假设我的目录结构如下：
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</div>其中各种python文件都包含称为root/
__init__.py
foo.py
dir1/
__init__.py
bar.py
dir2/
__init__.py
baz.py
的函数。
我想做的是写这样的东西：
hello_world当然，当我这样做时，会出现import root
root.dir1.foo.hello_world()
root.dir1.dir2.baz.hello_world()
错误。我认为，这是由于docs中的“ __init__.py陷阱”所致。如何操作初始化文件，使上面的代码运行？
AttributeError: module has no attribute参数，这对我的程序很重要。 我试图通过将__init__.py文件更改为namespace packages来解决此问题。不幸的是，这似乎不起作用。例如。添加：
__all__对各种init文件没有任何更改。同样，直接设置__path__ = __import__('pkgutil').extend_path(__path__, __name__)
属性也不能解决问题。例如。每个初始化文件中的设置__path__不能解决该错误。
这对我来说有点神秘，因为__path__ = ['<string name of child dir>']应该告诉python在哪里寻找子模块，而init文件应该按顺序运行，那么为什么不能弄清楚事物在哪里呢？
此问题似乎已在文档here
中进行了描述答案 0 :(得分：1)
一种解决方案是在每个初始化文件中定义from . import *属性之后放置__all__。例如。在dir1的init中：
__all__ = ['foo', 'dir2']
from . import *
这解决了用例，但似乎是错误的方式。打开以获得更好的答案。