How can module determine its own path?

[Stef Mientki]

Robert Kern wrote:
> On 2009-10-30 18:40 PM, Stef Mientki wrote:
>> Robert Kern wrote:
>>> On 2009-10-30 12:19 PM, kj wrote:
>>>> How can a module determine the path of the file that defines it?
>>>> (Note that this is, in the general case, different from sys.argv[0].)
>>>
>>> __file__
>>>
>> but for modules launched with execfile, __file__ doesn't exists.
>
> Modules launched from execfile() using a properly initialized 
> namespace dict do.
>
interesting,
but how do I configure a "properly initialized namespace dict"  (other 
than my current namespace) ?

thanks,
Stef Mientki