How to type annotate the splat/spread function

Question

Consider the common splat function, also known as spread, which takes a function of n arguments and wraps it in a function of one n-element tuple argument:

def splat(f: Callable[???, A]) -> Callable[[???], A]:
    def splatted(args: ???) -> A:
        return f(*args)
    return splatted

How do I annotate the type of parameter f and the return value? Is it even expressible in Python's type system?

Answer 1

Sure this is possible, but I'd advise sticking to positional-only parameters and strictly avoid keyword parameters (which won't work).

The idea is to employ a type variable tuple (TVT) to capture the function's n (positional) parameters, then unpack the parameters (represented by the TVT) into a single concrete tuple in the transformed function. The transformed function will then only have one parameter, which is the concrete tuple.

Demo (mypy Playground, Pyright playground):

import collections.abc as cx

class A: ...

def splat[*Ts](f: cx.Callable[[*Ts], A], /) -> cx.Callable[[tuple[*Ts]], A]:
    def splatted(args: tuple[*Ts], /) -> A:
        return f(*args)
    return splatted


@splat
def f(a: int, b: str, /) -> A:
    return A()

>>> f((1, ""))  # OK
>>> f(1, "")  # error, expected 1 tuple argument
>>> f((1, 1))  # error, second element of tuple should be `str`

The decorator usage is for demonstration purposes only - you don't need to use this as a decorator.