Question

我被告知要使用这段有趣的代码，但我的用例要求它比目前可能做的更多。

  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))
  }

一些用例样板：

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) (_) }

到目前为止一切正常。但是说我想做这样的事情：

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

  val filters = {

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

  list.filter { filters (_) }

所以现在我需要&&一个Option[(A) => Boolean]如果选项为无，则只需忽略过滤器。

如果我使用类似的东西：

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]
}

我遇到||的问题，其中一个过滤器设置为true。我可以通过将true更改为false来解决此问题，但&&存在相同的问题。

我也可以采用这种方法：

  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

但是当谓词不是选项的子类型时，必须将Predicate转换为OptionalPredicate，这似乎是错误的：

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

允许：

ifFruit && isShort

我觉得Optional可以在遵循DRY原则的同时与Predicate无缝结合。

Answer 1

见下文（UPD）

能够将Option[Filter]转换为Filter：

会很好

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]
}

然后我们想使用任何可以转换为谓词的类型：

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

（并将pred的用法更改为(pred:A=>Boolean)）

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

<强> UPD

Option[A=>Boolean]将是真正的过滤器类型。

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) )
}

Scala - 可选谓词

1 个答案: