python - Why is (numpy.nan, 1) == (numpy.nan, 1)? -

while numpy.nan not equal numpy.nan, , (float('nan'), 1) not equal float('nan', 1),

(numpy.nan, 1) == (numpy.nan, 1) 

what reason? python first check see if ids identical? if identity checked first when comparing items of tuple, why isn't checked when objects compared directly?

when numpy.nan == numpy.nan it's numpy deciding whether condition true or not. when compare tuples python checking if tuples have same objects do. can make numpy have decision turning tuples numpy arrays.

np.array((1, numpy.nan)) == np.array((1,numpy.nan)) >>array([ true, false], dtype=bool) 

the reason when == numpy objects you're calling numpy function __eq__() says nan != nan because mathematically speaking nan undetermined (could anything) makes sense nan != nan. when == tuples call tuples __eq__() function doesn't care mathematics , cares if python objects same or not. in case of (float('nan'),1)==(float('nan'),1) returns false because each call of float('nan') allocates memory in different place can check doing float('nan') float('nan').