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Scalaz(14)- Monad:函数组合-Kleisli to Reader

2015-11-25 21:12 323 查看
Monad Reader就是一种函数的组合。在scalaz里函数(function)本身就是Monad,自然也就是Functor和applicative。我们可以用Monadic方法进行函数组合:

import scalaz._
import Scalaz._
object decompose {
//两个测试函数
val f = (_: Int) + 3                              //> f  : Int => Int = <function1>
val g = (_: Int) * 5                              //> g  : Int => Int = <function1>
//functor
val h = f map g  // f andThen g                   //> h  : Int => Int = <function1>
val h1 = g map f  // f compose g                  //> h1  : Int => Int = <function1>
h(2)  //g(f(2))                                   //> res0: Int = 25
h1(2) //f(g(2))                                   //> res1: Int = 13
//applicative
val k = (f |@| g){_ + _}                          //> k  : Int => Int = <function1>
k(10) // f(10)+g(10)                              //> res2: Int = 63
//monad
val m = g.flatMap{a => f.map(b => a+b)}           //> m  : Int => Int = <function1>
val n = for {
a <- f
b <- g
} yield a + b                                     //> n  : Int => Int = <function1>
m(10)                                             //> res3: Int = 63
n(10)                                             //> res4: Int = 63
}


以上的函数f,g必须满足一定的条件才能实现组合。这个从f(g(2))或g(f(2))可以看出:必需固定有一个输入参数及输入参数类型和函数结果类型必需一致,因为一个函数的输出成为另一个函数的输入。在FP里这样的函数组合就是Monadic Reader。
但是FP里函数运算结果一般都是M[R]这样格式的,所以我们需要对f:A => M[B],g:B => M[C]这样的函数进行组合。这就是scalaz里的Kleisli了。Kleisli就是函数A=>M[B]的类封套,从Kleisli的类定义可以看出:scalaz/Kleisli.scala

final case class Kleisli[M[_], A, B](run: A => M[B]) { self =>
...
trait KleisliFunctions {
/**Construct a Kleisli from a Function1 */
def kleisli[M[_], A, B](f: A => M[B]): Kleisli[M, A, B] = Kleisli(f)
...


Kleisli的目的是把Monadic函数组合起来或者更形象说连接起来。Kleisli提供的操作方法如>=>可以这样理解:

(A=>M[B]) >=> (B=>M[C]) >=> (C=>M[D]) 最终运算结果M[D]

可以看出Kleisli函数组合有着固定的模式:

1、函数必需是 A => M[B]这种模式;只有一个输入,结果是一个Monad M[_]

2、上一个函数输出M[B],他的运算值B就是下一个函数的输入。这就要求下一个函数的输入参数类型必需是B

3、M必须是个Monad;这个可以从Kleisli的操作函数实现中看出:scalaz/Kleisli.scala

/** alias for `andThen` */
def >=>[C](k: Kleisli[M, B, C])(implicit b: Bind[M]): Kleisli[M, A, C] =  kleisli((a: A) => b.bind(this(a))(k.run))

def andThen[C](k: Kleisli[M, B, C])(implicit b: Bind[M]): Kleisli[M, A, C] = this >=> k

def >==>[C](k: B => M[C])(implicit b: Bind[M]): Kleisli[M, A, C] = this >=> kleisli(k)

def andThenK[C](k: B => M[C])(implicit b: Bind[M]): Kleisli[M, A, C] = this >==> k

/** alias for `compose` */
def <=<[C](k: Kleisli[M, C, A])(implicit b: Bind[M]): Kleisli[M, C, B] = k >=> this

def compose[C](k: Kleisli[M, C, A])(implicit b: Bind[M]): Kleisli[M, C, B] = k >=> this

def <==<[C](k: C => M[A])(implicit b: Bind[M]): Kleisli[M, C, B] = kleisli(k) >=> this

def composeK[C](k: C => M[A])(implicit b: Bind[M]): Kleisli[M, C, B] = this <==< k


