从另一个函数复制类型签名
Copy type signature from another function
假设我有一组如下所示的函数。 foo 有很多不同类型的参数，bar 将其所有参数传递给另一个函数。有没有办法让 mypy 理解 bar 与 foo 具有相同的类型，而无需显式复制整个参数列表？
def foo(a: int, b: float, c: str, d: bool, *e: str, f: str = "a", g: str = "b") -> str:
...
def bar(*args, **kwargs):
val = foo(*args, **kwargs)
...
return val
关于添加此功能的讨论很多here. For the straightforward case of passing all arguments you can use the recipe from this comment：
F = TypeVar('F', bound=Callable[..., Any])
class copy_signature(Generic[F]):
def __init__(self, target: F) -> None: ...
def __call__(self, wrapped: Callable[..., Any]) -> F: ...
def f(x: bool, *extra: int) -> str: ...
@copy_signature(f)
def test(*args, **kwargs):
return f(*args, **kwargs)
reveal_type(test) # Revealed type is 'def (x: bool, *extra: int) -> str'
假设我有一组如下所示的函数。 foo 有很多不同类型的参数，bar 将其所有参数传递给另一个函数。有没有办法让 mypy 理解 bar 与 foo 具有相同的类型，而无需显式复制整个参数列表？
def foo(a: int, b: float, c: str, d: bool, *e: str, f: str = "a", g: str = "b") -> str:
...
def bar(*args, **kwargs):
val = foo(*args, **kwargs)
...
return val
关于添加此功能的讨论很多here. For the straightforward case of passing all arguments you can use the recipe from this comment：
F = TypeVar('F', bound=Callable[..., Any])
class copy_signature(Generic[F]):
def __init__(self, target: F) -> None: ...
def __call__(self, wrapped: Callable[..., Any]) -> F: ...
def f(x: bool, *extra: int) -> str: ...
@copy_signature(f)
def test(*args, **kwargs):
return f(*args, **kwargs)
reveal_type(test) # Revealed type is 'def (x: bool, *extra: int) -> str'