%timeit 和变量的重新赋值

Question

令人惊讶的 ipython 魔法%timeit错误：

In[1]: a = 2

In[2]: %timeit a = 2 * a

Traceback (most recent call last):
  File "...\site-packages\IPython\core\interactiveshell.py", line 3326, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)
  File "<ipython-input-97-6f70919654d1>", line 1, in <module>
    get_ipython().run_line_magic('timeit', 'a = 2 * a')
  File "...\site-packages\IPython\core\interactiveshell.py", line 2314, in run_line_magic
    result = fn(*args, **kwargs)
  File "<...\site-packages\decorator.py:decorator-gen-61>", line 2, in timeit
  File "...\site-packages\IPython\core\magic.py", line 187, in <lambda>
    call = lambda f, *a, **k: f(*a, **k)
  File "...\site-packages\IPython\core\magics\execution.py", line 1158, in timeit
    time_number = timer.timeit(number)
  File "...\site-packages\IPython\core\magics\execution.py", line 169, in timeit
    timing = self.inner(it, self.timer)
  File "<magic-timeit>", line 1, in inner
UnboundLocalError: local variable 'a' referenced before assignment

所以%timeit不喜欢自我重新分配。为什么？无论如何要克服这个？

Answer 1

与底层timeit模块一样，定时语句被集成到执行计时的生成函数中。对 的赋值a导致函数有一个a局部变量，隐藏了全局变量。这和你做过的问题是一样的

a = 2

def f():
    a = 2 * a

f()

尽管生成的函数有更多的代码。