mypy字面量类型检查异常：分支类型推断不符预期求助

问题原因 #

你定义的函数返回两种元组的联合类型，但Python元组解构后，mypy无法跟踪a的布尔值与b的类型之间的关联——它只会把b推断为两种返回类型的交集，也就是Optional[Type[Temp]]，所以无论if a还是else分支，都会触发union-attr错误。

解决方案 #

方法1：用NamedTuple定义明确的返回类型（推荐） #

通过两个NamedTuple分别对应两种返回场景，让mypy能通过flag的值自动窄化关联字段的类型：

from typing import Literal, Optional, Type, Union, NamedTuple

class Temp:
    @staticmethod
    def pankaj():
        print("pankaj")

class TrueResult(NamedTuple):
    flag: Literal[True]
    cls: Type[Temp]

class FalseResult(NamedTuple):
    flag: Literal[False]
    cls: Optional[Type[Temp]]

def accepts_only_int(x: int) -> Union[TrueResult, FalseResult]:
    # 可根据实际逻辑返回TrueResult或FalseResult
    return FalseResult(False, Temp)

result = accepts_only_int(6)

if result.flag:
    print(result.cls.pankaj())
else:
    if result.cls is not None:
        print(result.flag, result.cls.pankaj())
    else:
        print(result.flag, "cls is None")

方法2：添加类型断言 #

在分支中用assert明确告诉mypy当前分支下b的类型，同时在else分支先检查b是否为None：

from typing import Literal, Optional, Type, Union

class Temp:
    @staticmethod
    def pankaj():
        print("pankaj")

def accepts_only_int(x: int) -> Union[tuple[Literal[True],Type[Temp]],tuple[Literal[False],Optional[Type[Temp]]]]:
    return (False,Temp)

a, b = accepts_only_int(6)

if a:
    assert b is not None, "当a为True时b不可能为None"
    print(b.pankaj())
else:
    if b is not None:
        print(a, b.pankaj())
    else:
        print(a, "b is None")

方法3：强制类型转换 #

使用cast手动指定b在对应分支的类型，适合不想修改返回类型定义的场景：

from typing import Literal, Optional, Type, Union, cast

class Temp:
    @staticmethod
    def pankaj():
        print("pankaj")

def accepts_only_int(x: int) -> Union[tuple[Literal[True],Type[Temp]],tuple[Literal[False],Optional[Type[Temp]]]]:
    return (False,Temp)

a, b = accepts_only_int(6)

if a:
    b = cast(Type[Temp], b)
    print(b.pankaj())
else:
    if b is not None:
        print(a, b.pankaj())
    else:
        print(a, "b is None")

内容的提问来源于stack exchange，提问作者Pankaj Singh Rana