拿操作函数>=>(andThen)举例:implicit b: Bind[M]明确了M必须是个Monad。

kleisli((a: A) => b.bind(this(a))(k.run))的意思是先运算M[A],接着再运算k,以M[A]运算结果值a作为下一个函数k.run的输入参数。整个实现过程并不复杂。
实际上Reader就是Kleisli的一个特殊案例:在这里kleisli的M[]变成了Id[],因为Id[A]=A >>> A=>Id[B] = A=>B,就是我们上面提到的Reader,我们看看Reader在scalaz里是如何定义的:scalar/package.scala
type ReaderT[F[_], E, A] = Kleisli[F, E, A]
val ReaderT = Kleisli
type =?>[E, A] = Kleisli[Option, E, A]
type Reader[E, A] = ReaderT[Id, E, A]

type Writer[W, A] = WriterT[Id, W, A]
type Unwriter[W, A] = UnwriterT[Id, W, A]

object Reader {
def apply[E, A](f: E => A): Reader[E, A] = Kleisli[Id, E, A](f)
}

object Writer {
def apply[W, A](w: W, a: A): WriterT[Id, W, A] = WriterT[Id, W, A]((w, a))
}

object Unwriter {
def apply[U, A](u: U, a: A): UnwriterT[Id, U, A] = UnwriterT[Id, U, A]((u, a))
}


type ReaderT[F[_], E, A] = Kleisli[F, E, A] >>> type Reader[E,A] = ReaderT[Id,E,A]
好了,说了半天还是回到如何使用Kleisli进行函数组合的吧:
//Kleisli款式函数kf,kg
val kf: Int => Option[String] = (i: Int) => Some((i + 3).shows)
//> kf  : Int => Option[String] = <function1>
val kg: String => Option[Boolean] = { case "3" => true.some; case _ => false.some }
//> kg  : String => Option[Boolean] = <function1>
//Kleisli函数组合操作
import Kleisli._
val kfg = kleisli(kf) >=> kleisli(kg)             //> kfg  : scalaz.Kleisli[Option,Int,Boolean] = Kleisli(<function1>)
kfg(1)                                            //> res5: Option[Boolean] = Some(false)
kfg(0)                                            //> res6: Option[Boolean] = Some(true)


例子虽然很简单,但它说明了很多重点:上一个函数输入的运算值是下一个函数的输入值 Int=>String=>Boolean。输出Monad一致统一,都是Option。
那么,Kleisli到底用来干什么呢?它恰恰显示了FP函数组合的真正意义:把功能尽量细分化,通过各种方式的函数组合实现灵活的函数重复利用。也就是在FP领域里,我们用Kleisli来组合FP函数。
下面我们就用scalaz自带的例子scalaz.example里的KleisliUsage.scala来说明一下Kleisli的具体使用方法吧:
下面是一组地理信息结构:
// just some trivial data structure ,
// Continents contain countries. Countries contain cities.
case class Continent(name: String, countries: List[Country] = List.empty)
case class Country(name: String, cities: List[City] = List.empty)
case class City(name: String, isCapital: Boolean = false, inhabitants: Int = 20)


分别是:洲(Continent)、国家(Country)、城市(City)。它们之间的关系是层级的:Continent(Country(City))
下面是一组模拟数据:
val data: List[Continent] = List(
Continent("Europe"),
Continent("America",
List(
Country("USA",
List(
City("Washington"), City("New York"))))),
Continent("Asia",
List(
Country("India",
List(City("New Dehli"), City("Calcutta"))))))


从上面的模拟数据也可以看出Continent,Country,City之间的隶属关系。我们下面设计三个函数分别对Continent,Country,City进行查找:
def continents(name: String): List[Continent] =
data.filter(k => k.name.contains(name))       //> continents: (name: String)List[Exercises.kli.Continent]
//查找名字包含A的continent
continents("A")                                 //> res7: List[Exercises.kli.Continent] = List(Continent(America,List(Country(U
//| SA,List(City(Washington,false,20), City(New York,false,20))))), Continent(A
//| sia,List(Country(India,List(City(New Dehli,false,20), City(Calcutta,false,2
//| 0))))))
//找到两个:List(America,Asia)
def countries(continent: Continent): List[Country] = continent.countries
//> countries: (continent: Exercises.kli.Continent)List[Exercises.kli.Country]
//查找America下的国家
val america =
Continent("America",
List(
Country("USA",
List(
City("Washington"), City("New York")))))
//> america  : Exercises.kli.Continent = Continent(America,List(Country(USA,Lis
//| t(City(Washington,false,20), City(New York,false,20)))))
countries(america)                              //> res8: List[Exercises.kli.Country] = List(Country(USA,List(City(Washington,f
//| alse,20), City(New York,false,20))))
def cities(country: Country): List[City] = country.cities
//> cities: (country: Exercises.kli.Country)List[Exercises.kli.City]
val usa = Country("USA",
List(
City("Washington"), City("New York")))
//> usa  : Exercises.kli.Country = Country(USA,List(City(Washington,false,20),
//| City(New York,false,20)))
cities(usa)                                     //> res9: List[Exercises.kli.City] = List(City(Washington,false,20), City(New Y
//| ork,false,20))


