Forced naming of parameters in Python

You can force people to use keyword arguments in Python3 by defining a function in the following way.

def foo(*, arg0="default0", arg1="default1", arg2="default2"):
    pass

By making the first argument a positional argument with no name you force everyone who calls the function to use the keyword arguments which is what I think you were asking about. In Python2 the only way to do this is to define a function like this

def foo(**kwargs):
    pass

That'll force the caller to use kwargs but this isn't that great of a solution as you'd then have to put a check to only accept the argument that you need.

In Python 3 - Yes, you can specify * in the argument list.

From docs:

Parameters after “*” or “*identifier” are keyword-only parameters and may only be passed used keyword arguments.

>>> def foo(pos, *, forcenamed):
...   print(pos, forcenamed)
... 
>>> foo(pos=10, forcenamed=20)
10 20
>>> foo(10, forcenamed=20)
10 20
>>> foo(10, 20)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() takes exactly 1 positional argument (2 given)

This can also be combined with **kwargs:

def foo(pos, *, forcenamed, **kwargs):

To complete example:

def foo(pos, *, forcenamed ):
    print(pos, forcenamed)

foo(pos=10, forcenamed=20)
foo(10, forcenamed=20)
# basically you always have to give the value!
foo(10)

output:

Traceback (most recent call last):
  File "/Users/brando/anaconda3/envs/metalearning/lib/python3.9/site-packages/IPython/core/interactiveshell.py", line 3444, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)
  File "<ipython-input-12-ab74191b3e9e>", line 7, in <module>
    foo(10)
TypeError: foo() missing 1 required keyword-only argument: 'forcenamed'

So you are forced to always give the value. If you don't call it you don't have to do anything else named argument forced.