On Thu, May 22, 2008 at 9:15 AM, Keith Goodman <[EMAIL PROTECTED]> wrote: > On Thu, May 22, 2008 at 9:08 AM, Keith Goodman <[EMAIL PROTECTED]> wrote: >> On Thu, May 22, 2008 at 8:59 AM, Kevin Jacobs <[EMAIL PROTECTED]> >> <[EMAIL PROTECTED]> wrote: >>> After poking around for a bit, I was wondering if there was a faster method >>> for the following: >>> >>> # Array of index values 0..n >>> items = numpy.array([0,3,2,1,4,2],dtype=int) >>> >>> # Count the number of occurrences of each index >>> counts = numpy.zeros(5, dtype=int) >>> for i in items: >>> counts[i] += 1 >>> >>> In my real code, 'items' contain up to a million values and this loop will >>> be in a performance critical area of code. If there is no simple solution, >>> I can trivially code this using the C-API. >> >> How big is n? If it is much smaller than a million then loop over that >> instead. > > Or how about using a list instead: > >>> items = [0,3,2,1,4,2] >>> uitems = frozenset(items) >>> count = [items.count(i) for i in uitems] >>> count > [1, 1, 2, 1, 1]
Oh, I see, so uitems should be range(n) _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
