

scala> val f = Map("a"->1, "b"->2)
f: scala.collection.immutable.Map[String,Int] = Map(a -> 1, b -> 2)

scala> val g = Map(1->'c', 3->'d')
g: scala.collection.immutable.Map[Int,Char] = Map(1 -> c, 3 -> d)

scala> (f andThen g).isDefinedAt("b")
res3: Boolean = true

scala> (f andThen g).lift("b")
java.util.NoSuchElementException: key not found: 2
    at scala.collection.MapLike$class.default(MapLike.scala:228)

Answer 1

这是一个比链接问题更简短的方法,取自这个帖子:

  val f = Map("a" -> 1, "b" -> 2)                 

  val g = Map(1 -> 'c', 3 -> 'd')                 

  def andThenPartial[A, B, C](pf1: PartialFunction[A, B], pf2: PartialFunction[B, C]): PartialFunction[A, C] = {
    Function.unlift(pf1.lift(_) flatMap pf2.lift)
  }                                               

  val h = andThenPartial(f, g)            //> h  : PartialFunction[String,Char]

  h.isDefinedAt("a")                      //> res2: Boolean = true
  h.isDefinedAt("b")                      //> res3: Boolean = false
  h.lift("a")                             //> res4: Option[Char] = Some(c)
  h.lift("b")                             //> res5: Option[Char] = None

当然,这也可以包含为隐式类:

  implicit class ComposePartial[A, B](pf: PartialFunction[A, B]) {
    def andThenPartial[C](that: PartialFunction[B, C]): PartialFunction[A, C] =
      Function.unlift(pf.lift(_) flatMap that.lift)
  }

  val h2 = f andThenPartial g         //> h2  : PartialFunction[String,Char]

  h2.isDefinedAt("a")                 //> res6: Boolean = true
  h2.isDefinedAt("b")                 //> res7: Boolean = false
  h2.lift("a")                        //> res8: Option[Char] = Some(c)
  h2.lift("b")                        //> res9: Option[Char] = None