On 10/02/2008, Matthew Brett <[EMAIL PROTECTED]> wrote: > > Ah, I see. You definitely do not want to reassign the .data buffer in > > this case. An out= parameter does not reassign the memory location > > that the array object points to. It should use the allocated memory > > that was already there. It shouldn't "copy" anything at all; > > otherwise, "median(x, out=out)" is no better than "out[:] = > > median(x)". Personally, I don't think that a function should expose an > > out= parameter unless if it can make good on that promise of memory > > efficency. > > I agree - but there are more efficient median algorithms out there > which can make use of the memory efficiently. I wanted to establish > the call signature to allow that. I don't feel strongly about it > though.
This is a startling claim! Are there really median algorithms that are faster for having the use of a single float as storage space? If it were permissible to mutilate the original array in-place, I can certainly see a good median algorithm (based on quicksort, perhaps) being faster, but modifying the input array is a different question from using an output array. I can also see that this could possibly be improved by using a for loop to iterate over the output elements, so that there was no need to duplicate the large input array, or perhaps a "blocked" iteration that duplicated arrays of modest size would be better. But how can a single float per data set whose median is being taken help? Anne _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
