问题描述
我正在尝试组织一些模块供我自己使用.我有这样的事情:
lib/__init__.py设置.py富/__init__.py一些对象.py酒吧/__init__.py别的东西.py
在 lib/__init__.py 中,我想定义一些在导入 lib 时要使用的类.但是,如果不将类分成文件并将它们导入__init__.py,我似乎无法弄清楚.
而不是说:
库/__init__.py设置.py助手类.py富/__init__.py一些对象.py酒吧/__init__.py别的东西.py从 lib.settings 导入值从 lib.helperclass 导入助手
我想要这样的东西:
库/__init__.py #Helper 在这个文件中定义设置.py富/__init__.py一些对象.py酒吧/__init__.py别的东西.py从 lib.settings 导入值从库导入助手
有可能吗,还是我必须将类分离到另一个文件中?
编辑
好的,如果我从另一个脚本导入 lib,我可以访问 Helper 类.如何从 settings.py 访问 Helper 类?
此处示例描述了包内引用.我引用子模块经常需要相互引用".就我而言,lib.settings.py 需要 Helper,而 lib.foo.someobject 需要访问 Helper,那么我应该在哪里定义 Helper 类?
'lib/ 的父目录必须在 sys.path 中.
您的lib/__init__.py"可能如下所示:
来自 .导入设置 # 或者只是在旧 Python 版本上导入设置"类助手(对象):经过
那么下面的例子应该可以工作:
从 lib.settings 导入值从库导入助手
对问题的编辑版本的回答:
__init__.py 定义了你的包从外部看起来如何.如果您需要在 settings.py 中使用 Helper,则在不同的文件中定义 Helper,例如 'lib/helper.py'.
<上一页>.|`-- import_submodule.py`--库|-- __init__.py|-- 富||-- __init__.py|`--someobject.py|-- 助手.py`-- 设置.py2个目录,6个文件命令:
$ python import_submodule.py
输出:
设置帮手lib.settings 中的助手某物lib.foo.someobject 中的助手# ./import_submodule.py导入 fnmatch,操作系统从 lib.settings 导入值从库导入助手打印对于 os.walk('.') 中的根目录、目录和文件:对于 fnmatch.filter(files, '*.py') 中的 f:print "# %s/%s" % (os.path.basename(root), f)打印打开(os.path.join(root, f)).read()打印# lib/helper.py打印帮手"类助手(对象):def __init__(self, module_name):print "Helper in", module_name# lib/settings.py打印设置"导入助手类值(对象):经过helper.Helper(__name__)# lib/__init__.py#from __future__ 导入 absolute_import导入设置、foo.someobject、助手助手 = helper.Helper# foo/someobject.py打印某物"从 .. 导入助手helper.Helper(__name__)# foo/__init__.py导入某个对象
I am trying to organize some modules for my own use. I have something like this:
lib/ __init__.py settings.py foo/ __init__.py someobject.py bar/ __init__.py somethingelse.py
In lib/__init__.py, I want to define some classes to be used if I import lib. However, I can't seem to figure it out without separating the classes into files, and import them in__init__.py.
Rather than say:
lib/ __init__.py settings.py helperclass.py foo/ __init__.py someobject.py bar/ __init__.py somethingelse.py from lib.settings import Values from lib.helperclass import Helper
I want something like this:
lib/ __init__.py #Helper defined in this file settings.py foo/ __init__.py someobject.py bar/ __init__.py somethingelse.py from lib.settings import Values from lib import Helper
Is it possible, or do I have to separate the class into another file?
EDIT
OK, if I import lib from another script, I can access the Helper class. How can I access the Helper class from settings.py?
The example here describes Intra-Package References. I quote "submodules often need to refer to each other". In my case, the lib.settings.py needs the Helper and lib.foo.someobject need access to Helper, so where should I define the Helper class?
'lib/'s parent directory must be in sys.path.
Your 'lib/__init__.py' might look like this:
from . import settings # or just 'import settings' on old Python versions class Helper(object): pass
Then the following example should work:
from lib.settings import Values from lib import Helper
Answer to the edited version of the question:
__init__.py defines how your package looks from outside. If you need to use Helper in settings.py then define Helper in a different file e.g., 'lib/helper.py'.
. | `-- import_submodule.py `-- lib |-- __init__.py |-- foo | |-- __init__.py | `-- someobject.py |-- helper.py `-- settings.py 2 directories, 6 files
The command:
$ python import_submodule.py
Output:
settings helper Helper in lib.settings someobject Helper in lib.foo.someobject # ./import_submodule.py import fnmatch, os from lib.settings import Values from lib import Helper print for root, dirs, files in os.walk('.'): for f in fnmatch.filter(files, '*.py'): print "# %s/%s" % (os.path.basename(root), f) print open(os.path.join(root, f)).read() print # lib/helper.py print 'helper' class Helper(object): def __init__(self, module_name): print "Helper in", module_name # lib/settings.py print "settings" import helper class Values(object): pass helper.Helper(__name__) # lib/__init__.py #from __future__ import absolute_import import settings, foo.someobject, helper Helper = helper.Helper # foo/someobject.py print "someobject" from .. import helper helper.Helper(__name__) # foo/__init__.py import someobject