[Numpy-discussion] min() of array containing NaN
Wed Aug 13 14:37:09 CDT 2008
>On Tue, Aug 12, 2008 at 19:28, Charles R Harris
>> On Tue, Aug 12, 2008 at 5:13 PM, Andrew Dalke <email@example.com>
>>> On Aug 12, 2008, at 9:54 AM, Anne Archibald wrote:
>>> > Er, is this actually a bug? I would instead consider the fact that
>>> > np.min() raises an exception a bug of sorts - the identity of min is
>>> > inf.
>>> Personally, I expect that if my array 'x' has a NaN then
>>> min(x) must be a NaN.
>> I suppose you could use
>> min(a,b) = (abs(a - b) + a + b)/2
>> which would have that effect.
>Or we could implement the inner loop of the minimum ufunc to return
>NaN if there is a NaN. Currently it just compares the two values
>(which causes the unpredictable results since having a NaN on either
>side of the < is always False). I would be amenable to that provided
>that the C isnan() call does not cause too much slowdown in the normal
While you're doing that, can you do it so that if keyword nan=False it
returns NaN if NaNs exist, and if keyword nan=True it ignores NaNs?
We can argue which should be the default (see my prior post). Both
are compatible with the current undefined behavior.
I assume that the fastest way to do it is two separate loops for the
separate cases, but it might be fast enough straight (with a
conditional in the inner loop), or with some other trick (macro magic,
function pointer, whatever).
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