On Thu, Aug 11, 2011 at 4:43 PM, Jose Borreguero <borregu...@gmail.com> wrote: > a = random.randn(3,3) > b = a.reshape(1,3,3).repeat(50,axis=0) > scipy.linalg.block_diag( *b ) >
slightly simpler, but equivalent, code: b = [a]*50 scipy.linalg.block_diag( *b) Cheers, f _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
