How to locate the file a code object come from?

Question

I need to know where a code object comes from ; its module. So I (naively) tried :

os.path.abspath(code.co_filename)

But that may or may not work, (I think it's because abspath depends on cwd)

Any way to get the full path of a code object's module ?

EDIT :
The functions from the inspect module : getfile, getsourcefile, getmodule, get only the file name, not its path (same thing as co_filename). Maybe they use abspath.

What exactly are you looking for? What output do you get from your example code, and what are you expecting? — Shawn Chin, Aug 19 '11 at 15:20
@Shaw Chin : I get only the name of the file, and I need it's full path. — dugres, Aug 19 '11 at 15:27
Sounds like a problem with the way the object was compiled. See http://bugs.python.org/issue1665 — Shawn Chin, Aug 19 '11 at 15:41
Sarcasm, great. I was simply wondering how you installed it and if it was a pre-compiled object. — Shawn Chin, Aug 19 '11 at 15:48
@Shaw Chin : Thanks for pointing out issue1665, that helped. — dugres, Sep 16 '11 at 08:36

score 3 · Answer 1 · answered Aug 19 '11 at 14:47

3

import inspect
print inspect.getfile(inspect)

answered Aug 19 '11 at 14:47

Mariy

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score 1 · Answer 2 · answered Aug 19 '11 at 14:50

1

The inspect.getsourcefile() function is what you need, it returns the relative path to the file where the object's source can be found.

answered Aug 19 '11 at 14:50

mdeous

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inspect.getsourcefile(os.path) returned me '/usr/lib/python2.7/posixpath.py' (it seems the returned path is either relative or absolute depending on the context). – mdeous Aug 19 '11 at 15:21
yes, it depends on the context (like abspath), and that's the problem, I couln't rely on it. – dugres Aug 19 '11 at 15:24
`os.path.realpath(inspect.getsourcefile(obj))` seems to do what you want – mdeous Aug 19 '11 at 15:27
realpath seems to depend also on the context – dugres Aug 19 '11 at 15:33

Brent Newey · Answer 3 · 2011-08-19T15:19:10.093

If you have access to the module, try module.__file__.

>>> import xml
>>> xml.__file__
'/usr/local/lib/python2.6/dist-packages/PyXML-0.8.4-py2.6-linux-i686.egg/_xmlplus/__init__.pyc'

If you don't, something like this should work:

>>> import xml.etree.ElementTree as ET
>>> thing = ET.Element('test')
>>> __import__(thing.__module__).__file__
'/usr/local/lib/python2.6/dist-packages/PyXML-0.8.4-py2.6-linux-i686.egg/_xmlplus/__init__.pyc'

In this case, we are using the fact that import can be called on the string version of the module, which returns the actual module object, and then calling __file__ on that.

How to locate the file a code object come from?

3 Answers3