Not quite. Bincount is fine if you have a set of approximately sequential numbers. But if you don't....
>>> a = numpy.array((1,500,1000)) >>> a array([ 1, 500, 1000]) >>> b = numpy.bincount(a) >>> b array([0, 1, 0, ..., 0, 0, 1]) >>> len(b) 1001 -Mark On Wed, Jun 1, 2011 at 9:32 AM, Skipper Seabold <jsseab...@gmail.com> wrote: > On Wed, Jun 1, 2011 at 11:31 AM, Mark Miller <markperrymil...@gmail.com> > wrote: >> I'd love to see something like a "count_unique" function included. The >> numpy.unique function is handy, but it can be a little awkward to >> efficiently go back and get counts of each unique value after the >> fact. >> > > Does bincount do what you're after? > > [~/] > [1]: x = np.random.randint(1,6,5) > > [~/] > [2]: x > [2]: array([1, 3, 4, 5, 1]) > > [~/] > [3]: np.bincount(x) > [3]: array([0, 2, 0, 1, 1, 1]) > > Skipper > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
