changing namespace of current python script (class with module name)

Question

i have a 2.6 python script and library in the following directory structure:

+ bin
\- foo.py
+ lib
\+ foo
 \- bar.py

i would like users to run bin/foo.py to instantiate the classes within lib/foo.py. to achieve this, in my bin/foo.py script i have the following code:

from __future__ import absolute_import
import foo
klass = foo.bar.Klass()

however, this results in:

AttributeError: 'module' object has no attribute 'bar'

ie it thinks that foo is itself rather than the library foo - renaming bin/foo.py to bin/foo-script.py works as expected.

is there a way i can keep the bin/foo.py script and import lib/foo.py?

Have you tried import foo as bar?

Steve Peak
– Steve Peak

2013-01-24 19:48:58 +00:00
Commented Jan 24, 2013 at 19:48 — Steve Peak
– Steve Peak, Commented Jan 24, 2013 at 19:48

Thomas · Accepted Answer · 2013-01-24 19:51:37Z

2

The current directory is on the path by default, so you need to remove that before you import the other foo module:

import sys
sys.path = [dir for dir in sys.path if dir != '']

Alternatively, prepend the lib directory so that it takes precedence:

import sys
sys.path = ['../lib'] + sys.path

answered Jan 24, 2013 at 19:51

Thomas

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Comments

Manmohan Singh · Accepted Answer · 2013-01-24 19:57:06Z

0

If you just write import foo, it will definitely load the foo module in the current scope. Assuming lib and foo as packages, won't you need to write something like this in order to make it work?

import lib.foo.bar as foobar
klass = foobar.Klass()

answered Jan 24, 2013 at 19:57

Manmohan Singh

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