What is the __main__.py file for, what sort of code should I put into it, and when should I have one?
406
votes
6 Answers
396
votes
Often, a Python program is run by naming a .py file on the command line:
$ python my_program.py
You can also create a directory or zipfile full of code, and include a __main__.py. Then you can simply name the directory or zipfile on the command line, and it executes the __main__.py automatically:
$ python my_program_dir
$ python my_program.zip
# Or, if the program is accessible as a module
$ python -m my_program
You'll have to decide for yourself whether your application could benefit from being executed like this.
Note that a __main__ module usually doesn't come from a __main__.py file. It can, but it usually doesn't. When you run a script like python my_program.py, the script will run as the __main__ module instead of the my_program module. This also happens for modules run as python -m my_module, or in several other ways.
If you saw the name __main__ in an error message, that doesn't necessarily mean you should be looking for a __main__.py file.
133
votes
What is the __main__.py file for?
When creating a Python module, it is common to make the module execute some functionality (usually contained in a main function) when run as the entry point of the program. This is typically done with the following common idiom placed at the bottom of most Python files:
if __name__ == '__main__':
# execute only if run as the entry point into the program
main()
You can get the same semantics for a Python package with __main__.py, which might have the following structure:
.
└── demo
├── __init__.py
└── __main__.py
To see this, paste the below into a Python 3 shell:
from pathlib import Path
demo = Path.cwd() / 'demo'
demo.mkdir()
(demo / '__init__.py').write_text("""
print('demo/__init__.py executed')
def main():
print('main() executed')
""")
(demo / '__main__.py').write_text("""
print('demo/__main__.py executed')
from demo import main
main()
""")
We can treat demo as a package and actually import it, which executes the top-level code in the __init__.py (but not the main function):
>>> import demo
demo/__init__.py executed
When we use the package as the entry point to the program, we perform the code in the __main__.py, which imports the __init__.py first:
$ python -m demo
demo/__init__.py executed
demo/__main__.py executed
main() executed
You can derive this from the documentation. The documentation says:
__main__— Top-level script environment
'__main__'is the name of the scope in which top-level code executes. A module’s__name__is set equal to'__main__'when read from standard input, a script, or from an interactive prompt.A module can discover whether or not it is running in the main scope by checking its own
__name__, which allows a common idiom for conditionally executing code in a module when it is run as a script or withpython -mbut not when it is imported:if __name__ == '__main__': # execute only if run as a script main()For a package, the same effect can be achieved by including a
__main__.pymodule, the contents of which will be executed when the module is run with-m.
Zipped
You can also zip up this directory, including the __main__.py, into a single file and run it from the command line like this - but note that zipped packages can't execute sub-packages or submodules as the entry point:
from pathlib import Path
demo = Path.cwd() / 'demo2'
demo.mkdir()
(demo / '__init__.py').write_text("""
print('demo2/__init__.py executed')
def main():
print('main() executed')
""")
(demo / '__main__.py').write_text("""
print('demo2/__main__.py executed')
from __init__ import main
main()
""")
Note the subtle change - we are importing main from __init__ instead of demo2 - this zipped directory is not being treated as a package, but as a directory of scripts. So it must be used without the -m flag.
Particularly relevant to the question - zipapp causes the zipped directory to execute the __main__.py by default - and it is executed first, before __init__.py:
$ python -m zipapp demo2 -o demo2zip
$ python demo2zip
demo2/__main__.py executed
demo2/__init__.py executed
main() executed
Note again, this zipped directory is not a package - you cannot import it either.
37
votes
__main__.py is used for python programs in zip files. The __main__.py file will be executed when the zip file in run. For example, if the zip file was as such:
test.zip
__main__.py
and the contents of __main__.py was
import sys
print "hello %s" % sys.argv[1]
Then if we were to run python test.zip world we would get hello world out.
So the __main__.py file run when python is called on a zip file.
17
votes
Some of the answers here imply that given a "package" directory (with or without an explicit __init__.py file), containing a __main__.py file, there is no difference between running that directory with the -m switch or without.
The big difference is that without the -m switch, the "package" directory is first added to the path (i.e. sys.path), and then the files are run normally, without package semantics.
Whereas with the -m switch, package semantics (including relative imports) are honoured, and the package directory itself is never added to the system path.
This is a very important distinction, both in terms of whether relative imports will work or not, but more importantly in terms of dictating what will be imported in the case of unintended shadowing of system modules.
Example:
Consider a directory called PkgTest with the following structure
:~/PkgTest$ tree
.
├── pkgname
│ ├── __main__.py
│ ├── secondtest.py
│ └── testmodule.py
└── testmodule.py
where the __main__.py file has the following contents:
:~/PkgTest$ cat pkgname/__main__.py
import os
print( "Hello from pkgname.__main__.py. I am the file", os.path.abspath( __file__ ) )
print( "I am being accessed from", os.path.abspath( os.curdir ) )
from testmodule import main as firstmain; firstmain()
from .secondtest import main as secondmain; secondmain()
(with the other files defined similarly with similar printouts).
If you run this without the -m switch, this is what you'll get. Note that the relative import fails, but more importantly note that the wrong testmodule has been chosen (i.e. relative to the working directory):
:~/PkgTest$ python3 pkgname
Hello from pkgname.__main__.py. I am the file ~/PkgTest/pkgname/__main__.py
I am being accessed from ~/PkgTest
Hello from testmodule.py. I am the file ~/PkgTest/pkgname/testmodule.py
I am being accessed from ~/PkgTest
Traceback (most recent call last):
File "/usr/lib/python3.6/runpy.py", line 193, in _run_module_as_main
"__main__", mod_spec)
File "/usr/lib/python3.6/runpy.py", line 85, in _run_code
exec(code, run_globals)
File "pkgname/__main__.py", line 10, in <module>
from .secondtest import main as secondmain
ImportError: attempted relative import with no known parent package
Whereas with the -m switch, you get what you (hopefully) expected:
:~/PkgTest$ python3 -m pkgname
Hello from pkgname.__main__.py. I am the file ~/PkgTest/pkgname/__main__.py
I am being accessed from ~/PkgTest
Hello from testmodule.py. I am the file ~/PkgTest/testmodule.py
I am being accessed from ~/PkgTest
Hello from secondtest.py. I am the file ~/PkgTest/pkgname/secondtest.py
I am being accessed from ~/PkgTest
Note: In my honest opinion, running without -m should be avoided. In fact I would go further and say that I would create any executable packages in such a way that they would fail unless run via the -m switch.
In other words, I would only import from 'in-package' modules explicitly via 'relative imports', assuming that all other imports represent system modules. If someone attempts to run your package without the -m switch, the relative import statements will throw an error, instead of silently running the wrong module.
