a = np.empty(3*n.size, np.int) a[::3]=n a[1::3]=m a[2::3]=o or np.array(zip(n,m,o)).ravel()
but the first solution is faster, even if you have to write more :D Emmanuelle On Sun, Jun 14, 2009 at 04:11:29PM +0200, Robert wrote: > whats the right way to efficiently weave arrays like this ? : > >>> n > array([1, 2, 3, 4]) > >>> m > array([11, 22, 33, 44]) > >>> o > array([111, 222, 333, 444]) > => > [ 1, 11, 111, 2, 22, 222, 3, 33, 333, 4, 44, 444] > _______________________________________________ > 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
