python - python total_ordering : why __lt__ and __eq__ instead of __le__?

Question

In Python3, the functools.total_ordering decorator allows one to only overload __lt__ and __eq__ to get all 6 comparison operators.

I don't get why one has to write two operators when one would be enough, namely __le__ or __ge__, and all others would be defined accordingly :

a < b   <=>   not (b <= a)
a > b   <=>   not (a <= b)
a == b   <=>   (a <= b) and (b <= a)
a != b   <=>   (a <= b) xor (b <= a)

Is that just because xor operator does not exists natively?

Answer 1

文档声明您必须定义 、 、 或 中的一个__lt__()，__le__()但__gt__()只__ge__()应提供一个__eq__()方法。

换句话说，该__eq__方法是可选的。

该total_ordering实现不需要您指定__eq__方法；它只测试__lt__()、__le__()、__gt__()或__ge__()方法。它提供了多达 3 种缺失的特殊方法，基于这 4 种方法之一。

你不能仅仅 __le__基于或者__ge__因为你不能假设你可以交换a和b; if bis a different typeb.__le__可能无法实现，因此a < b <=> not (b <= a)无法保证您的地图。实现使用(a <= b) and (a != b)if__le__未定义但__lt__已定义。

完整的映射表是：

比较	可用的	选择
a > b	a < b	(not a < b) and (a != b)
	a <= b	(not a <= b)
	a >= b	(a >= b) and (a != b)
a <= b	a < b	(a < b) or (a == b)
	a > b	(not a > b)
	a >= b	(not a >= b) or (a == b)
a < b	a <= b	(a <= b) and (a != b)
	a > b	(not a > b) and (a != b)
	a >= b	(not a >= b)
a >= b	a < b	(not a < b)
	a <= b	(not a <= b) or (a == b)
	a > b	(a > b) or (a == b)

该__eq__方法是可选的，因为基础object对象为您定义了一个；只有当它们是同一个对象时，两个实例才被认为是相等的；ob1 == ob2只有当ob1 is ob2是True。见do_richcompare()函数object.c；请记住，==代码中的运算符是比较指针。