[Numpy-discussion] comparing arrays with NaN in them.
Fri Aug 24 10:06:35 CDT 2007
There may be multiple nan-s, but what Chris did is simply create one
with the same nan's
>>> a = N.array((1,2,3,N.nan))
>>> b = N.array((1,2,3,N.nan))
I think these should be the same.
Can anybody give me a good reason why they shouldn't, because it could
confuse a lot of people?
ps. I have to admit though, that matlab does the same thing. nan==nan
On Aug 24, 4:51 am, Warren Focke <fo...@slac.stanford.edu> wrote:
> On Thu, 23 Aug 2007, Christopher Barker wrote:
> > but that feels like a kludge. maybe some sort of "TheseArrays are binary
> > equal" would be useful.
> But there are multiple possible NaNs, so you couldn't rely on the bits
> Maybe something with masked arrays?
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