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). -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco ------------------------------------------------------------------------- Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642 _______________________________________________ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion
