有人能为我提供一个导入整个模块目录的好方法吗? 我有一个这样的结构:

/Foo
    bar.py
    spam.py
    eggs.py

我尝试通过添加__init__.py并从Foo import *将其转换为一个包,但它没有按我希望的方式工作。


当前回答

包含一个目录下的所有文件:

专为那些无法上手的新手准备的。

Make a folder /home/el/foo and make a file main.py under /home/el/foo Put this code in there: from hellokitty import * spam.spamfunc() ham.hamfunc() Make a directory /home/el/foo/hellokitty Make a file __init__.py under /home/el/foo/hellokitty and put this code in there: __all__ = ["spam", "ham"] Make two python files: spam.py and ham.py under /home/el/foo/hellokitty Define a function inside spam.py: def spamfunc(): print("Spammity spam") Define a function inside ham.py: def hamfunc(): print("Upgrade from baloney") Run it: el@apollo:/home/el/foo$ python main.py spammity spam Upgrade from baloney

其他回答

注意你的__init__.py定义了__all__。模块-包文档说

The __init__.py files are required to make Python treat the directories as containing packages; this is done to prevent directories with a common name, such as string, from unintentionally hiding valid modules that occur later on the module search path. In the simplest case, __init__.py can just be an empty file, but it can also execute initialization code for the package or set the __all__ variable, described later. ... The only solution is for the package author to provide an explicit index of the package. The import statement uses the following convention: if a package’s __init__.py code defines a list named __all__, it is taken to be the list of module names that should be imported when from package import * is encountered. It is up to the package author to keep this list up-to-date when a new version of the package is released. Package authors may also decide not to support it, if they don’t see a use for importing * from their package. For example, the file sounds/effects/__init__.py could contain the following code: __all__ = ["echo", "surround", "reverse"] This would mean that from sound.effects import * would import the three named submodules of the sound package.

2017年更新:你可能想用importlib代替。

通过添加__init__.py使Foo目录成为一个包。在__init__.py中添加:

import bar
import eggs
import spam

因为你希望它是动态的(这可能是也可能不是一个好主意),用list dir列出所有的py-files,然后像这样导入它们:

import os
for module in os.listdir(os.path.dirname(__file__)):
    if module == '__init__.py' or module[-3:] != '.py':
        continue
    __import__(module[:-3], locals(), globals())
del module

然后,从你的代码这样做:

import Foo

您现在可以使用

Foo.bar
Foo.eggs
Foo.spam

从Foo导入*不是一个好主意,有几个原因,包括名称冲突,使其难以分析代码。

我自己也厌倦了这个问题,所以我写了一个名为automodinit的包来解决它。你可以从http://pypi.python.org/pypi/automodinit/上得到它。

用法是这样的:

将automodinit包包含到setup.py依赖项中。 像这样替换所有__init__.py文件:

__all__ = ["I will get rewritten"]
# Don't modify the line above, or this line!
import automodinit
automodinit.automodinit(__name__, __file__, globals())
del automodinit
# Anything else you want can go after here, it won't get modified.

就是这样!从现在开始导入一个模块将设置__all__为 模块中的.py[co]文件列表,也将导入每个文件 就好像你输入了:

for x in __all__: import x

因此,“from M import *”的效果与“import M”完全匹配。

automodinit从ZIP档案内部运行,因此是ZIP安全的。

尼尔

我已经为此创建了一个模块,它不依赖于__init__.py(或任何其他辅助文件),并让我只输入以下两行:

import importdir
importdir.do("Foo", globals())

请随意重用或贡献:http://gitlab.com/aurelien-lourot/importdir

看看标准库中的pkgutil模块。只要目录中有__init__.py文件,它就会让你做你想做的事情。__init__.py文件可以为空。