python-3.x - 如何在泛型类型上声明 Python 约束以支持 __lt__？

Question

在下面的 Python 3.5 代码中，我想使用小于运算符 ( < ) 来比较两个通用值。如何声明 T 上的约束以支持 __lt__ ?

from typing import *
import operator 

T = TypeVar('T')

class MyList(Generic[T]):
    class Node:
        def __init__(self, k:T) -> None:
            self.key = k 
            self.next = None  # type: Optional[MyList.Node]

    def __init__(self) -> None:
        self.root = None # type: Optional[MyList.Node]

    def this_works(self, val:T) -> bool:
        return self.root.key == val 

    def not_works(self, val:T) -> bool:
        return operator.lt(self.root.key, val)

我正在使用 Mypy键入 check 并且它在 not_works 上失败消息:

$ mypy test.py
test.py: note: In member "not_works" of class "MyList":
test.py:20: error: Unsupported left operand type for < ("T")

其他语言支持对 T 的约束。

在 C# 中: class MyList<T> where T:IComparable<T>
在 Java 中: class MyList<T extends Comparable<? super T>>

Answer 1

你可以通过传递一个额外的参数来实现你的目标 bound至 TypeVar ，如 PEP484 中所述:

A type variable may specify an upper bound using bound=<type>. This means that an actual type substituted (explicitly or implicitly) for the type variable must be a subtype of the boundary type. A common example is the definition of a Comparable type that works well enough to catch the most common errors:

来自上述 PEP 的示例代码:

from typing import TypeVar

class Comparable(metaclass=ABCMeta):
    @abstractmethod
    def __lt__(self, other: Any) -> bool: ...
    ... # __gt__ etc. as well

CT = TypeVar('CT', bound=Comparable)

def min(x: CT, y: CT) -> CT:
    if x < y:
        return x
    else:
        return y

min(1, 2) # ok, return type int
min('x', 'y') # ok, return type str

在mypy的最新版本(用0.521验证)中，正确处理了上述场景。