将空元组分配给专用变量时的类型不兼容 [英] Incompatible types when assigning an empty tuple to a specialised variable

问题描述

我有一个变量 path，它应该是一个字符串元组.我想从它设置为一个空元组开始，但 mypy 抱怨.

I have a variable path, which should be a tuple of strings. I want to start with it set to an empty tuple, but mypy complains.

path: Tuple[str] = ()

错误是:

赋值中的类型不兼容(表达式类型为Tuple[]"，变量类型为Tuple[str]")

Incompatible types in assignment (expression has type "Tuple[]", variable has type "Tuple[str]")

如何将空元组分配给类型化变量?

How can I assign an empty tuple to a typed variable?

我想这样做的原因是:我想动态地建立元组，元组(与列表不同)可以用作字典键.例如(不是我实际在做的):

The reason I want to do this is: I want to build up the tuple dynamically, and tuples (unlike lists) can be used as dictionary keys. For example (not what I'm actually doing):

for line in fileob:
    path += (line,)
some_dict[path] = some_object

这很有效，只是 mypy 不喜欢上面的类型声明.我可以使用列表，然后将其转换为元组，但这会使代码复杂化.

This works well, except that mypy doesn't like the type declaration above. I could use a list and then convert it to a tuple, but it complicates the code.

推荐答案

您可以定义一个可变长度、同质元组，例如这个:

You can define a variable length, homogenous tuple like this:

Tuple[str, ...]

您还可以使用 <创建a 或 b"类型变量代码>typing.Union:

You can also create a "a or b" type variable with typing.Union:

from typing import Union

path: Union[Tuple[()], Tuple[str]] = ()

旧答案:

通过尝试将空元组分配给您输入的不允许空元组的变量，您实际上错过了输入变量的要点.

By trying to assign an empty tuple to a variable that you have typed to not allow empty tuples, you're effectivly missing the point of typing variables.

我假设您尝试分配的空元组只是一个标记值，并且您打算稍后在代码中重新分配变量(分配给仅包含字符串的元组.)

I assume that the empty tuple you are trying to assign is just a token value, and that you intend to reassign the variable later in the code (to a tuple that contains only strings.)

在这种情况下，只需让标记值是一个包含字符串的元组:

In that case, simply let the token value be a tuple that contains a string:

path: Tuple[str] = ('token string')

这应该会停止错误消息.

This should stop the error message.

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