In Python 3, modules can be namespace modules without an __init__.py
(as per PEP 420) or as a regular module (i.e. '[modules] packages as they are implemented in Python 3.2 and earlier' - PEP 420) that have an __init__.py
or are a single .py
file.
How can you tell the difference between a namespace module and an 'ordinary' module?
(I am using Python 3.5.3)
e.g.
Namespace module named mod
prints out as:
(Pdb) mod
<module 'mymodule' (namespace)>
and ordinary modules print out as:
(Pdb) mod
<module 'mymodule' from '/path/to/mymodule/__init__.py'>
Namespace packages have a __path__
, and either __file__
set to None
or no __file__
attribute. (__file__
is set to None
on Python 3.7 and later; previously, it was unset.)
if hasattr(mod, __path__) and getattr(mod, '__file__', None) is None:
print("It's a namespace package.")
In contrast, modules that aren't packages don't have a __path__
, and packages that aren't namespace packages have __file__
set to the location of their __init__.py
.
From the Python 3.8 documentation, __file__
is:
Name of the place from which the module is loaded, e.g. “builtin” for built-in modules and the filename for modules loaded from source. Normally “origin” should be set, but it may be None (the default) which indicates it is unspecified (e.g. for namespace packages).
Also, the correct answer should be:
is_namespace = (
lambda module: hasattr(module, "__path__")
and getattr(module, "__file__", None) is None
)
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