Checking for NaN presence in a container

Question

NaN is handled perfectly when I check for its presence in a list or a set. But I don\'t understand how. [UPDATE: no it\'s not; it is reported as present if the identical ins

Answer 1

I can't repro you tuple/set cases using float('nan') instead of NaN.

So i assume that it worked only because id(NaN) == id(NaN), i.e. there is no interning for NaN objects:

>>> NaN = float('NaN')
>>> id(NaN)
34373956456
>>> id(float('NaN'))
34373956480

And

>>> NaN is NaN
True
>>> NaN is float('NaN')
False

I believe tuple/set lookups has some optimization related to comparison of the same objects.

Answering your question - it seam to be unsafe to relay on in operator while checking for presence of NaN. I'd recommend to use None, if possible.

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Just a comment. __eq__ has nothing to do with is statement, and during lookups comparison of objects' ids seem to happen prior to any value comparisons:

>>> class A(object):
...     def __eq__(*args):
...             print '__eq__'
...
>>> A() == A()
__eq__          # as expected
>>> A() is A()
False           # `is` checks only ids
>>> A() in [A()]
__eq__          # as expected
False
>>> a = A()
>>> a in [a]
True            # surprise!