How python deals with module and package having the same name?

问题

Suppose I have a module foo.py and a package foo/. If I call

import foo

which one will be loaded? How can I specify I wand to load the module, or the package?

回答1:

I believe the package will always get loaded. You can't work around this, as far as I know. So change either the package or the module name. Docs: http://docs.python.org/tutorial/modules.html#the-module-search-path

回答2:

Actually, it is possible (this code is not well tested, but seems to work).

File foo.py

print "foo module loaded"

File foo/__init__.py

print "foo package loaded"

File test1.py

import foo

File test2.py

import os, imp

def import_module(dir, name):
    """ load a module (not a package) with a given name 
        from the specified directory 
    """
    for description in imp.get_suffixes():
        (suffix, mode, type) = description
        if not suffix.startswith('.py'): continue
        abs_path = os.path.join(dir, name + suffix)
        if not os.path.exists(abs_path): continue
        fh = open(abs_path)
        return imp.load_module(name, fh, abs_path, (description))

import_module('.', 'foo')

Running

$ python test1.py 
foo package loaded

$ python test2.py 
foo module loaded

回答3:

Maybe you want to move your classes from foo.py module to __init__.py.

This way you'll be able to import them from the package as well as importing optional subpackages:

File foo/__init__.py:

class Bar(object):
...

File mymodule.py:

from foo import Bar
from foo.subfoo import ...

Nonetheless I would like somebody to double-check this approach and let me know if it's correct or the __init__ module shouldn't be used like that.

来源：https://stackoverflow.com/questions/6393861/how-python-deals-with-module-and-package-having-the-same-name

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