How can module determine its own path?

[kj]

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].)

TIA!

kynn

 

[Robert Kern]

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__

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

 

[AK Eric]

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__

Also:

import inspect
print inspect.getsourcefile(lambda:None)

 

[kj]

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__

Also:

import inspect
print inspect.getsourcefile(lambda:None)

Thank you both!

kynn

 

[Stef Mientki]

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.

cheers,
Stef

 

[Robert Kern]

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.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

 

[Stef Mientki]

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

 

[Robert Kern]

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) ?

filename = '...'
ns = dict(
__file__=filename,
__name__='whatever_you_need_it_to_be',
__builtins__=__builtins__,
)
execfile(filename, ns)


If you need to access variables in the current namespace:

filename = '...'
global_ns = globals().copy()
global_ns['__file__'] = filename
global_ns['__name__'] = 'whatever_you_need_it_to_be'
local_ns = locals().copy()
execfile(filename, global_ns, local_ns)

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

 

[Steven D'Aprano]

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.

If you execute a file with execfile, it isn't a module. It's a string
read from a file being execute, which is not the same thing.

Hint: in Python 3, execfile is gone (thank goodness!). To get the same
result, you use exec(file('myfile').read()).

 

[Steven D'Aprano]

Stef Mientki wrote: ....

filename = '...'
ns = dict(
__file__=filename,
__name__='whatever_you_need_it_to_be', __builtins__=__builtins__,
)
execfile(filename, ns)

Or better still, don't use execfile at all, and just use the import
system the way it's supposed to be used. If you find yourself doing by
hand what the Python compiler normally does for you, it's a good sign
that you're probably doing it wrong.

 

[Dave Angel]

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.

cheers,
Stef

The way I read the docs, execfile() doesn't create a module, so this is
irrelevant. Effectively it adds to the current module, or to whatever
the global() and local() dictionary define.

I haven't experimented with it (as the doc says, it's used rarely), so
if I'm wrong, please correct me.

DaveA

 

[Gabriel Genellina]

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.

Remember that execfile just executes the file contents, it does not create
a new module nor interacts with the module management.
"modules launched with execfile" has no meaning.