Scala - Optional Predicate

问题

I was informed to use this interesting piece of code, but my use case requires it to do a bit more than is currently possible.

  implicit class Predicate[A](val pred: A => Boolean) {
    def apply(x: A) = pred(x)

    def &&(that: A => Boolean) = new Predicate[A](x => pred(x) && that(x))
    def ||(that: A => Boolean) = new Predicate[A](x => pred(x) || that(x))
    def unary_! = new Predicate[A](x => !pred(x))
  }

Some use case boilerplate:

type StringFilter = (String) => Boolean

def nameFilter(value: String): StringFilter = 
    (s: String) => s == value

def lengthFilter(length: Int): StringFilter = 
    (s: String) => s.length == length


val list = List("Apple", "Orange", "Meat")
val isFruit = nameFilter("Apple") || nameFilter("Orange")
val isShort = lengthFilter(5)

list.filter { (isFruit && isShort) (_) }

So far everything works ok. But say that I want to do something like this:

  val nameOption: Option[String]
  val lengthOption: Option[Int]

  val filters = {

    nameOption.map((name) =>
      nameFilter(name)
    ) &&
    lengthOption.map((length) =>
      lengthFilter(length)
    )
  }

  list.filter { filters (_) }

So now I need to && an Option[(A) => Boolean] If the Option is None then simply disregard the filter.

If I use something like:

def const(res:Boolean)[A]:A=>Boolean = a => res

implicit def optToFilter[A](optFilter:Option[A => Boolean]):A => Boolean = optFilter match {
  case Some(filter) => filter
  case None => const(true)[A]
}

I have the problem of || with one filter set to true. I can solve this by changing true to false but then the same problem exists with &&.

I could also take this approach:

  implicit def optionalPredicate[A](pred: Option[A => Boolean]): OptionalPredicate[A] = new OptionalPredicate(pred)

  class OptionalPredicate[A](val pred: Option[A => Boolean]) {
    def apply(x: A) = pred match {
      case Some(filter) => filter(x)
      case None => trueFilter(x)
    }

    def &&(that: Option[A => Boolean]) = Some((x: A) =>
      pred.getOrElse(trueFilter)(x) && that.getOrElse(trueFilter)(x))

    def ||(that: Option[A => Boolean]) = Some((x: A) =>
      pred.getOrElse(falseFilter)(x) || that.getOrElse(falseFilter)(x))
  }

  def trueFilter[A]: A => Boolean = const(res = true)

  def falseFilter[A]: A => Boolean = const(res = false)

  def const[A](res: Boolean): A => Boolean = a => res

But something seems inherently wrong with having to convert the Predicate into an OptionalPredicate when the predicate is not the child type of an Option:

  implicit def convertSimpleFilter[A](filter: A => Boolean) = Some(filter)

Which allows:

ifFruit && isShort

I have the feeling that the Optional should tie in seamlessly with Predicate while following DRY principles.

回答1:

See below (UPD)

It would be good to be able to convert an Option[Filter] to Filter:

def const(res:Boolean)[A]:A=>Boolean = a => res

implicit def optToFilter[A](optFilter:Option[A => Boolean]):A => Boolean = optFilter match {
  case Some(filter) => filter
  case None => const(true)[A]
}

Then we would like to use any type that can be converted to a predicate:

implicit class Predicate[A, P <% A=>Boolean](val pred: P)

(and change usages of pred to (pred:A=>Boolean))

def &&[P2:A=>Boolean](that: P2) = new Predicate[A](x => (pred:A=>Boolean)(x) && (that:A=>Boolean)(x))

UPD

Option[A=>Boolean] will be the real filter type.

implicit def convertSimpleFilter[A](filter:A=>Boolean)=Some(filter)

implicit class Predicate[A](val pred: Option[A=>Boolean]){
  def &&(that:Option[A=>Boolean]) = Some((x:A) => pred.getOrElse(const(true))(x) && that.getOrElse(const(true))(x) )
  def ||(that:Option[A=>Boolean]) = Some((x:A) => pred.getOrElse(const(false))(x) || that.getOrElse(const(false))(x) )
}

来源：https://stackoverflow.com/questions/28659506/scala-optional-predicate

标签

scala

implicit-conversion

implicit