python float('nan') vs math.nan

Asked May 16 '22 at 20:18

Active May 17 '22 at 18:52

Viewed 35 times

I read Why does float('nan') == float('nan') return False? and accept that float('nan) != float('nan'). I modified my code to successfully use math.isnan().

But if float('nan') cannot be found in lists, why can math.nan be found in lists?

$ python
Python 3.6.4 |Anaconda, Inc.| (default, Jan 16 2018, 18:10:19)
[GCC 7.2.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import math
>>> math.nan in [ 1.0, 2.0, float('nan'), 5.0 ]
False
>>> float('nan') in [ 1.0, 2.0, float('nan'), 5.0 ]
False
>>> math.nan in [ 1.0, 2.0, math.nan, 5.0 ]
True

edited May 17 '22 at 18:52

Kevin M. Mansour

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asked May 16 '22 at 20:18

Donald McLachlan

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Because it's checked for by _identity_ - `math.nan is math.nan`. `float('nan')`, by contrast, creates a new object each time it's called. – jonrsharpe May 16 '22 at 20:21
Not obvious but makes sense. Thanks! – Donald McLachlan May 16 '22 at 20:28
Note that you can get the same result by defining `nan = float('nan')` and using that (which is what is essentially happening when you use `math.nan`) – Pranav Hosangadi May 16 '22 at 20:39

python float('nan') vs math.nan

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