Why does collections.Counter treat numpy.nan as equal?

Question

I am surprised by the following behavior:

>>> import numpy as np
>>> from collections import Counter
>>> my_list = [1,2,2, np.nan, np.nan]
>>> Counter(my_list)
Counter({nan: 2, 2: 2, 1: 1}) # Counter treats np.nan as equal and 
                              # tells me that I have two of them
>>> np.nan == np.nan          # However, np.nan's are not equal  
False

What is going on here?

When I use float('nan') instead of np.nan, I get the behavior I expect:

>>> my_list = [1,2,2, float('nan'), float('nan')]
>>> Counter(my_list)
Counter({2: 2, nan: 1, 1: 1, nan: 1}) # two different nan's
>>> float('nan') == float('nan')
False

I am using python 2.7.3 and numpy 1.8.1.

Edit:

If I do:

>>> a = 300
>>> b = 300
>>> a is b
False
>>> Counter([a, b])
Counter({300: 2})

So, Counter or any python dict considers two objects X and Y not the same if:

X == Y -> False

and

X is Y -> False

correct?

Answer 1

This isn't about numpy.nan vs. float("nan"), it's that you've got two separate float nans.

>>> np.nan is np.nan
True
>>> float("nan") is float("nan")
False

and so

>>> Counter([1,2,2, np.nan, np.nan])
Counter({nan: 2, 2: 2, 1: 1})
>>> Counter([1,2,2, float("nan"), float("nan")])
Counter({2: 2, nan: 1, 1: 1, nan: 1})

but

>>> f = float("nan")
>>> Counter([1,2,2, f, f])
Counter({nan: 2, 2: 2, 1: 1})