On Sat, 4 Jul 2020 at 19:58, Christopher Barker <python...@gmail.com> wrote:
> Hmm.
>
>
> Since NaN is neither greater than nor less that anything, it seems the
> only correct answer to any
> Min,max,clamp involving a NaN is NaN.
>
>
Simplifying the signature, in Python we have:
def min(*iterable):
iterator = iter(iterable)
minimum = next(iterable)
for item in iterator:
if item < minimum:
minimum = item
return minimum
Due to this, min(0, float('nan')) == 0 and same for max. I would hence
expect clamp to behave similarly.
As a side note, the documentation actually underspecifies as in these kind
of situations the iterable overall does not have a well-defined "minimum"
(in a mathematical sense it simply does not have one):
- how you look at the list (forward or backwards)
- how you approach the comparison (item < minimum or not(minimum <= item))
change the behaviour:
- max({0}, {0, 2}, {0, 1}, {1}) = {0, 2} v.s. {0, 1} when you reverse the
iterable
- min(0, float('nan')) = 0 v.s. float('nan') when you change the comparison
Should this be clarified/specified in the docs?
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