MyPy - “赋值中的类型不兼容(表达式的类型为 None，变量的类型为 ...)"； [英] MyPy - "Incompatible types in assignment (expression has type None, variable has type ...)"

问题描述

我有下面的函数，给定一个 'a-02/b-03/foobarbaz_c-04' 形式的字符串，将提取 a 之后的数字em>、b 和 c.问题是，对于我的用例，输入字符串可能不包含 c，这样就没有要提取的数字.

I've got the following function, which given a string of the form 'a-02/b-03/foobarbaz_c-04', will extract the digits after a, b and c. The issue is that, for my use case, the input strings may not contain c, such that there will be no digits to extract.

代码如下:

from typing import Tuple, Optional


def regex_a_b_c(name: str) -> Tuple[int, int, Optional[int]]:
        a_b_info = re.search('a-(\d\d)/b-(\d\d)/', name)
        a, b = [int(a_b_info.group(x)) for x in range(1, 3)]
        c_info = re.search('c-(\d\d)', name)
        if c_info:
            c = int(c_info.group(1))
        else:
            c = None   
        return a, b, c

我遇到的问题是，尽管试图明确最后一个返回参数是一个 Optional[int]，但我无法让我的 linter 停止抱怨变量 c.

The issue I have is that, despite trying to make it clear that the last return argument is an Optional[int], I can't get my linter to stop complaining about the variable c.

我在 c = None 行收到一条警告，内容为:

I get a warning at the line c = None that says:

赋值中的类型不兼容(表达式的类型为 None，变量有类型 int)

Incompatible types in assignment (expression has type None, variable has type int)

我该如何解决这个问题?

How can I solve the issue?

推荐答案

当你第一次使用一个变量时，mypy 基本上会根据它看到的第一个赋值来推断它的类型.

When you use a variable for the first time, mypy will essentially infer its type based on the very first assignment it sees.

所以在这种情况下，行 c = int(_info.group(1)) 首先出现，因此 mypy 决定类型必须是 int.然后，当它看到 c = None 时，它随后会抱怨.

So in this case, the line c = int(_info.group(1)) appears first, so mypy decides that the type must be int. It then subsequently complains when it sees c = None.

解决此限制的一种方法是仅使用预期类型向前声明变量.如果您使用的是 Python 3.6+ 并且可以使用变量注释，您可以这样做:

One way of working around this limitation is to just forward-declare the variable with the expected type. If you are using Python 3.6+ and can use variable annotations, you can do so like this:

c: Optional[int]
if c_info:
    c = int(c_info.group(1))
else:
    c = None

或者更简洁，像这样:

c: Optional[int] = None
if c_info:
    c = int(c_info.group(1))

如果您需要支持旧版本的 Python，您可以使用基于注释的语法对类型进行注释，如下所示:

If you need to support older versions of Python, you can annotate the type using the comment-based syntax, like so:

c = None  # type: Optional[int]
if c_info:
    c = int(c_info.group(1))

rje 的建议:

if c_info:
    c = int(c_info.group(1))
    return a, b, c
else:
    return a, b, None

...也是合理的.

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