为什么嵌套函数比unnested版本慢得多？

Question

我正试图在斐波那契中包含备忘录,因为我正试图挑选这个函数.但是,在我的测试中,似乎嵌套函数明显变慢了,但是只有当我使用memoized函数时才会看到与其他版本的fibonacci相关的函数.

我所有的测试:https://gist.github.com/dasickis/4733353

#!/usr/bin/env python

memo = {0: 0, 1: 1}

# Contract: [int > 0] -> [int > 0]
def fibonacci(n):
    """ Return the `x`th number in the fibonacci series. """
    if not n in memo:
        memo[n] = fibonacci(n - 1) + fibonacci(n - 2)
    return memo[n]

#--------------------------#

# Contract: [int > 0] -> [int > 0]
def fibonacci_nested(n):
    memo = {0: 0, 1: 1}

    def fib(n):
        """ Return the `x`th number in the fibonacci series. """
        if not n in memo:
            memo[n] = fib(n - 1) + fib(n - 2)
        return memo[n]

    return fib(n)

#--------------------------#

import timeit
stmt = "assert fib(20) == 6765"

print "fibonacci"
print timeit.timeit(stmt, setup="from __main__ import fibonacci as fib")
print 

print "fibonacci_nested"
print timeit.timeit(stmt, setup="from __main__ import fibonacci_nested as fib")

输出:

fibonacci
0.263559103012

fibonacci_nested
11.4014730453

Answer 1

您不会memo在运行之间清除字典,而不会嵌套不公平的优势.第一次timeit运行fib它将填充memodict然后后续运行重新使用它.

嵌套函数memo每次都设置一个新的空.