Comparing lists containing NaNs

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Chapters
00:00 Comparing Lists Containing Nans
00:56 Accepted Answer Score 4
01:35 Thank you

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Full question
https://stackoverflow.com/questions/3916...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #list #comparison #nan

#avk47

ACCEPTED ANSWER

Score 4

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