每次调用函数时,如果对于给定的参数值它的结果尚未被记忆化,则我希望将结果放入内存表中。一个列用于存储结果,其他列用于存储参数值。
如何最好地实现这个功能?参数具有多种类型,包括一些枚举。
在C#中,我通常会使用 DataTable。在Scala中是否有相当的东西?
如何最好地实现这个功能?参数具有多种类型,包括一些枚举。
在C#中,我通常会使用 DataTable。在Scala中是否有相当的东西?
mutable.Map[TupleN[A1, A2, ..., AN], R]
,以下定义(基于michid's blog中的记忆化代码)允许你轻松地对具有多个参数的函数进行记忆化。例如:import Memoize._
def reallySlowFn(i: Int, s: String): Int = {
Thread.sleep(3000)
i + s.length
}
val memoizedSlowFn = memoize(reallySlowFn _)
memoizedSlowFn(1, "abc") // returns 4 after about 3 seconds
memoizedSlowFn(1, "abc") // returns 4 almost instantly
定义:
/**
* 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) {
import scala.collection.mutable
// map that stores (argument, result) pairs
private[this] val vals = mutable.Map.empty[T, R]
// Given an argument x,
// If vals contains x return vals(x).
// Otherwise, update vals so that vals(x) == f(x) and return f(x).
def apply(x: T): R = vals getOrElseUpdate (x, f(x))
}
object Memoize {
/**
* Memoize a unary (single-argument) function.
*
* @param f the unary function to memoize
*/
def memoize[T, R](f: T => R): (T => R) = new Memoize1(f)
/**
* Memoize a binary (two-argument) function.
*
* @param f the binary function to memoize
*
* This works by turning a function that takes two arguments of type
* T1 and T2 into a function that takes a single argument of type
* (T1, T2), memoizing that "tupled" function, then "untupling" the
* memoized function.
*/
def memoize[T1, T2, R](f: (T1, T2) => R): ((T1, T2) => R) =
Function.untupled(memoize(f.tupled))
/**
* Memoize a ternary (three-argument) function.
*
* @param f the ternary function to memoize
*/
def memoize[T1, T2, T3, R](f: (T1, T2, T3) => R): ((T1, T2, T3) => R) =
Function.untupled(memoize(f.tupled))
// ... more memoize methods for higher-arity functions ...
/**
* Fixed-point combinator (for memoizing recursive functions).
*/
def Y[T, R](f: (T => R) => T => R): (T => R) = {
lazy val yf: (T => R) = memoize(f(yf)(_))
yf
}
}
Memoize.Y
)使得可以对递归函数进行记忆化:val fib: BigInt => BigInt = {
def fibRec(f: BigInt => BigInt)(n: BigInt): BigInt = {
if (n == 0) 1
else if (n == 1) 1
else (f(n-1) + f(n-2))
}
Memoize.Y(fibRec)
}
[1] WeakHashMap不适合用作缓存。请参见http://www.codeinstructions.com/2008/09/weakhashmap-is-not-cache-understanding.html和这个相关问题。
anovstrup提出的使用可变Map的版本基本上与C#中的相同,因此易于使用。
但是,如果您希望还可以使用更多的函数式风格。它使用不可变映射(Maps),作为一种累加器。在示例中,将元组(而不是Int)用作键与可变情况完全相同。
def fib(n:Int) = fibM(n, Map(0->1, 1->1))._1
def fibM(n:Int, m:Map[Int,Int]):(Int,Map[Int,Int]) = m.get(n) match {
case Some(f) => (f, m)
case None => val (f_1,m1) = fibM(n-1,m)
val (f_2,m2) = fibM(n-2,m1)
val f = f_1+f_2
(f, m2 + (n -> f))
}
当然,这有点复杂,但这是一个很有用的技巧(请注意,上面的代码旨在清晰易懂,而不是速度快)。
def MyFunction(dt : DateTime, param : Int) : Double
{
val argsTuple = (dt, param)
if(Memo.contains(argsTuple)) Memo(argsTuple) else Memoize(dt, param, MyRawFunction(dt, param))
}
def MyRawFunction(dt : DateTime, param : Int) : Double
{
1.0 // 这里进行了繁重的计算/查询
}
def Memoize(dt : DateTime, param : Int, result : Double) : Double
{
Memo += (dt, param) -> result
result
}
val Memo = new scala.collection.mutable.HashMap[(DateTime, Int), Double]
运行得非常完美。如果我错过了什么,请指出。
以下线程安全代码创建了一个记忆化的 斐波那契
函数,初始化了几个线程(从 'a' 到 'd' 命名),并对其进行调用。多次尝试代码(在REPL中),可以很容易地看到 f(2) set
被打印了不止一次。这意味着线程A已经启动了 f(2)
的计算,但线程B完全不知道,并开始自己的复制计算。这种无知在缓存构建阶段是如此普遍,因为所有线程都看不到已建立的子解决方案,都会进入 else
子句。
object ScalaMemoizationMultithread {
// do not use case class as there is a mutable member here
class Memo[-T, +R](f: T => R) extends (T => R) {
// don't even know what would happen if immutable.Map used in a multithreading context
private[this] val cache = new java.util.concurrent.ConcurrentHashMap[T, R]
def apply(x: T): R =
// no synchronized needed as there is no removal during memoization
if (cache containsKey x) {
Console.println(Thread.currentThread().getName() + ": f(" + x + ") get")
cache.get(x)
} else {
val res = f(x)
Console.println(Thread.currentThread().getName() + ": f(" + x + ") set")
cache.putIfAbsent(x, res) // atomic
res
}
}
object Memo {
def apply[T, R](f: T => R): T => R = new Memo(f)
def Y[T, R](F: (T => R) => T => R): T => R = {
lazy val yf: T => R = Memo(F(yf)(_))
yf
}
}
val fibonacci: Int => BigInt = {
def fiboF(f: Int => BigInt)(n: Int): BigInt = {
if (n <= 0) 1
else if (n == 1) 1
else f(n - 1) + f(n - 2)
}
Memo.Y(fiboF)
}
def main(args: Array[String]) = {
('a' to 'd').foreach(ch =>
new Thread(new Runnable() {
def run() {
import scala.util.Random
val rand = new Random
(1 to 2).foreach(_ => {
Thread.currentThread().setName("Thread " + ch)
fibonacci(5)
})
}
}).start)
}
}
foldLeft
(s)。 def fibo(n: Int) = (1 to n).foldLeft((0, 1)) {
(acc, i) => (acc._2, acc._1 + acc._2)
}._1
最长递增子序列的示例
def longestIncrSubseq[T](xs: List[T])(implicit ord: Ordering[T]) = {
xs.foldLeft(List[(Int, List[T])]()) {
(memo, x) =>
if (memo.isEmpty) List((1, List(x)))
else {
val resultIfEndsAtCurr = (memo, xs).zipped map {
(tp, y) =>
val len = tp._1
val seq = tp._2
if (ord.lteq(y, x)) { // current is greater than the previous end
(len + 1, x :: seq) // reversely recorded to avoid O(n)
} else {
(1, List(x)) // start over
}
}
memo :+ resultIfEndsAtCurr.maxBy(_._1)
}
}.maxBy(_._1)._2.reverse
}