Scalaz（6）－ typeclass：Functor－just map-阿里云开发者社区

Functor是范畴学（Category theory）里的概念。不过无须担心，我们在scala FP编程里并不需要先掌握范畴学知识的。在scalaz里，Functor就是一个普通的typeclass，具备map over特性。我的理解中，Functor的主要用途是在FP过程中更新包嵌在容器（高阶类）F[T]中元素T值。典型例子如：List[String], Option[Int]等。我们曾经介绍过FP与OOP的其中一项典型区别在于FP会尽量避免中间变量（temp variables）。FP的变量V是以F[V]这种形式存在的，如：List[Int]里一个Int变量是包嵌在容器List里的。所以FP需要特殊的方式来更新变量V，这就是Functor map over的意思。scalaz提供了Functor typeclass不但使用户能map over自定义的高阶类型F[T]，并且用户通过提供自定义类型的Functor实例就可以免费使用scalaz Functor typeclass提供的一系列组件函数（combinator functions）。

scalaz中Functor的trait是这样定义的：scalaz/Functor.scala

trait Functor[F[_]] extends InvariantFunctor[F] { self =>
  ////
  import Liskov.<~<

  /** Lift `f` into `F` and apply to `F[A]`. */
  def map[A, B](fa: F[A])(f: A => B): F[B]

...

任何类型的实例只需要实现这个抽象函数map就可以使用scalaz Functor typeclass的这些注入方法了：scalaz/syntax/FunctorSyntax.scala

final class FunctorOps[F[_],A] private[syntax](val self: F[A])(implicit val F: Functor[F]) extends Ops[F[A]] {
  ////
  import Leibniz.===
  import Liskov.<~<

  final def map[B](f: A => B): F[B] = F.map(self)(f)
  final def distribute[G[_], B](f: A => G[B])(implicit D: Distributive[G]): G[F[B]] = D.distribute(self)(f)
  final def cosequence[G[_], B](implicit ev: A === G[B], D: Distributive[G]): G[F[B]] = D.distribute(self)(ev(_))
  final def cotraverse[G[_], B, C](f: F[B] => C)(implicit ev: A === G[B], D: Distributive[G]): G[C] = D.map(cosequence)(f)
  final def ∘[B](f: A => B): F[B] = F.map(self)(f)
  final def strengthL[B](b: B): F[(B, A)] = F.strengthL(b, self)
  final def strengthR[B](b: B): F[(A, B)] = F.strengthR(self, b)
  final def fpair: F[(A, A)] = F.fpair(self)
  final def fproduct[B](f: A => B): F[(A, B)] = F.fproduct(self)(f)
  final def void: F[Unit] = F.void(self)
  final def fpoint[G[_]: Applicative]: F[G[A]] = F.map(self)(a => Applicative[G].point(a))
  final def >|[B](b: => B): F[B] = F.map(self)(_ => b)
  final def as[B](b: => B): F[B] = F.map(self)(_ => b)
  final def widen[B](implicit ev: A <~< B): F[B] = F.widen(self)
  ////
}

以上的注入方法中除了map外其它方法的应用场景我还没有确切的想法，不过这不会妨碍我们示范它们的用法。Functor必须遵循一些定律：

1、map(fa)(x => x) === fa

2、map(map(fa)(f1))(f2) === map(fa)(f2 compose f1)

scalaz/Functor.scala

trait FunctorLaw extends InvariantFunctorLaw {
    /** The identity function, lifted, is a no-op. map(fa)(x => x*/
    def identity[A](fa: F[A])(implicit FA: Equal[F[A]]): Boolean = FA.equal(map(fa)(x => x), fa)

    /**
     * A series of maps may be freely rewritten as a single map on a
     * composed function.
     */
    def composite[A, B, C](fa: F[A], f1: A => B, f2: B => C)(implicit FC: Equal[F[C]]): Boolean = FC.equal(map(map(fa)(f1))(f2), map(fa)(f2 compose f1))
  }

我们可以用List来证明：map(fa)(x => x) === fa

scala> List(1,2,3).map(x => x) assert_=== List(1,2,3)

scala> List(1,2,3).map(identity) assert_=== List(1,2)
java.lang.RuntimeException: [1,2,3] ≠ [1,2]
  at scala.sys.package$.error(package.scala:27)
  at scalaz.syntax.EqualOps.assert_$eq$eq$eq(EqualSyntax.scala:16)
  ... 43 elided

map(map(fa)(f1))(f2) === map(fa)(f2 compose f1)

scala> Functor[List].map(List(1,2,3).map(i => i + 1))(i2 => i2 * 3) assert_=== List(1,2,3).map(((i2:Int) => i2 * 3) compose ((i:Int) => i + 1))

scala> Functor[List].map(List(1,2,3).map(i => i + 1))(i2 => i2 * 3) assert_=== List(1,2,3).map(((i:Int) => i + 1) compose ((i2:Int) => i2 * 3))
java.lang.RuntimeException: [6,9,12] ≠ [4,7,10]
  at scala.sys.package$.error(package.scala:27)
  at scalaz.syntax.EqualOps.assert_$eq$eq$eq(EqualSyntax.scala:16)
  ... 43 elided

