导入模块，因为它是更高的级别 [英] Importing module as it was a level higher
问题描述
我具有通过setuptools获得的目录结构:
I have this directory structure obtained with setuptools:
root/
A/
__init__.py
1.py
2.py
B/
__init__.py
3.py
__init__.py
包装部分如下:
packages=['root', 'root.A', 'root.B', ],
导入我使用的内部.py文件的内容:
to import the content of the inner .py files i use:
from root.A import 1
from root.B.3 import a_func
现在，如果我想直接从根模块导入a_func，我会将以下行添加到根目录的init文件中
now, if i wanted to import a_func directly from the root module, i would add the following line to the init file in the root directory
# to allow root.a_func access
from .B.3 import a_func
但是有没有办法导入整个模块而不是单个特定项目(同时保留相同的目录结构)?
but is there a way to import an entire module instead of a single specific item (while preserving the same directory structure) ?
from root import 1
from root.3 import a_func
换句话说，是否可以在导入期间隐藏中间级别模块的访问权限?
in other words, is it possible to hide the access of an intermediate level module during import?
我已经尝试将以下行添加到根目录中的init文件中，但是它不起作用.
I have already tried to add the following lines to the init file in the root dir, but it does not work.
from .A import *
from .B import *
有什么建议吗?
推荐答案
修改 __ path __ 变量可能会有所帮助.
Modifying the __path__ variable might help.
# root/__init__.py
import os
__path__.append(os.path.join(os.path.dirname(__file__), 'A'))
# root/A/__init__.py
# empty
# root/A/one.py
def first_func():
print("first_func belongs to:", __name__, "in:", __file__)
这允许执行以下操作:
This allows to do the following:
$ python3 -c "from root import one; one.first_func()"
first_func belongs to: root.one in: /home/sinoroc/workspace/root/.venv/lib/python3.6/site-packages/root-0.0.0.dev0-py3.6.egg/root/A/one.py
以及:
$ python3 -c "from root.A import one; one.first_func()"
first_func belongs to: root.A.one in: /home/sinoroc/workspace/root/.venv/lib/python3.6/site-packages/root-0.0.0.dev0-py3.6.egg/root/A/one.py
请参阅:
See:
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