从continents,countries,cities这三个函数运算结果可以看出它们都可以独立运算。这三个函数的款式如下:
String => List[Continent]
Continent => List[Country]
Country => List[City]
无论函数款式或者类封套(List本来就是Monad)都适合Kleisli。我们可以用Kleisli把这三个局部函数用各种方法组合起来实现更广泛功能:
val allCountry = kleisli(continents) >==> countries
//> allCountry  : scalaz.Kleisli[List,String,Exercises.kli.Country] = Kleisli(<
//| function1>)
val allCity = kleisli(continents) >==> countries >==> cities
//> allCity  : scalaz.Kleisli[List,String,Exercises.kli.City] = Kleisli(<functi
//| on1>)
allCountry("Amer")                              //> res10: List[Exercises.kli.Country] = List(Country(USA,List(City(Washington,
//| false,20), City(New York,false,20))))
allCity("Amer")                                 //> res11: List[Exercises.kli.City] = List(City(Washington,false,20), City(New
//| York,false,20))


还有个=<<符号挺有意思:
def =<<(a: M[A])(implicit m: Bind[M]): M[B] = m.bind(a)(run)


意思是用包嵌的函数flatMap一下输入参数M[A]:
allCity =<< List("Amer","Asia")                 //> res12: List[Exercises.kli.City] = List(City(Washington,false,20), City(New
//| York,false,20), City(New Dehli,false,20), City(Calcutta,false,20))


那么如果我想避免使用List(),用Option[List]作为函数输出可以吗?Option是个Monad,第一步可以通过。下一步是把函数款式对齐了:
List[String] => Option[List[Continent]]
List[Continent] => Option[List[Country]]
List[Country] => Option[List[City]]
下面是这三个函数的升级版:
//查找Continent List[String] => Option[List[Continent]]
def maybeContinents(names: List[String]): Option[List[Continent]] =
names.flatMap(name => data.filter(k => k.name.contains(name))) match {
case h :: t => (h :: t).some
case _ => none
}                                             //> maybeContinents: (names: List[String])Option[List[Exercises.kli.Continent]]
//|
//测试运行
maybeContinents(List("Amer","Asia"))            //> res13: Option[List[Exercises.kli.Continent]] = Some(List(Continent(America,
//| List(Country(USA,List(City(Washington,false,20), City(New York,false,20))))
//| ), Continent(Asia,List(Country(India,List(City(New Dehli,false,20), City(Ca
//| lcutta,false,20)))))))
//查找Country  List[Continent] => Option[List[Country]]
def maybeCountries(continents: List[Continent]): Option[List[Country]] =
continents.flatMap(continent => continent.countries.map(c => c)) match {
case h :: t => (h :: t).some
case _ => none
}                                             //> maybeCountries: (continents: List[Exercises.kli.Continent])Option[List[Exer
//| cises.kli.Country]]
//查找City  List[Country] => Option[List[Country]]
def maybeCities(countries: List[Country]): Option[List[City]] =
countries.flatMap(country => country.cities.map(c => c)) match {
case h :: t => (h :: t).some
case _ => none
}                                             //> maybeCities: (countries: List[Exercises.kli.Country])Option[List[Exercises.
//| kli.City]]

val maybeAllCities = kleisli(maybeContinents) >==> maybeCountries >==> maybeCities
//> maybeAllCities  : scalaz.Kleisli[Option,List[String],List[Exercises.kli.Cit
//| y]] = Kleisli(<function1>)
maybeAllCities(List("Amer","Asia"))             //> res14: Option[List[Exercises.kli.City]] = Some(List(City(Washington,false,2
//| 0), City(New York,false,20), City(New Dehli,false,20), City(Calcutta,false,
//| 20)))


我们看到,只要Monad一致,函数输入输出类型匹配,就能用Kleisli来实现函数组合。
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