2013/7/8 Issam <[email protected]>: > On 7/8/2013 12:53 PM, Lars Buitinck wrote: >> cost = np.sum(np.einsum('ij,ji->i', diff, diff.T)) / (2 * n_samples) > Thanks for all the remarks! > > I found out that the `einsum` can be replaced simply by 'cost = > np.sum(diff**2)/ (2 * n_samples)' which is faster and more readable.
For vectors x, note that np.sum(x ** 2) == np.dot(x, x), and the latter can be a lot faster. -- Lars Buitinck Scientific programmer, ILPS University of Amsterdam ------------------------------------------------------------------------------ See everything from the browser to the database with AppDynamics Get end-to-end visibility with application monitoring from AppDynamics Isolate bottlenecks and diagnose root cause in seconds. Start your free trial of AppDynamics Pro today! http://pubads.g.doubleclick.net/gampad/clk?id=48808831&iu=/4140/ostg.clktrk _______________________________________________ Scikit-learn-general mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/scikit-learn-general
