On 11/11/06, Robert Kern <[EMAIL PROTECTED]> wrote:
I think the problem is that the max and min functions use the first value in the array as the starting point. That could be fixed by using the first non-nan and returning nan if there aren't any "real" numbers. But it probably isn't worth the effort as the behavior becomes more complicated. A better rule of thumb is to note that comparisons involving nans are basically invalid because nans aren't comparable -- the comparison violates trichotomy. Don't really know what to do about that.
Chuck
Keith Goodman wrote:
> x.min() and x.max() depend on the ordering of the elements:
>
>>> x = M.matrix([[ M.nan, 2.0, 1.0]])
>>> x.min()
> nan
>
>>> x = M.matrix([[ 1.0, 2.0, M.nan]])
>>> x.min()
> 1.0
>
> If I were to try the latter in ipython, I'd assume, great, min()
> ignores NaNs. But then the former would be a bug in my program.
>
> Is this related to how sort works?
Not really. sort() is a more complicated algorithm that does a number of
different comparisons in an order that is difficult to determine beforehand.
x.min() should just be a straight pass through all of the elements. However, the
core problem is the same: a < nan, a > nan, a == nan are all False for any a.
Barring a clever solution (at least cleverer than I feel like being
immediately), the way to solve this would be to check for nans in the array and
deal with them separately (or simply ignore them in the case of x.min()).
However, this checking would slow down the common case that has no nans (sans
nans, if you will).
I think the problem is that the max and min functions use the first value in the array as the starting point. That could be fixed by using the first non-nan and returning nan if there aren't any "real" numbers. But it probably isn't worth the effort as the behavior becomes more complicated. A better rule of thumb is to note that comparisons involving nans are basically invalid because nans aren't comparable -- the comparison violates trichotomy. Don't really know what to do about that.
Chuck
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