让我们考虑以下最小示例:
class Dummy:
def __init__(self, v1, v2, v3):
self.v1 = v1
self.v2 = v2
self.v3 = v3
def __key(self):
return (self.v1, self.v2, self.v3)
def __hash__(self):
return hash(self.__key())
def __eq__(self, other):
""" == comparison method."""
return isinstance(self, type(other)) and self.__key() == other.__key()
def __ne__(self, other):
""" != comparison method."""
return not self.__eq__(self, other)
D1 = Dummy(1, 2, 3)
D2 = Dummy(1, 4, 5)
如果我尝试 D1 == D2,我会得到 False。但是,如果我尝试 D1 != D2,我会得到:
D1 != D2
Traceback (most recent call last):
File "<ipython-input-3-82e7c8b040e3>", line 1, in <module>
D1 != D2
File "<ipython-input-1-34c16f7f1c83>", line 19, in __ne__
return not self.__eq__(self, other)
TypeError: __eq__() takes 2 positional arguments but 3 were given
我一直将 __ne__() 定义为 not self.__eq__()。直到现在我才遇到任何问题,而且我不明白为什么它不起作用...
请您参考如下方法:
def __ne__(self, other):
""" != comparison method."""
return not self.__eq__(self, other)
你不应该显式地将 self 传递给 self.__eq__ 就像你不将 self 传递给 self._key( ):
def __ne__(self, other):
""" != comparison method."""
return not self.__eq__(other)
