Why does np.float('nan') cause problems with dicts, but math.nan does not?

Question

As in this question that shows how using nan from numpy causes problems with a dict, why does math.nan behave differently? (Python3)

import math
import numpy as np

d = {math.nan:'baz', math.nan:'bip' }
print(d)
e = {np.float('nan'):'foo', np.float('nan'):'bar' }
print(e)

Output

{nan: 'bip'}
{nan: 'foo', nan: 'bar'}

Moreover, both below are true: `math.nan is math.nan` `np.float('nan') is not np.float('nan')` — Chris, Oct 31 '19 at 20:51
@wim: On every single platform where `numpy` is an official distribution — Eric, Dec 02 '19 at 20:30

wim · Accepted Answer · 2019-10-31T20:58:24.990

It's because math.nan is a module level attribute, so you get the identical object both times. CPython dicts have a shortcut^† for identity checks

>>> any(k == math.nan for k in d)
False
>>> math.nan in d
True

With np.float('nan') you have a function call, returning different instance each time. Using Python's built-in float('nan') would be similar again, this is not really related to numpy.

>>> np.float('nan') is np.float('nan')
False
>>> math.nan is math.nan
True
>>> float('nan') is float('nan')
False

^†_{Lists, tuples, etc also have this. See Making an object x such that “x in [x]” returns False for more about that.}

Why does np.float('nan') cause problems with dicts, but math.nan does not?

1 Answers1