[Numpy-discussion] min() of array containing NaN

Andrew Dalke dalke@dalkescientific....
Tue Aug 12 18:13:35 CDT 2008

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.

That'll break consistency with the normal 'max'
function in Python.

> Really nanmin of an array containing only nans should be the same
> as an empty array; both should be infinity.

One thing I expect is that if min(x) exists then there is
some i where x[i] "is" min(x) .  Returning +inf for min([NaN])
breaks that.

However, my expectation doesn't hold true for Python.  If
I use Python's object identity test 'is' then object
identity is lost in numpy.min, although it is preserved
under Python's min:

 >>> import numpy as np
 >>> x = [200, 300]
 >>> np.min(x)
 >>> np.min(x) is x[0]
 >>> min(x) is x[0]

and if I use '==' for equality testing then my
expectation will fail if isnan(x[i]) because
then x[i] != x[i].

 >>> import numpy as np
 >>> np.nan
 >>> np.nan == np.nan

So when I say "is" I means "acts the same as
except for in some strange corner cases".

Or to put it another way, it should be possible
to implement a hypothetical 'argnanmin' just
like there is an 'argmin' which complements 'min'.

> I guess this is a problem for types that don't have an infinity
> (returning maxint is a poor substitute), but what is the correct
> behaviour here?

"Doctor, doctor it hurts when I do this."
"Well, don't do that."

Raise an exception.  Refuse the temptation to guess.
Force the user to handle this case as appropriate.

Personally, I expect that if my array 'x' has a NaN then
min(x) must be a NaN.


More information about the Numpy-discussion mailing list