为什么我要在python包的init中使用uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu？

我已经读过了，我知道它只影响来自。。。import*语句，但我无法找出真正的用例。既然我可以简单地避免在

\uuuuu init\uuuu

名称空间中导入这些名称，为什么我要在

\uuuuuuu all\uuuuuu

中重复导出的名称

例如：

mypackage/\uuuuu init\uuuuuu.py

from a import A

A = "A"
A1 = "A1"

B = "B"    

mypackage/a.py

from a import A

A = "A"
A1 = "A1"

B = "B"    

mypackage/b.py

from a import A

A = "A"
A1 = "A1"

B = "B"    

然后在python中：

>>> from mypackage import *
>>> A
'A'
>>> 
>>> A1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'A1' is not defined
>>> b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'b' is not defined

来自mypackage导入的

>>*
>>>A
“A”
>>> 
>>>A1
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“A1”
>>>b
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“b”

如您所见，

A

位于命名空间中，但

A1

和

b

不在命名空间中。为什么我必须定义

\uuuuuuuuuuuuuuuuuuu=[“A”]

？

在包的

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuupy.py中定义
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

from package import *

这记录在

注意：如果您没有在包中定义
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

如果未定义
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
，则sound.effects import*中的
不会将包sound.effects中的所有子模块导入到
当前名称空间；它只确保包sound.effects
已导入（可能在中运行任何初始化代码）
然后导入包中定义的任何名称。这包括明确定义的任何名称（和子模块）
通过
\uuuu init\uuuuu.py
加载）。它还包括包的任何子模块
由以前的导入语句显式加载的。考虑
此代码：

对此的“天真”解释可以是：

如果您不定义
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu；来自包导入的
*
将导入来自该包的所有内容以及在中导入的任何内容，该包的
\uuuuu init\uuuuuuuuuuuuy

看看我的例子：我没有使用
\uuuu all\uuu
，但我仍然可以从mypackage import*
在
中显示或隐藏名称。那么为什么要使用它呢？@Zac您可以限制
*
附带的内容。在您的示例中，它将带来所有内容。@PeterWood:不是所有内容，而是只有在
\uuuuu init\uuuuuuuuu
中导入的内容（有问题的更新示例）“如果您不定义
\uuuuuuu all\uuuuuuuu
，来自包导入的
将带来该包和所有子包中的所有内容。”并非所有内容，而是仅导入在
\uuuu init\uuuu
@Zac中的内容。您可能需要初始化包，并需要导入标识符才能完成此操作。但您可能不想将其与软件包一起导出。因此，您可以使用
\uuuuuu all\uuuuu
来控制它。