Calling a helper function from inside a class variable

Question

Consider the following code:

# foo.py

class A:
    def _foo():
        print('hello world')
    
    bar = {'foo': _foo}

    def run_bar(self):
        self.bar['foo']()

def main():
    A().run_bar()

if __name__ == '__main__':
    raise SystemExit(main())

It runs just fine with Python 3.9.5:

python3.9 foo.py
> hello world

but mypy will give me the following error:

mypy foo.py
> foo.py:2: error: Method must have at least one argument
> Found 1 error in 1 file (checked 1 source file)

Is there a way to tell mypy that this function will only ever get called as a class variable ? If no, is this bad practice? I know I could simply add # type: ignore, but that seems too hacky.

You could just define `_foo` outside your class, if it isn't really a class method or instance method. — khelwood, Aug 08 '21 at 10:34

Alex Waygood · Accepted Answer · 2021-08-08T16:14:10.657

Two ways of doing this:

1. Use a staticmethod

Either like this:

from typing import Callable, Any

class A:
    bar: dict[str, Callable[..., Any]] = {}

    def __init__(self):
        self.bar.update({'foo': self._foo})

    @staticmethod
    def _foo():
        print('hello world')

    def run_bar(self):
        self.bar['foo']()
    

def main():
    A().run_bar()
  

if __name__ == '__main__':
    raise SystemExit(main())

Or like this:

class A:
    @staticmethod
    def _foo():
        print('hello world')
 
    def run_bar(self):
        getattr(self, '_foo')()
        

def main():
    A().run_bar()


if __name__ == '__main__':
    raise SystemExit(main())

2. Put the function outside of the class

def _foo():
    print('hello world')


class A:
    bar = {'foo': _foo}
   
    def run_bar(self):
        self.bar['foo']()
    

def main():
    A().run_bar()
  

if __name__ == '__main__':
    raise SystemExit(main())

Calling a helper function from inside a class variable

1 Answers1