Fix Python – Execution of Python code with -m option or not

Question

Asked By – prosseek

The python interpreter has -m module option that “Runs library module module as a script”.

With this python code a.py:

if __name__ == "__main__":
    print __package__
    print __name__

I tested python -m a to get

"" <-- Empty String
__main__

whereas python a.py returns

None <-- None
__main__

To me, those two invocation seems to be the same except __package__ is not None when invoked with -m option.

Interestingly, with python -m runpy a, I get the same as python -m a with python module compiled to get a.pyc.

What’s the (practical) difference between these invocations? Any pros and cons between them?

Also, David Beazley’s Python Essential Reference explains it as “The -m option runs a library module as a script which executes inside the __main__ module prior to the execution of the main script“. What does it mean?

Now we will see solution for issue: Execution of Python code with -m option or not


Answer

When you use the -m command-line flag, Python will import a module or package for you, then run it as a script. When you don’t use the -m flag, the file you named is run as just a script.

The distinction is important when you try to run a package. There is a big difference between:

python foo/bar/baz.py

and

python -m foo.bar.baz

as in the latter case, foo.bar is imported and relative imports will work correctly with foo.bar as the starting point.

Demo:

$ mkdir -p test/foo/bar
$ touch test/foo/__init__.py
$ touch test/foo/bar/__init__.py
$ cat << EOF > test/foo/bar/baz.py 
> if __name__ == "__main__":
>     print __package__
>     print __name__
> 
> EOF
$ PYTHONPATH=test python test/foo/bar/baz.py 
None
__main__
$ PYTHONPATH=test python -m foo.bar.baz 
foo.bar
__main__

As a result, Python has to actually care about packages when using the -m switch. A normal script can never be a package, so __package__ is set to None.

But run a package or module inside a package with -m and now there is at least the possibility of a package, so the __package__ variable is set to a string value; in the above demonstration it is set to 'foo.bar', for plain modules not inside a package it is set to an empty string.

As for the __main__ module, Python imports scripts being run as it would import regular modules. A new module object is created to hold the global namespace and is stored in sys.modules['__main__']. This is what the __name__ variable refers to, it is a key in that structure.

For packages, you can create a __main__.py module inside and have that run when running python -m package_name; in fact that is the only way you can run a package as a script:

$ PYTHONPATH=test python -m foo.bar
python: No module named foo.bar.__main__; 'foo.bar' is a package and cannot be directly executed
$ cp test/foo/bar/baz.py test/foo/bar/__main__.py
$ PYTHONPATH=test python -m foo.bar
foo.bar
__main__

So, when naming a package for running with -m, Python looks for a __main__ module contained in that package and executes that as a script. Its name is then still set to '__main__' and the module object is still stored in sys.modules['__main__'].

This question is answered By – Martijn Pieters

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0