如何在泛型類型上聲明Python約束來支持__lt__？

Question

在下面的Python 3.5代碼中，我想使用小於運算符（<）來比較兩個通用值。我如何在T上聲明一個約束來支持__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鍵入檢查和它的失敗上not_works與消息：

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

其他語言對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下面的場景的最新版本（驗證0.521）正確處理。

---

嗯，我一直在尋找同樣的問題，事實證明，你可以通過額外的參數bound到TypeVar實現自己的目標，如PEP484描述：從提到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:

示例代碼：

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