CODEWITHSUNDEEP
pythonpython-3.xcomparison-operators
Guillaume Lemaître
Guillaume Lemaître

Reputation: 1280

python total_ordering : why __lt__ and __eq__ instead of __le__?

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?

Upvotes: 23

Views: 9065

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1123860

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.

Upvotes: 33

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