from pack.mod import f
How to get from object f information about import - 'pack.mod'
I can get it using f.__module__
but if function def in module where i get this attribute (f.__module__
) it return '__main__'
. But i need real path here - 'pack.mod'
I found this way to get this information:
inspect.getmodule(f).__file__
then i can sub start path from sys.path
, replace /
on .
and get path like - 'pack.mod'
But may be exist some more conveni开发者_运维知识库ent way?
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
.
You want the __name__
attribute from __module__
:
In [16]: import inspect
In [17]: inspect.getmodule(MyObject).__name__
Out[17]: 'lib.objects.MyObject'
I think that Will's answer above https://stackoverflow.com/a/18712381/3166043 is incorrect, as commented:
cd /tmp/
mkdir test
cd test
mkdir lib
echo "class MyObject: pass" > lib/objects.py
touch lib/__init__.py
export PYTHONPATH=.
python3 -c 'import inspect; from lib.objects import MyObject; print(inspect.getmodule(MyObject).__name__)'
outputs:
lib.objects
and if I replace the from lib.objects import MyObject
with from lib.objects import MyObject as MyObjectClass; MyObject = MyObjectClass()
, I get the same results.
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