％timeit和变量的重新分配 [英] %timeit and re-assignment of variable

问题描述

令人惊讶的ipython magic %timeit错误:

Surprising ipython magic %timeit error:

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

因此，%timeit不喜欢自我重新分配.为什么?无论如何要克服这个问题?

So %timeit doesn't like self re-assignment. Why? Anyway to overcome this?

推荐答案

与基础的timeit模块一样，定时语句被集成到执行定时的生成函数中.分配给a会使函数具有a局部变量，从而隐藏全局变量.就像你做完一样的问题

As with the underlying timeit module, the timed statement is integrated into a generated function that performs the timing. The assignment to a causes the function to have an a local variable, hiding the global. It's the same issue as if you had done

a = 2

def f():
    a = 2 * a

f()

尽管生成的函数具有更多的代码.

although the generated function has more code than that.

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