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 ==

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