Forcing python function parameters to not have order

Question

I have a python function that takes a large amount of parameters :

def func(p1=0, p2=0, p3=0, p4=0, p5=0, ..., pN=0) -> None: pass

I wanted to force the user to set the parameters as keyword arguments.

I thought about one solution that seems off to me:

def func(*_, p1=0, p2=0, p3=0, p4=0, p5=0, ..., pN=0) -> None: pass

I can even raise an error if unwanted ordered arguments were given and even separate ordered arguments from unordered arguments:

def func(p1, p2, p3, *unwanted, p4=0, p5=0, p6=0, ..., pN=0) -> None:
    if unwanted: raise TypeError

I haven't seen anyone do this, is there a problem with this?

I am not quite sure whether I understood the question, but try func(p1, p2, p3, *, p4=0, ..., pN=0) — j1-lee
– j1-lee, Commented Nov 13, 2022 at 19:24
You don't actually need the _; * alone is sufficient to mark the "boundary" between ordinary parameters and keyword-only parameters. (*_ is the same as *args: it's a parameter named _ that gathers any positional arguments not assigned to other parameters. It can also mark the boundary between ordinary and keyword-only parameters.) — chepner
– chepner, Commented Nov 13, 2022 at 19:32

mozway · Accepted Answer · 2022-11-13 19:30:32Z

2

This is already a standard method. It is defined in PEP3102

It's used in many libraries.

To give you one example: in pandas' drop function, all parameters after * are keywords only:

DataFrame.drop(labels=None, *, axis=0, index=None, columns=None, level=None, inplace=False, errors='raise')

Note that you don't need the _ if you just want to ignore the parameters, a bare * is sufficient.

answered Nov 13, 2022 at 19:30

mozway

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