How to load all modules in a folder?

List all python (.py) files in the current folder and put them as __all__ variable in __init__.py

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')]

Add the __all__ Variable to __init__.py containing:

__all__ = ["bar", "spam", "eggs"]

See also http://docs.python.org/tutorial/modules.html

Update in 2017: you probably want to use importlib instead.

Make the Foo directory a package by adding an __init__.py. In that __init__.py add:

import bar
import eggs
import spam

Since you want it dynamic (which may or may not be a good idea), list all py-files with list dir and import them with something like this:

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

Then, from your code do this:

import Foo

You can now access the modules with

Foo.bar
Foo.eggs
Foo.spam

etc. from Foo import * is not a good idea for several reasons, including name clashes and making it hard to analyze the code.

Expanding on Mihail's answer, I believe the non-hackish way (as in, not handling the file paths directly) is the following:

1. create an empty __init__.py file under Foo/
2. Execute

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')

You'll get:

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