Comparing lists containing NaNs

Question

I am trying to compare two different lists to see if they are equal, and was going to remove NaNs, only to discover that my list comparisons still work, despite NaN == NaN -> False.

Could someone explain why the following evaluate True or False, as I am finding this behavior unexpected. Thanks,

I have read the following which don't seem to resolve the issue:

• Why in numpy nan == nan is False while nan in [nan] is True?
• Why is NaN not equal to NaN? [duplicate]

(Python 2.7.3, numpy-1.9.2)

I have marked surprising evaluations with a * at the end

>>> nan = np.nan
>>> [1,2,3]==[3]
False
>>> [1,2,3]==[1,2,3]
True
>>> [1,2,nan]==[1,2,nan]
True ***
>>> nan == nan
False
>>> [nan] == [nan]
True ***
>>> [nan, nan] == [nan for i in range(2)]
True ***
>>> [nan, nan] == [float(nan) for i in range(2)]
True ***
>>> float(nan) is (float(nan) + 1)
False
>>> float(nan) is float(nan)
True ***

Answer 1

To understand what happens here, simply replace nan = np.nan by foo = float('nan'), you will get exactly the same result, why?

>>> foo = float('nan')
>>> foo is foo # This is obviously True! 
True
>>> foo == foo # This is False per the standard (nan != nan).
False
>>> bar = float('nan') # foo and bar are two different objects.
>>> foo is bar
False
>>> foo is float(foo) # "Tricky", but float(x) is x if type(x) == float.
True

Now think that numpy.nan is just a variable name that holds a float('nan').

Now why [nan] == [nan] is simply because list comparison first test identity equality between items before equality for value, think of it as:

def equals(l1, l2):
    for u, v in zip(l1, l2):
        if u is not v and u != v:
            return False
    return True