注意：compose对f1,f2的施用是互换的。

针对我们自定义的类型，我们只要实现map函数就可以得到这个类型的Functor实例。一旦实现了这个类型的Functor实例，我们就可以使用以上scalaz提供的所有Functor组件函数了。

我们先试着创建一个类型然后推算它的Functor实例：

1 case class Item3[A](i1: A, i2: A, i3: A)
2 val item3Functor = new Functor[Item3] {
3     def map[A,B](ia: Item3[A])(f: A => B): Item3[B] = Item3(f(ia.i1),f(ia.i2),f(ia.i3))
4 }                                                 //> item3Functor  : scalaz.Functor[scalaz.functor.Item3] = scalaz.functor$$anonf
5                                                   //| un$main$1$$anon$1@5e265ba4

scalaz同时在scalaz-tests下提供了一套scalacheck测试库。我们可以对Item3的Functor实例进行测试：

1 scala> functor.laws[Item3].check
2 <console>:27: error: could not find implicit value for parameter af: org.scalacheck.Arbitrary[Item3[Int]]
3               functor.laws[Item3].check
4                           ^

看来我们需要提供自定义类型Item3的随意产生器（Generator）：

scala> implicit def item3Arbi[A](implicit a: Arbitrary[A]): Arbitrary[Item3[A]] = Arbitrary {
     | def genItem3: Gen[Item3[A]]  = for {
     | b <- Arbitrary.arbitrary[A]
     | c <- Arbitrary.arbitrary[A]
     | d <- Arbitrary.arbitrary[A]
     | } yield Item3(b,c,d)
     | genItem3
     | }
item3Arbi: [A](implicit a: org.scalacheck.Arbitrary[A])org.scalacheck.Arbitrary[Item3[A]]

scala> functor.laws[Item3].check
+ functor.invariantFunctor.identity: OK, passed 100 tests.
+ functor.invariantFunctor.composite: OK, passed 100 tests.
+ functor.identity: OK, passed 100 tests.
+ functor.composite: OK, passed 100 tests.

Item3的Functor实例是合理的。

实际上map就是(A => B) => (F[A] => F[B])，就是把(A => B)升格（lift）成（F[A] => F[B]）:

case class Item3[A](i1: A, i2: A, i3: A)
implicit val item3Functor = new Functor[Item3] {
    def map[A,B](ia: Item3[A])(f: A => B): Item3[B] = Item3(f(ia.i1),f(ia.i2),f(ia.i3))
}                                                 //> item3Functor  : scalaz.Functor[scalaz.functor.Item3] = scalaz.functor$$anonf
                                                  //| un$main$1$$anon$1@5e265ba4
val F = Functor[Item3]                            //> F  : scalaz.Functor[scalaz.functor.Item3] = scalaz.functor$$anonfun$main$1$$
                                                  //| anon$1@5e265ba4
F.map(Item3("Morning","Noon","Night"))(_.length)  //> res0: scalaz.functor.Item3[Int] = Item3(7,4,5)
F.apply(Item3("Morning","Noon","Night"))(_.length)//> res1: scalaz.functor.Item3[Int] = Item3(7,4,5)
F(Item3("Morning","Noon","Night"))(_.length)      //> res2: scalaz.functor.Item3[Int] = Item3(7,4,5)
F.lift((s: String) => s.length)(Item3("Morning","Noon","Night"))
                                                  //> res3: scalaz.functor.Item3[Int] = Item3(7,4,5)

