python - 如何注释 Optional[int] 的包装器？难度 : when it's a parameter

Question

我似乎无法为接收以下值的变量找到可接受的注释:a) 函数返回 int或 b) None .问题在于函数返回的值是父函数的可选关键字参数，所以之前声明为Optional[int] .但是，运行时赋值保证函数永远不会返回 None .
如果我删除注释，mypy接受它为黄金。但我更喜欢使用一些(可接受的)注释。一分钱，一磅...
我的代码:

from typing import Optional, Callable


def myfun(p1: Optional[int] = None):
    # mypy complains about this
    dyn_p1:Optional[Callable[[], int]] = (lambda: p1) if p1 else None

    # ... but has no problem with this
    # dyn_p1 = (lambda: p1) if p1 else None

    otherFun(dyn_p1)


# I expect the parameter annotation here to be checked at the point of invocation above.
def otherFun(dyn_p1: Optional[Callable[[], int]]):
    pass

这是 mypy错误:

PS $ mypy .\prompt_toolkit\shortcuts\test1.py
prompt_toolkit\shortcuts\test1.py:6: error: Incompatible types in assignment (expression has type "Optional[Callable[[], Optional[int]]]", variable has type "Optional[Callable[[], int]]")
prompt_toolkit\shortcuts\test1.py:6: error: Incompatible return value type (got "Optional[int]", expected "int")
Found 2 errors in 1 file (checked 1 source file)

# comment out the first dyn_p1 and uncomment the second, run again:
PS $ mypy .\prompt_toolkit\shortcuts\test1.py
Success: no issues found in 1 source file

Answer 1

它只需要更改如下

from typing import Optional, Callable


def myfun(p1: Optional[int] = None):
    dyn_p1: Optional[Callable[[], Optional[int]]] = (lambda: p1) if p1 else None
    otherFun(dyn_p1)


def otherFun(dyn_p1: Optional[Callable[[], Optional[int]]]):
    pass

这是因为您已经明确地将 p1 定义为一个可选的 int，因此如果 p1 要由 lambda 返回，那么 dyn_p1 必须能够返回一个可选的 int，而不仅仅是一个 int。
现在运行 mypy

(venv) ➜  pythonProject mypy okay.py
Success: no issues found in 1 source file