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] _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
