列出当前文件夹中的所有python (.py)文件，并将它们作为__init__.py中的__all__变量

from os.path import dirname, basename, isfile, join
import glob
modules = glob.glob(join(dirname(__file__), "*.py"))
__all__ = [ basename(f)[:-3] for f in modules if isfile(f) and not f.endswith('__init__.py')]

列出当前文件夹中的所有python (.py)文件，并将它们作为__init__.py中的__all__变量

from os.path import dirname, basename, isfile, join
import glob
modules = glob.glob(join(dirname(__file__), "*.py"))
__all__ = [ basename(f)[:-3] for f in modules if isfile(f) and not f.endswith('__init__.py')]

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

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。

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.

何时从。Import *不够好，这是对ted的回答的改进。具体来说，这种方法不需要使用__all__。

"""Import all modules that exist in the current directory."""
# Ref https://stackoverflow.com/a/60861023/
from importlib import import_module
from pathlib import Path

for f in Path(__file__).parent.glob("*.py"):
    module_name = f.stem
    if (not module_name.startswith("_")) and (module_name not in globals()):
        import_module(f".{module_name}", __package__)
    del f, module_name
del import_module, Path

请注意，globals()中没有module_name是为了避免在已经导入模块时重新导入模块，因为这可能存在循环导入的风险。