问题描述

这是我目前拥有的代码:

Here is the code as I currently have it:

def F(n):
    t=time.time()
    if n==0:
        return (0)
    elif n==1:
        return (1)
    else:
        return (F(n-1)+F(n-2))
    t1==time.time()
    return t
    F_time==t1-t
    print ('It took',F_time,'seconds to sort',n,'values using recursion')

我正在尝试打印n个斐波那契数列并取前后的时间，但是由于return语句，它不会接受return之后的变量，也不接受下面的print语句.

I'm trying to print n numbers of the Fibonacci sequence and take the time before and after, but because of the return statements, it won't accept the variable after the return, nor the following print statement.

推荐答案

为了真正执行函数内部 return 语句之后的事情，请使用 try-finally 像这样:

In order to really execute things after the return statement(s) inside a function use try-finally like this:

# platform independent high-resolution clock
from timeit import default_timer as timer  

def F(n):
    t = timer()
    try:
        if n == 0:
            return 0
        elif n == 1:
            return 1
        else:
            return (F(n - 1) + F(n - 2))
    finally:
        t1 = timer()
        F_time = t1 - t
        print(u"F(%s) took %.2fµs seconds" % (n, F_time * 1e6))

if __name__ == '__main__':
    print("RESULT:", F(4))

输出:

F(1) took 0.92µs seconds
F(0) took 0.84µs seconds
F(2) took 429.78µs seconds
F(1) took 0.84µs seconds
F(3) took 520.12µs seconds
F(1) took 0.67µs seconds
F(0) took 0.75µs seconds
F(2) took 90.85µs seconds
F(4) took 700.82µs seconds
RESULT: 3

注意:在这个例子中，n > 1 的时间当然包括在嵌套递归调用中打印到标准输出的时间，然后在那个简单的例子中支配时间.

Note: in that example here the timings for n > 1 of course include time for printing to stdout in nested recursive calls, which then dominate timing in that trivial example.

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