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')]

我也遇到过这个问题，这是我的解决方案:

import os

def loadImports(path):
    files = os.listdir(path)
    imps = []

    for i in range(len(files)):
        name = files[i].split('.')
        if len(name) > 1:
            if name[1] == 'py' and name[0] != '__init__':
               name = name[0]
               imps.append(name)

    file = open(path+'__init__.py','w')

    toWrite = '__all__ = '+str(imps)

    file.write(toWrite)
    file.close()

这个函数创建一个名为__init__.py的文件(在提供的文件夹中)，其中包含一个__all__变量，该变量保存文件夹中的每个模块。

例如，我有一个名为Test的文件夹 它包含:

Foo.py
Bar.py

所以在脚本中，我想把模块导入，我会写:

loadImports('Test/')
from Test import *

这将从Test中导入所有内容，Test中的__init__.py文件现在将包含:

__all__ = ['Foo','Bar']

扩展Mihail的回答，我认为非黑客的方式(即不直接处理文件路径)如下:

在Foo/下创建一个空的__init__.py文件 执行

import pkgutil
import sys


def load_all_modules_from_dir(dirname):
    for importer, package_name, _ in pkgutil.iter_modules([dirname]):
        full_package_name = '%s.%s' % (dirname, package_name)
        if full_package_name not in sys.modules:
            module = importer.find_module(package_name
                        ).load_module(full_package_name)
            print module


load_all_modules_from_dir('Foo')

你会得到:

<module 'Foo.bar' from '/home/.../Foo/bar.pyc'>
<module 'Foo.spam' from '/home/.../Foo/spam.pyc'>

Anurag Uniyal给出了改进建议!

#!/usr/bin/python
# -*- encoding: utf-8 -*-

import os
import glob

all_list = list()
for f in glob.glob(os.path.dirname(__file__)+"/*.py"):
    if os.path.isfile(f) and not os.path.basename(f).startswith('_'):
        all_list.append(os.path.basename(f)[:-3])

__all__ = all_list  

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.

Anurag的例子有几个更正:

import os, glob

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