You're just trying to do this...correct? >>> import numpy >>> items = numpy.array([0,3,2,1,4,2],dtype=int) >>> unique = numpy.unique(items) >>> unique array([0, 1, 2, 3, 4]) >>> counts=numpy.histogram(items,unique) >>> counts (array([1, 1, 2, 1, 1]), array([0, 1, 2, 3, 4])) >>> counts[0] array([1, 1, 2, 1, 1]) >>>
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. > _______________________________________________ > Numpy-discussion mailing list > Numpy-discussion@scipy.org > http://projects.scipy.org/mailman/listinfo/numpy-discussion >
_______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
