Get function import path

Question

from pack.mod import f

How to get from object f information about import - \'pack.mod\'

I can get it using f.__module__

Answer 1

What inspect.getmodule(f) does internally, per inspect.py's sources, is essentially sys.modules.get(object.__module__) -- I wouldn't call using that code directly "more convenient", though (beyond the "essentially" part, inspect has a lot of useful catching and correction of corner cases).

Why not call directly inspect.getsourcefile(f)?

Edit: reading between the lines it seems the OP is trying to do something like

python /foo/bar/baz/bla.py

and within bla.py (which is thus being run as __main__) determine "what from or import statement could a different main script use to import this function from within me?".

Problem is, the question is ill-posed, because there might not be any such path usable for the purpose (nothing guarantees the current main script's path is on sys.path when that different main script gets run later), there might be several different ones (e.g. both /foo/bar and /foo/bar/baz might be on sys.path and /foo/bar/baz/__init__.py exist, in which case from baz.bla import f and from bla import f might both work), and nothing guarantees that some other, previous sys.path item might not "preempt" the import attempt (e.g., say /foo/bar/baz is on sys.path, but before it there's also /fee/fie/foo, and a completely unrelated file /fee/fie/foo/bla.py also exists -- etc, etc).

Whatever the purpose of this kind of discovery attempt, I suggest finding an alternative architecture -- e.g., one where from baz.bla import f is actually executed (as the OP says at the start of the question), so that f.__module__ is correctly set to baz.bla.