Python 自动导出所有函数（与手动指定所有函数相比）

我有一个

helpers.py

文件，它定义了大约30个要导出的helper函数，如下所示：

from helpers import *

---

为了能够做到这一点，我将所有30个函数都添加到

\uuuuuuuuuuuuuu

变量中。我可以自动导出所有函数，而不必指定每个函数吗？

是的，通过简单地不指定

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

如果您没有定义
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

如果您有一些想要保持私有的函数，那么可以在它们的名称前面加下划线。根据我的测试，这将阻止通过
import*

例如，在helper.py中：

def _HiddenFunc():
    return "Something"

def AnActualFunc():
    return "Hello"

然后：

来自帮助器导入的
>>*
>>>AnActualFunc（）
“你好”
>>>_HiddenFunc（）
回溯（最近一次呼叫最后一次）：
文件“”，第1行，在
名称错误：未定义名称“\u HiddenFunc”
我想在helpers.py中有一些未导出的方法，否则
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu>将没有用……如果不定义
\uuuuuuuuuuuu。事实上，如果需要，您可以在模块中以更精细的方式确定
\uuuuu all\uuuu
的内容，也就是说，您不局限于一条语句。
def _HiddenFunc():
    return "Something"

def AnActualFunc():
    return "Hello"

>>> from helper import *
>>> AnActualFunc()
'Hello'
>>> _HiddenFunc()
Traceback (most recent call last):
   File "<stdin>", line 1, in <module>
NameError: name '_HiddenFunc' is not defined