In Python3,
a = b = 3
a is None == b is None
returns False, but
(a is None) == (b is None)
returns True. So I would assume based on this example alone, == has precedence over is.
However,
a = b = None
a is None == b is None
returns True. And
(a is None) == (b is None)
returns True. But
a is (None == b) is None
returns False. In this case, it would seem as if is has precedence over ==.
To give another example, and this expression isn't meant to do anything, but bear with me please. If I say
None is None == None
it returns True. But both of the following return False.
None is (None == None)
(None is None) == None
So clearly, Python isn't evaluating these with some strict precedence, but I'm confused what is going on. How is it evaluating this expression with 2 different operators, but differently from either order?
What you see here is operator chaining and there is no precedence involved at all!
Python supports expressions like
1 < a < 3
To test that a number is in between 1 and 3; it's equal to (1 < a) and (a < 3) except that a is only evaluated once.
Unfortunately that also means that e.g.
None is None == None
actually means
(None is None) and (None == None)
which is of course True, and the longer example you started with
a = b = 3
a is None == b is None
means
(a is None) and (None == b) and (b is None)
which can only be True if both a and b are None.
Documentation here, see the bit about chaining.
Very useful sometimes but it also pops up when you least expect it!