If we were at so or ask.scipy I would vote for Mark's solution :) Usually in cases like yours, I tend to use the shortest version of the solutions.
On Tue, Jun 8, 2010 at 2:08 PM, Andreas Hilboll <li...@hilboll.de> wrote: > Hi, > > > newtimes = [times[idx[x][y]] for x in range(2) for y in range(2)] > > np.array(newtimes).reshape(2,2) > > array([[104, 102], > > [103, 101]]) > > Great, thanks a lot! > > Cheers, > > Andreas. > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > -- Gökhan
_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
