记忆处理程序 [英] Memoization Handler

问题描述

创建像下面这样的类可以为您处理记忆化过程是良好实践"吗?记忆化的好处是如此之大(在某些情况下，像这样的情况下，它从501003下降到1507函数调用，从我的计算机上的1.409下降到0.006秒的CPU时间)，看来这样的类很有用.

Is it "good practice" to create a class like the one below that can handle the memoization process for you? The benefits of memoization are so great (in some cases, like this one, where it drops from 501003 to 1507 function calls and from 1.409 to 0.006 seconds of CPU time on my computer) that it seems a class like this would be useful.

但是，我只阅读了关于eval()用法的否定评论. 鉴于这种方法提供的灵活性，这种用法是否可以原谅?

However, I've read only negative comments on the usage of eval(). Is this usage of it excusable, given the flexibility this approach offers?

这可以自动保存任何返回值，但会损失副作用.谢谢.

This can save any returned value automatically at the cost of losing side effects. Thanks.

import cProfile

class Memoizer(object):
    """A handler for saving function results."""
    def __init__(self):
        self.memos = dict()
    def memo(self, string):
        if string in self.memos:
            return self.memos[string]
        else:
            self.memos[string] = eval(string)
            self.memo(string)

def factorial(n):
    assert type(n) == int
    if n == 1:
        return 1
    else:
        return n * factorial(n-1) 

# find the factorial of num
num = 500
# this many times
times = 1000

def factorialTwice():
    factorial(num)
    for x in xrange(0, times):
        factorial(num)
    return factorial(num)

def memoizedFactorial():
    handler = Memoizer()
    for x in xrange(0, times):
        handler.memo("factorial(%d)" % num)
    return handler.memo("factorial(%d)" % num)


cProfile.run('factorialTwice()')

cProfile.run('memoizedFactorial()')

推荐答案

您可以记住，而不必求助于eval.

You can memoize without having to resort to eval.

一个(非常基本的)记忆符:

A (very basic) memoizer:

def memoized(f):
    cache={}
    def ret(*args):
        if args in cache:
            return cache[args]
        else:
            answer=f(*args)
            cache[args]=answer
            return answer
    return ret

@memoized
def fibonacci(n):
    if n==0 or n==1:
        return 1
    else:
        return fibonacci(n-1)+fibonacci(n-2)

print fibonacci(100)

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