Python: How can I choose which module to import when they are named the same

Question

Lets say I'm in a file called openid.py and I do :

from openid.consumer.discover import discover, DiscoveryFailure

I have the openid module on my pythonpath but the interpreter seems to be trying to use my openid.py file. How can I get the library version?

(Of course, something other than the obvious 'rename your file' answer would be nice).

Answer 1

Rename it. This is the idea behind name spaces. your openid could be a sub-module in your top module project. your email will clash with top module email in stdlib.

because your openid is not universal, it provides a special case for your project.

Answer 2

You can use relative or absolute imports (depending on the specifics of your situation), which are covered in PEP 328 most recently. Of course, seriously, you should not be creating naming conflicts like this and should rename your file.

Answer 3

Thats the reason absolute imports have been chosen as the new default behaviour. However, they are not yet the default in 2.6 (maybe in 2.7...). You can get their behaviour now by importing them from the future:

from __future__ import absolute_import

You can find out more about this in the PEP metnioned by Nick or (easier to understand, I think) in the document "What's New in Python 2.5".

Answer 4

You could try shuffling sys.path, to move the interesting directories to the front before doing the import.

Answer 5

I won't get into the polemics on renaming and instead focus on showing you how to do what you want (whether it's "good for you" or not;-). The solution is not difficult...

Just set __path__! A little demonstration:

$ mkdir /tmp/modules /tmp/packages
$ mkdir /tmp/packages/openid
$ echo 'print "Package!"' > /tmp/packages/openid/__init__.py
$ gvim /tmp/modules/openid.py
$ PYTHONPATH='/tmp/modules:/tmp/packages' python -c'import openid'
Module!
Package!

this shows a module openid managing to import a homonymous package even though the module's path comes earlier in sys.path, and sys.modules['openid'] is clearly already set at that time. And all the "secret" is in openid.py's simple code...:

print "Module!"
__path__ = ['/tmp/packages']
import openid

without the __path__ assignment, of course, it would only emit Module!.

Also works for importing submodules within the package, of course. Do:

$ echo 'print "Submod!"' > /tmp/packages/openid/submod.py

and change openid.py's last line to

from openid import submod

and you'll see:

$ PYTHONPATH='/tmp/modules:/tmp/packages' python -c'import openid'
Module!
Package!
Submod!
$