Question

我想在scala中编写一个memoize函数，无论函数对象是什么，它都可以应用于任何函数对象。我希望以一种允许我使用memoize的单个实现的方式这样做。我对语法很灵活，但理想情况下，memoize出现在非常接近函数声明的地方，而不是函数之后。我还想避免首先声明原始函数，然后再回忆一下memoized版本的第二个声明。

所以一些理想的语法可能就是这样：

def slowFunction(<some args left intentionally vague>) = memoize {
  // the original implementation of slow function
}

甚至可以接受：

def slowFUnction = memoize { <some args left intentionally vague> => {
  // the original implementation of slow function
}}

我已经看到了这样做的方法，其中必须为每个arity函数重新定义memoize，但我想避免这种方法。原因是我需要实现几十个类似于memoize的函数（即其他装饰器），并且要求每个arity函数复制每个函数太多了。

一种做memoize的方法，需要你重复memoize声明（所以它没有用）是What type to use to store an in-memory mutable data table in Scala?。

Answer 1

您可以使用类型方法来处理arity问题。你仍然需要处理你想要支持的每个功能，但不是每个arity / decorator组合：

/**
 * A type class that can tuple and untuple function types.
 * @param [U] an untupled function type
 * @param [T] a tupled function type
 */
sealed class Tupler[U, T](val tupled: U => T, 
                          val untupled: T => U)

object Tupler {
   implicit def function0[R]: Tupler[() => R, Unit => R] =
      new Tupler((f: () => R) => (_: Unit) => f(),
                 (f: Unit => R) => () => f(()))
   implicit def function1[T, R]: Tupler[T => R, T => R] = 
      new Tupler(identity, identity)
   implicit def function2[T1, T2, R]: Tupler[(T1, T2) => R, ((T1, T2)) => R] = 
      new Tupler(_.tupled, Function.untupled[T1, T2, R]) 
   // ... more tuplers
}

然后您可以按如下方式实现装饰器：

/**
 * A memoized unary function.
 *
 * @param f A unary function to memoize
 * @param [T] the argument type
 * @param [R] the return type
 */
class Memoize1[-T, +R](f: T => R) extends (T => R) {
   // memoization implementation
}

object Memoize {
   /**
    * Memoize a function.
    *
    * @param f the function to memoize
    */
   def memoize[T, R, F](f: F)(implicit e: Tupler[F, T => R]): F = 
      e.untupled(new Memoize1(e.tupled(f)))
}

你的“理想”语法不起作用，因为编译器会假设传递给memoize的块是一个0参数的词法闭包。但是，您可以使用后一种语法：

// edit: this was originally (and incorrectly) a def
lazy val slowFn = memoize { (n: Int) => 
   // compute the prime decomposition of n
}

修改

为了消除许多用于定义新装饰器的样板，您可以创建一个特征：

trait FunctionDecorator { final def apply[T, R, F](f: F)(implicit e: Tupler[F, T => R]): F = e.untupled(decorate(e.tupled(f))) protected def decorate[T, R](f: T => R): T => R }

这允许您将Memoize装饰器重新定义为

object Memoize extends FunctionDecorator { /** * Memoize a function. * * @param f the function to memoize */ protected def decorate[T, R](f: T => R) = new Memoize1(f) }

不是在Memoize对象上调用memoize方法，而是直接应用Memoize对象：

// edit: this was originally (and incorrectly) a def lazy val slowFn = Memoize(primeDecomposition _)

或

lazy val slowFn = Memoize { (n: Int) => // compute the prime decomposition of n }

Answer 2

库

使用Scalaz的scalaz.Memo

手册

以下是类似于Aaron Novstrup的回答和this blog的解决方案，除了一些更正/改进，简洁且更容易满足人们的复制和粘贴需求：）

import scala.Predef._

class Memoized[-T, +R](f: T => R) extends (T => R) {

  import scala.collection.mutable

  private[this] val vals = mutable.Map.empty[T, R]

  def apply(x: T): R = vals.getOrElse(x, {
      val y = f(x)
      vals += ((x, y))
      y
    })
}

