Python类型：通过基类参数缩小输出子类实例的范围

Question

如果没有例子，我发现这很难描述。

from typing import List, Type, Optional, cast

class Base():
    kind = 'base'
class Child1(Base):
    kind = 'child1'
class Child2(Base):
    kind = 'child2'
class Child3(Base):
    kind = 'child3'

def find_in_list(lst: List[Base], SearchClass: Type[Base]): # Return Optional[??]
    for obj in lst: # type: Base
        if obj.kind == SearchClass.kind:
          return obj
    return None

lst = [Child1(), Child2(), Child3()]
def func2(c: Child2) -> None:
    assert isinstance(c, Child2)
res2_opt: Optional[Child2] = find_in_list(lst, Child2)
if res2_opt:
    func2(res2_opt)


def func3(c: Child3) -> None:
    assert isinstance(c, Child3)
res3_opt: Optional[Child3] = find_in_list(lst, Child2) # Should be Error!
if res3_opt:
    func3(res3_opt) # Is AssertionError

因为find_in_list 的返回类型太开放，mypy 不认为这有什么问题，但它在运行时正确地命中了断言错误。我试过-> Optional[SearchClass]，但它（正确地）没有将其识别为一种类型。

应该可以缩小find_in_list的返回类型的范围，以与SearchClass相同或相同的方式被参数化，这样如果你传入Child2作为参数，你可以将输出限制为Optional[Child2]。如何做到这一点？

Answer 1

您应该使用TypeVar 的通用函数并转换返回值：

from typing import TypeVar, cast
T = TypeVar('T', bound=Base)
def find_in_list(lst: List[Base], SearchClass: Type[T]) -> Optional[T]:
    for obj in lst:
        if obj.kind == SearchClass.kind:
            return cast(T, obj)
    return None

然后：

res3_opt: Optional[Child3] = find_in_list(lst, Child2)
# Mypy: Incompatible types in assignment (expression has type "Optional[Child2]", 
# variable has type "Optional [Child3]")