in operator, float("NaN") and np.nan

Question

I used to believe that in operator in Python checks the presence of element in some collection using equality checking ==, so element in some_list is roughly equivalent to any(x == element for x in some_list). For example:

True in [1, 2, 3] # True because True == 1 

or

1 in [1., 2., 3.] # also True because 1 == 1. 

However, it is well-known that NaN is not equal to itself. So I expected that float("NaN") in [float("NaN")] is False. And it is False indeed.

However, if we use numpy.nan instead of float("NaN"), the situation is quite different:

import numpy as np np.nan in [np.nan, 1, 2] # True 

But np.nan == np.nan still gives False!

How is it possible? What's the difference between np.nan and float("NaN")? How does in deal with np.nan?

Answer 1

To check if the item is in the list, Python tests for object identity first, and then tests for equality only if the objects are different.¹

float("NaN") in [float("NaN")] is False because two different NaN objects are involved in the comparison. The test for identity therefore returns False, and then the test for equality also returns False since NaN != NaN.

np.nan in [np.nan, 1, 2] however is True because the same NaN object is involved in the comparison. The test for object identity returns True and so Python immediately recognises the item as being in the list.

The __contains__ method (invoked using in) for many of Python's other builtin Container types, such as tuples and sets, is implemented using the same check.

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¹ At least this is true in CPython. Object identity here means that the objects are found at the same memory address: the contains method for lists is performed using PyObject_RichCompareBool which quickly compares object pointers before a potentially more complicated object comparison. Other Python implementations may differ.