On Thu, May 22, 2008 at 9:22 AM, Robin <[EMAIL PROTECTED]> wrote: > On Thu, May 22, 2008 at 4:59 PM, 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. > > I would use bincount: > count = bincount(items) > should be all you need:
I guess bincount is *little* faster: >> items = mp.random.randint(0, 100, (1000000,)) >> timeit mp.bincount(items) 100 loops, best of 3: 4.05 ms per loop >> items = items.tolist() >> timeit [items.count(i) for i in range(100)] 10 loops, best of 3: 2.91 s per loop _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
