定义仿函数时如何抽象第二个类型参数？

Question

使用 Scala 和 Cats 学习惯用的 FP。我有以下代码...

package org.economicsl.mechanisms

import cats._
import cats.implicits._

trait Preference[-A] {
  self =>

  def compare(a1: A, a2: A): Int

  final def ordering[A1 <: A]: Ordering[A1] = {
    new Ordering[A1] {
      def compare(a1: A1, a2: A1): Int = {
        self.compare(a1, a2)
      }
    }
  }

}


object Preference {

  implicit val contravariant: Contravariant[Preference] = {
    new Contravariant[Preference] {
      def contramap[A, B](fa: Preference[A])(f: B => A): Preference[B] = {
        new Preference[B] {
          def compare(b1: B, b2: B): Int = {
            fa.compare(f(b1), f(b2))
          }
        }
      }
    }
  }

  /** Defines a preference for a particular alternative. */
  def particular[A](alternative: A): Preference[A] = {
    new Preference[A] {
      def compare(a1: A, a2: A): Int = {
        if ((a1 != alternative) & (a2 == alternative)) {
          -1
        } else if ((a1 == alternative) & (a2 != alternative)) {
          1
        } else {
          0
        }
      }
    }
  }

}

/** Base trait defining a generic social welfare function.
  *
  * A social welfare function aggregates the preferences of individual agents
  * into a common preference ordering.
  */
trait SocialWelfareFunction[-CC <: Iterable[P], +P <: Preference[A], A]
  extends (CC => P)


/** Companion object for the `SocialWelFareFunction` trait. */
object SocialWelfareFunction {

  val setInvariant: Invariant[({ type F[A] = SocialWelfareFunction[Set[Preference[A]], Preference[A], A] })#F] = {
    new Invariant[({ type F[A] = SocialWelfareFunction[Set[Preference[A]], Preference[A], A] })#F] {
      def imap[A, B](fa: SocialWelfareFunction[Set[Preference[A]], Preference[A], A])(f: A => B)(g: B => A): SocialWelfareFunction[Set[Preference[B]], Preference[B], B] = {
        new SocialWelfareFunction[Set[Preference[B]], Preference[B], B] {
          def apply(preferences: Set[Preference[B]]): Preference[B] = {
            fa(preferences.map(pb => pb.contramap(f))).contramap(g)
          }
        }
      }
    }
  }

  val seqInvariant: Invariant[({ type F[A] = SocialWelfareFunction[Seq[Preference[A]], Preference[A], A] })#F] = {
    new Invariant[({ type F[A] = SocialWelfareFunction[Seq[Preference[A]], Preference[A], A] })#F] {
      def imap[A, B](fa: SocialWelfareFunction[Seq[Preference[A]], Preference[A], A])(f: A => B)(g: B => A): SocialWelfareFunction[Seq[Preference[B]], Preference[B], B] = {
        new SocialWelfareFunction[Seq[Preference[B]], Preference[B], B] {
          def apply(preferences: Seq[Preference[B]]): Preference[B] = {
            fa(preferences.map(pb => pb.contramap(f))).contramap(g)
          }
        }
      }
    }
  }
}

在 SocialWelfareFunction 的伴随对象中，我定义了两个不变仿函数。这些实现几乎相同：第一个函子使用 Set 来存储 Preference[A] 个实例，第二个函子使用 Seq.

有没有办法抽象存储首选项的集合类型？如果解决方案需要，我愿意包括其他类型级别的依赖项。

Answer 1

你可以这样做：

  def invariant[C[X] <: Iterable[X] : Functor]: Invariant[({ type F[A] = SocialWelfareFunction[C[Preference[A]], Preference[A], A] })#F] = {
    new Invariant[({ type F[A] = SocialWelfareFunction[C[Preference[A]], Preference[A], A] })#F] {
      def imap[A, B](fa: SocialWelfareFunction[C[Preference[A]], Preference[A], A])(f: A => B)(g: B => A): SocialWelfareFunction[C[Preference[B]], Preference[B], B] = {
        new SocialWelfareFunction[C[Preference[B]], Preference[B], B] {
          def apply(preferences: C[Preference[B]]): Preference[B] = {
            fa(Functor[C].map(preferences)(pb => pb.contramap(f))).contramap(g)
          }
        }
      }
    }
  }

然后对于 cats 具有 Functor 的类型，您可以这样做：

  val listInvariant = invariant[List]

令人讨厌的是，他们似乎没有为 Set 和 Seq 内置的 Functors，但他们很容易定义自己：

  private[this] implicit val setFunctor = new Functor[Set] {
    override def map[A, B](s: Set[A])(f: A => B): Set[B] = s.map(f)
  }

  val setInvariant = invariant[Set]

定义仿函数时如何抽象第二个类型参数？

How to abstract over a second type parameter when defining functors?

functional-programming

scala

functor

scala-cats