numpy.isnan(value) not the same as value == numpy.nan?

Question

Why am I getting the following:

>>> v
nan
>>> type(v)
<type 'numpy.float64'>
>>> v == np.nan
False
>>> np.isnan(v)
True

I would have thought the two should be equivalent?

Also see: stackoverflow.com/a/1573715/325565 (not directly python-related, but written by a member of the IEEE-754 committee that defined why this is the way it is) — Joe Kington
– Joe Kington, Commented Apr 9, 2015 at 1:27
I guess it makes sense that two undefined values cannot be compared as identical because they are by definition undefined. Just a little confusing when you get a nan != nan error the first time! — Alexander McFarlane
– Alexander McFarlane, Commented Apr 9, 2015 at 1:31

user2357112 · Accepted Answer · 2015-04-09 01:24:43Z

35

nan != nan. That's just how equality comparisons on nan are defined. It was decided that this result is more convenient for numerical algorithms than the alternative. This is specifically why isnan exists.

answered Apr 9, 2015 at 1:24

user2357112

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2 Comments

Alexander McFarlane Over a year ago

Thanks! That thoroughly confused me for a while when debugging :)

jedwards Over a year ago

By the way, this is likely due to IEEE 754 which states: Four mutually exclusive relations are possible: less than, equal, greater than, and unordered. The last case arises when at least one operand is NaN. Every NaN shall compare unordered with everything, including itself. Edit: Looks like I was beaten by ~2 hours by a comment with a link that goes into more detail -- see Joe Kington's comment to the question.

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