虽然函数升格（function lifting (A => B) => (F[A] => F[B])是Functor的主要功能，但我们说过：一旦能够获取Item3类型的Functor实例我们就能免费使用所有的注入方法：

scalaz提供了Function1的Functor实例。Function1 Functor的map就是 andThen 也就是操作方调换的compose：

scala> (((_: Int) + 1) map((k: Int) => k * 3))(2)
res20: Int = 9

scala> (((_: Int) + 1) map((_: Int) * 3))(2)
res21: Int = 9

scala> (((_: Int) + 1) andThen ((_: Int) * 3))(2)
res22: Int = 9

scala> (((_: Int) * 3) compose ((_: Int) + 1))(2)
res23: Int = 9

我们也可以对Functor进行compose：

scala> val f = Functor[List] compose Functor[Item3]
f: scalaz.Functor[[α]List[Item3[α]]] = scalaz.Functor$$anon$1@647ce8fd

scala> val item3 = Item3("Morning","Noon","Night")
item3: Item3[String] = Item3(Morning,Noon,Night)

scala> f.map(List(item3,item3))(_.length)
res25: List[Item3[Int]] = List(Item3(7,4,5), Item3(7,4,5))

反过来操作：

1 scala> val f1 = Functor[Item3] compose Functor[List]
2 f1: scalaz.Functor[[α]Item3[List[α]]] = scalaz.Functor$$anon$1@5b6a0166
3 
4 scala> f1.map(Item3(List("1"),List("22"),List("333")))(_.length)
5 res26: Item3[List[Int]] = Item3(List(1),List(2),List(3))

我们再试着在Item3类型上调用那些免费的注入方法：

scala> item3.fpair
res28: Item3[(String, String)] = Item3((Morning,Morning),(Noon,Noon),(Night,Night))

scala> item3.strengthL(3)
res29: Item3[(Int, String)] = Item3((3,Morning),(3,Noon),(3,Night))

scala> item3.strengthR(3)
res30: Item3[(String, Int)] = Item3((Morning,3),(Noon,3),(Night,3))

scala> item3.fproduct(_.length)
res31: Item3[(String, Int)] = Item3((Morning,7),(Noon,4),(Night,5))

scala> item3 as "Day"
res32: Item3[String] = Item3(Day,Day,Day)

scala> item3 >| "Day"
res33: Item3[String] = Item3(Day,Day,Day)

scala> item3.void
res34: Item3[Unit] = Item3((),(),())

我现在还没有想到这些函数的具体用处。不过从运算结果来看，用这些函数来产生一些数据模型用在游戏或者测试的模拟（simulation）倒是可能的。

scalaz提供了许多现成的Functor实例。我们先看看一些简单直接的实例：

scala> Functor[List].map(List(1,2,3))(_ + 3)
res35: List[Int] = List(4, 5, 6)

scala> Functor[Option].map(Some(3))(_ + 3)
res36: Option[Int] = Some(6)

scala> Functor[java.util.concurrent.Callable]
res37: scalaz.Functor[java.util.concurrent.Callable] = scalaz.std.java.util.concurrent.CallableInstances$$anon$1@4176ab89

scala> Functor[Stream]
res38: scalaz.Functor[Stream] = scalaz.std.StreamInstances$$anon$1@4f5374b9

scala> Functor[Vector]
res39: scalaz.Functor[Vector] = scalaz.std.IndexedSeqSubInstances$$anon$1@4367920a

对那些多个类型变量的类型我们可以采用部分施用方式：即type lambda来表示。一个典型的类型：Either[E,A]，我们可以把Left[E]固定下来: Either[String, A]，我们可以用type lambda来这样表述：

1 scala> Functor[({type l[x] = Either[String,x]})#l].map(Right(3))(_ + 3)
2 res41: scala.util.Either[String,Int] = Right(6)

如此这般我可以对Either类型进行map操作了。

函数类型的Functor是针对返回类型的：

scala> Functor[({type l[x] = String => x})#l].map((s: String) => s + "!")(_.length)("Hello")
res53: Int = 6

scala> Functor[({type l[x] = (String,Int) => x})#l].map((s: String, i: Int) => s.length + i)(_ * 10)("Hello",5)
res54: Int = 100

scala> Functor[({type l[x] = (String,Int,Boolean) => x})#l].map((s: String,i: Int, b: Boolean)=> s + i.toString + b.toString)(_.toUpperCase)("Hello",3,true)
res56: String = HELLO3TRUE

tuple类型的Functor是针对最后一个元素类型的：

cala> Functor[({type l[x] = (String,x)})#l].map(("a",1))(_ + 2)
res57: (String, Int) = (a,3)

scala> Functor[({type l[x] = (String,Int,x)})#l].map(("a",1,"b"))(_.toUpperCase)
res58: (String, Int, String) = (a,1,B)

scala> Functor[({type l[x] = (String,Int,Boolean,x)})#l].map(("a",1,true,Item3("a","b","c")))(i => i.map(_.toUpperCase))
res62: (String, Int, Boolean, Item3[String]) = (a,1,true,Item3(A,B,C))

Scalaz（6）－ typeclass：Functor－just map

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Scalaz（6）－ typeclass：Functor－just map

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