// TODO Use macros
// See si9n.com/treehugger/
// http://stackoverflow.com/questions/11400705/code-generation-with-scala
object Tupler {
  implicit def t0t[R]: (() => R) => (Unit) => R = (f: () => R) => (_: Unit) => f()

  implicit def t1t[T, R]: ((T) => R) => (T) => R = identity

  implicit def t2t[T1, T2, R]: ((T1, T2) => R) => ((T1, T2)) => R = (_: (T1, T2) => R).tupled

  implicit def t3t[T1, T2, T3, R]: ((T1, T2, T3) => R) => ((T1, T2, T3)) => R = (_: (T1, T2, T3) => R).tupled

  implicit def t0u[R]: ((Unit) => R) => () => R = (f: Unit => R) => () => f(())

  implicit def t1u[T, R]: ((T) => R) => (T) => R = identity

  implicit def t2u[T1, T2, R]: (((T1, T2)) => R) => ((T1, T2) => R) = Function.untupled[T1, T2, R]

  implicit def t3u[T1, T2, T3, R]: (((T1, T2, T3)) => R) => ((T1, T2, T3) => R) = Function.untupled[T1, T2, T3, R]
}

object Memoize {
  final def apply[T, R, F](f: F)(implicit tupled: F => (T => R), untupled: (T => R) => F): F =
    untupled(new Memoized(tupled(f)))

  //I haven't yet made the implicit tupling magic for this yet
  def recursive[T, R](f: (T, T => R) => R) = {
    var yf: T => R = null
    yf = Memoize(f(_, yf))
    yf
  }
}

object ExampleMemoize extends App {

  val facMemoizable: (BigInt, BigInt => BigInt) => BigInt = (n: BigInt, f: BigInt => BigInt) => {
    if (n == 0) 1
    else n * f(n - 1)
  }

  val facMemoized = Memoize1.recursive(facMemoizable)

  override def main(args: Array[String]) {
    def myMethod(s: Int, i: Int, d: Double): Double = {
      println("myMethod ran")
      s + i + d
    }

    val myMethodMemoizedFunction: (Int, Int, Double) => Double = Memoize(myMethod _)

    def myMethodMemoized(s: Int, i: Int, d: Double): Double = myMethodMemoizedFunction(s, i, d)

    println("myMemoizedMethod(10, 5, 2.2) = " + myMethodMemoized(10, 5, 2.2))
    println("myMemoizedMethod(10, 5, 2.2) = " + myMethodMemoized(10, 5, 2.2))

    println("myMemoizedMethod(5, 5, 2.2) = " + myMethodMemoized(5, 5, 2.2))
    println("myMemoizedMethod(5, 5, 2.2) = " + myMethodMemoized(5, 5, 2.2))

    val myFunctionMemoized: (Int, Int, Double) => Double = Memoize((s: Int, i: Int, d: Double) => {
      println("myFunction ran")
      s * i + d + 3
    })

    println("myFunctionMemoized(10, 5, 2.2) = " + myFunctionMemoized(10, 5, 2.2))
    println("myFunctionMemoized(10, 5, 2.2) = " + myFunctionMemoized(10, 5, 2.2))

    println("myFunctionMemoized(7, 6, 3.2) = " + myFunctionMemoized(7, 6, 3.2))
    println("myFunctionMemoized(7, 6, 3.2) = " + myFunctionMemoized(7, 6, 3.2))
  }
}

运行ExampleMemoize时，您将获得：

myMethod ran
myMemoizedMethod(10, 5, 2.2) = 17.2
myMemoizedMethod(10, 5, 2.2) = 17.2
myMethod ran
myMemoizedMethod(5, 5, 2.2) = 12.2
myMemoizedMethod(5, 5, 2.2) = 12.2
myFunction ran
myFunctionMemoized(10, 5, 2.2) = 55.2
myFunctionMemoized(10, 5, 2.2) = 55.2
myFunction ran
myFunctionMemoized(7, 6, 3.2) = 48.2
myFunctionMemoized(7, 6, 3.2) = 48.2

Answer 3

我在想你可以做这样的事情，而不是使用DynamicProxy进行实际的实现。

def memo[T<:Product, R, F <: { def tupled: T => R }](f: F )(implicit m: Manifest[F]):F

因为becuase函数缺少一个常见的超类型，我们使用结构类型来查找任何可以被置换的东西（Function2-22，你仍然需要特殊情况下的Function1）。

我将Manifest放在那里，这样你就可以从函数特征F

构造DynamicProxy

Tupling也应该有助于memoization，例如你简单地将元组放在Map [T，R]

中

Answer 4

这是有效的，因为K可以是元组类型，因此memo(x,y,z) { function of x, y, z }有效：

import scala.collection.mutable

def memo[K,R](k: K)(f: => R)(implicit m: mutable.Map[K,R]) = m.getOrElseUpdate(k, f)

隐含的是我能够看到干净利落地来的唯一方式：

implicit val fibMap = new mutable.HashMap[Int,Int]
def fib(x: Int): Int = memo(x) {
    x match {
        case 1 => 1
        case 2 => 1
        case n => fib(n - 2) + fib(n - 1)
    }
}

感觉应该可以以某种方式结束自动HashMap[K,R]，这样您就不必明确地fibMap（并重新描述类型）。

scala：memoize函数，无论函数有多少参数？

4 个答案:

库

手册