如何输入一个接受类或类实例并返回该类实例的函数？

Question

我想输入一个可以采用类类型或类实例的函数，然后返回同一类的实例。例如：

from typing import Type, TypeVar, Union

DT = TypeVar("DT")

def tmap(dest_type_or_obj: Union[DT, Type[DT]]) -> DT:
    if isinstance(dest_type_or_obj, type):
        dest_obj = dest_type_or_obj()
    else:
        dest_obj = dest_type_or_obj
    return dest_obj

class Dest:
    pass

instance_dest = tmap(Dest())  # Works fine

type_dest = tmap(Dest)  # [arg-type] mypy(error)
# Argument 2 to "map" of "ObjectMapper" has incompatible type "Type[Dest]"; expected "Type[<nothing>]"

Answer 1

虽然技术上等效，但 MyPy 需要 使用 overload< /code>：

@overload
def tmap(dest_type_or_obj: Type[DT]) -> DT:
    ...
@overload
def tmap(dest_type_or_obj: DT) -> DT:
    ...

这使得 MyPy 能够在实例和类型情况下正确推断类型变量。

reveal_type(tmap(Dest()))  # note: Revealed type is "__main__.Dest*"

reveal_type(tmap(Dest))    # note: Revealed type is "__main__.Dest*"

Answer 2

您应该将 Dest 类绑定到 TypeVar，以防止 Type[] 错误：

from typing import Type, TypeVar, Union


class Dest:
    pass


DT = TypeVar("DT", bound=Dest)


def tmap(dest_type_or_obj: Union[DT, Type[DT]]) -> DT:
    if isinstance(dest_type_or_obj, type):
        dest_obj = dest_type_or_obj()
    else:
        dest_obj = dest_type_or_obj
    return dest_obj


instance_dest = tmap(Dest())
print((type(instance_dest)))
type_dest = tmap(Dest) 
print((type(type_dest)))
# output
<class '__main__.Dest'>
<class '__main__.Dest'>