如何在泛型类型上声明Python约束来支持__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。我如何在T上声明一个约束来支持__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) 

我使用Mypy键入检查和它的失败上not_works与消息：

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

其他语言对T.

支持限制在C＃：class MyList<T> where T:IComparable<T>

在Java中：class MyList<T extends Comparable<? super T>>

Answer 1

更新 - 在mypy下面的场景的最新版本（验证0.521）正确处理。

---

嗯，我一直在寻找同样的问题，事实证明，你可以通过额外的参数bound到TypeVar实现自己的目标，如PEP484描述：从提到PEP

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:

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