我想在Python 3.5中使用timeit来测量两个函数.第一个依赖于导入数学,第二个依赖于数学导入log2.我虽然可以通过在调用timeit.repeat时将适当的import语句作为设置字符串传递来处理此问题.但是我得到NameError：未定义名称“ math”.
我不想将函数作为它们的名称字符串传递.当函数不依赖于导入时,此代码在这里有效,但我需要用math.log2调用

期待您的回答.

def spam_log_v1():
    for _ in range(1000):
        math.log2(2)


def spam_log_v2():
    for _ in range(1000):
        log2(2)


if __name__ == '__main__':

    import timeit

    repeat = 3
    number = 1000
    unit = "usec"
    unittosec = {"usec": 1e6, "msec": 1000, "sec": 1}
    tests = [(spam_log_v1, 'import math'),
             (spam_log_v2, 'from math import log2')]

    for fct, setup in tests:
        res = timeit.repeat(fct, setup=setup, repeat=repeat, number=number)
        print("%s: %d loops, best of %d: %.3g %s per loop" %
              (fct.__name__, number, repeat,
               min(res) / number * unittosec[unit], unit))

解决方法:

由timeit.repeat运行的The inner function看起来像这样：

def inner(_it, _timer{init}):
    {setup}
    _t0 = _timer()
    for _i in _it:
        {stmt}
    _t1 = _timer()
    return _t1 - _t0

其中{setup}替换为设置字符串,{stmt}替换为语句字符串.请注意,安装程序是从内部内部运行的.因此,如果在安装程序中放置导入语句,则模块名称将成为inner的局部变量.稍后,当从stmt调用spam_log_v1时,spam_log_v1无法使用该模块名称,因为该模块名称不在全局范围内.

解决方法是仅在脚本的全局级别导入数学和/或从数学导入log2.

import math
from math import log2

def spam_log_v1():
    for _ in range(1000):
        math.log2(2)

def spam_log_v2():
    for _ in range(1000):
        log2(2)

if __name__ == '__main__':
    import timeit
    repeat = 3
    number = 1000
    unit = "usec"
    unittosec = {"usec": 1e6, "msec": 1000, "sec": 1}
    tests = [
        (spam_log_v1, ''),
        (spam_log_v2, '')]

    for fct, setup in tests:
        res = timeit.repeat(fct, setup=setup, repeat=repeat, number=number)
        print("%s: %d loops, best of %d: %.3g %s per loop" %
              (fct.__name__, number, repeat,
               min(res) / number * unittosec[unit], unit))