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python - 如何使用带有可选第一个参数的输入提示

问题描述

介绍

fun(start, stop, divisors)我有一个称为 as或的函数fun(stop, divisors)。

  1. 我想以这个特定的顺序调用参数。
  2. 我想以一种不会给出任何类型提示错误的方式来实现它。

如果我放宽这两个限制中的任何一个，实施起来就很容易。例如，可以像这样放宽对订单的限制：

def fun(stop: int, divisors: List[int], start: int=0) -> int:
    ...
    return 0

fun(5, [1, 2, 3])
fun(5, [1, 2, 3], start=2)

或者在下面的文件中显示了类型提示的缓和，但参数的顺序正确。

问题

我应该如何编写我的代码，以便我的函数参数是：

  1. 静态类型；
  2. 按特定顺序：(start[optional], stop, divisors) ;
  3. 给出 0 个 mypy 错误？

尝试

import typing
from typing import Union, Optional

Start = int
Stop = int
Divisor = int
Divisors = list[Divisor]


@typing.overload
def fun1(x: Start, y: Stop, divisors: Divisors) -> int:
    ...


@typing.overload
def fun1(y: Stop, divisors: Divisors) -> int:
    ...


def fun1(*args) -> int:
    if (arglen := len(args)) not in [2, 3]:
        raise TypeError("Function expected 2 or 3 arguments, got", arglen)
    if arglen == 2:
        args = [0] + list(args)
    start, stop, divisors = args
    return 0


def fun2(x, y, divisors) -> int:
    if divisors is None:
        start, stop, divisors = 0, x, y
    else:
        start, stop = x, y
    return 0


def fun3(*args) -> int:
    if (arglen := len(args)) == 3:
        start: Stop = args[0]
        stop: int = args[1]
        divisors: Divisors = args[2]
    elif arglen == 2:
        start: Stop = 0
        stop: int = args[0]
        divisors: Divisors = args[1]
    else:
        raise TypeError(
            f"Too {'few' if arglen == 0 else 'many'} values to unpack (2-3), got",
            arglen,
        )
    return 0


def fun4(*args) -> int:
    if (arglen := len(args)) == 3:
        pass
    elif arglen == 2:
        args = [0] + list(args)
    else:
        raise TypeError(
            f"Too {'few' if arglen == 0 else 'many'} values to unpack (2-3), got",
            arglen,
        )
    start: Stop = args[0]
    stop: int = args[1]
    divisors: Divisors = args[2]
    return 0

mypy 的输出：

typing_test_PE_001.py:20: error: Overloaded function implementation does not accept all possible arguments of signature 1
typing_test_PE_001.py:20: error: Overloaded function implementation does not accept all possible arguments of signature 2
typing_test_PE_001.py:24: error: Incompatible types in assignment (expression has type "List[int]", variable has type "Tuple[Any, ...]")
typing_test_PE_001.py:43: error: Name "start" already defined on line 39
typing_test_PE_001.py:44: error: Name "stop" already defined on line 40
typing_test_PE_001.py:45: error: Name "divisors" already defined on line 41
typing_test_PE_001.py:58: error: Incompatible types in assignment (expression has type "List[int]", variable has type "Tuple[Any, ...]")
Found 7 errors in 1 file (checked 1 source file)

标签： pythonpython-3.xtype-hintingmypypython-typing

解决方案

让我们看看slice是如何提示的：

class slice(object):
    start: Any
    step: Any
    stop: Any
    @overload
    def __init__(self, stop: Any) -> None: ...
    @overload
    def __init__(self, start: Any, stop: Any, step: Any = ...) -> None: ...
    __hash__: None  # type: ignore
    def indices(self, len: SupportsIndex) -> Tuple[int, int, int]: ...

和range

class range(Sequence[int]):
    start: int
    stop: int
    step: int
    @overload
    def __init__(self, stop: SupportsIndex) -> None: ...
    @overload
    def __init__(self, start: SupportsIndex, stop: SupportsIndex, step: SupportsIndex = ...) -> None: ...
    [...]

基本上，您分别提示两个重载版本，一个带有可选的第一个参数，一个没有。（所以基本上，你的尝试#2。）

from typing import overload, List

@overload
def fun(start: int, stop: int, divisors: List[int]):
    ...

@overload
def fun(stop: int, divisors: List[int]):
    ...

def fun(start, stop, divisors=None):
    if divisors is None:
        divisors = stop
        stop = start
        start = 0

    ...

fun(1, 2, [1,2,3])  # OK
fun(2, [1,2,3])  # OK

如果你愿意，你也可以让两个变体只接受位置参数：

# e.g.
def fun(start, stop, divisors=None, /):
    ...

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