从子目录中的不同文件导入类

尽管此处使用的名称与问题的目录结构中显示的名称不同，但是您可以使用我对名为Namespacing和classes的问题的答案。在

__init__.py

usepackage.py

package

subdir

Class1

myclass01

from package import *print Class1print Class2print Class3

修订（更新）：

糟糕，抱歉，我的其他答案中的代码并未完全满足您的要求-
它只会自动导入任何程序包子模块的名称。为了使其也从每个子模块导入命名属性，需要多几行代码。这是软件包

__init__.py

def _import_package_files():    """ Dynamically import all the public attributes of the python modules in this        file's directory (the package directory) and return a list of their names.    """    import os    exports = []    globals_, locals_ = globals(), locals()    package_path = os.path.dirname(__file__)    package_name = os.path.basename(package_path)    for filename in os.listdir(package_path):        modulename, ext = os.path.splitext(filename)        if modulename[0] != '_' and ext in ('.py', '.pyw'): subpackage = '{}.{}'.format(package_name, modulename) # pkg relative module = __import__(subpackage, globals_, locals_, [modulename]) modict = module.__dict__ names = (modict['__all__'] if '__all__' in modict else          [name for name in modict if name[0] != '_'])  # all public exports.extend(names) globals_.update((name, modict[name]) for name in names)    return exportsif __name__ != '__main__':    __all__ = ['__all__'] + _import_package_files()  # '__all__' in __all__

或者，您可以将以上内容放入包目录中单独的.py模块文件中（例如），

_import_package_files.py

__init__.py

if __name__ != '__main__':    from ._import_package_files import *  # defines __all__    __all__.remove('__all__')  # prevent export (optional)

无论您以什么名字命名文件，它都应该以

_

import