2014-04-08 14:52 GMT+02:00 Nathaniel Smith <n...@pobox.com>: > On Tue, Apr 8, 2014 at 9:58 AM, Björn Lindqvist <bjou...@gmail.com> wrote: >> 2014-04-07 3:41 GMT+02:00 Nathaniel Smith <n...@pobox.com>: >>> So, I guess as far as I'm concerned, this is ready to go. Feedback welcome: >>> http://legacy.python.org/dev/peps/pep-0465/ >> >> Couldn't you please have made your motivation example actually runnable? >> >> import numpy as np >> from numpy.linalg import inv, solve >> >> # Using dot function: >> S = np.dot((np.dot(H, beta) - r).T, >> np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r)) >> >> # Using dot method: >> S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r) >> >> Don't keep your reader hanging! Tell us what the magical variables H, >> beta, r and V are. And why import solve when you aren't using it? >> Curious readers that aren't very good at matrix math, like me, should >> still be able to follow your logic. Even if it is just random data, >> it's better than nothing! > > There's a footnote that explains the math in more detail and links to > the real code this was adapted from. And solve is used further down in > the section. But running it is really what you want, just insert: > > beta = np.random.randn(10) > H = np.random.randn(2, 10) > r = np.random.randn(2) > V = np.random.randn(10, 10) > > Does that help? ;-)
Thanks! Yes it does help. Then I can see that this expression: np.dot(H, beta) - r Evaluates to a vector. And a vector transposed is the vector itself. So the .T part in this expression np.dot(H, beta) - r).T is unnecessary, isn't it? -- mvh/best regards Björn Lindqvist _______________________________________________ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
