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

The documentation states you must define one of __lt__(), __le__(), __gt__(), or __ge__(), but only should supply an __eq__() method.

In other words, the __eq__ method is optional.

The total_ordering implementation does not require you to specify an __eq__ method; it only tests for the __lt__(), __le__(), __gt__(), or __ge__() methods. It supplies up to 3 missing special methods based of one of those 4.

You can't base the order on just __le__ or __ge__ because you can't assume that you can swap a and b; if b is a different type b.__le__ might not be implemented and so your a < b <=> not (b <= a) map can't be guaranteed. The implementation uses (a <= b) and (a != b) if __le__ is not defined but __lt__ has been.

The full table of mappings is:

comparison	available	alternative
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)

The __eq__ method is optional because the base object object defines one for you; two instances are considered equal only if they are the same object; ob1 == ob2 only if ob1 is ob2 is True. See the do_richcompare() function in object.c; remember that the == operator in the code there is comparing pointers.