Using lambda to define a function based on another: How do I keep my code generic?

Question

I have some python code that defines a new function based on an old one. It looks like this

def myFunction(a: int, b: int, c: int):
    # Do stuff

myNewFunction = lambda a, b: myFunction(a, b, 0)

My new function is the same as the old function, but sets the last argument to 0.

My question: Say I did not know the function took three arguments. Can I make the above solution more generic? An invalid solution with the right intention might be something like:

def myFunction(a: int, b: int, c: int):
    # Do stuff

func_args = myFunction.__code__.co_varnames
func_args = func_args[:-1]

myNewFunction = lambda *func_args : myFunction(*func_args, 0)

I don't have co_argvars 'co_argvars' in dir(myFunction.__code__) results to False. — DiMithras
– DiMithras, Commented Dec 3, 2022 at 20:33
Yes, it unfortunately doesn't exist. But I can see how it makes the question confusing. I have edited accordingly. — DJames
– DJames, Commented Dec 3, 2022 at 20:39

S.B · Accepted Answer · 2022-12-03 20:58:33Z

1

You're almost correct, you can use functools.partial this way(instead of lambda):

from functools import partial

def myFunction(a: int, b: int, c: int):
    print(a, b, c)

last_param_name = myFunction.__code__.co_varnames[-1]
new_func = partial(myFunction, **{last_param_name: 0})

new_func(10, 20)

Technically partial is not a function it's a class but I don't think this is what you concern about.

A pure Python (roughly)equivalent of the original partial exists in documentation if you want it to be a function type:

def partial(func, /, *args, **keywords):
    def newfunc(*fargs, **fkeywords):
        newkeywords = {**keywords, **fkeywords}
        return func(*args, *fargs, **newkeywords)
    newfunc.func = func
    newfunc.args = args
    newfunc.keywords = keywords
    return newfunc

edited Dec 3, 2022 at 20:58

answered Dec 3, 2022 at 20:50

S.B

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