I would use this:
>>> from pkg_resources import resource_filename
>>> resource_filename('zope.interface', '')
'/location/to/env/zope.interface-4.1.0-py2.7-linux-i686.egg/zope/interface'
>>>


2014-04-21 13:20 GMT-03:00, Chris Rossi <[email protected]>:
> The answer possibly depends on why you want to know the path and what you
> want to do with it.
>
> Chris
>
>
> On Mon, Apr 21, 2014 at 12:11 PM, Seth <[email protected]> wrote:
>
>> What's the "proper" way of getting the abspath for my Pyramid project?
>> Obviously I could set a "%(here)s" config in one of the .inis or use an
>> os.path.abspath(os.path.dirname(__file__)) call in my __init__.py, but I
>> was wondering if there was a more "Pyramidic" way of doing this that
>> works
>> pretty much anywhere?
>>
>> Thanks,
>> Seth
>>
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