On Wed, Dec 24, 2014 at 3:25 PM, Neal Becker <[email protected]> wrote: > What would be the most efficient way to compute: > > c[j] = \sum_i (a[i] * b[i,j]) > > where a[i] is a 1-d vector, b[i,j] is a 2-d array?
I think this formula is just np.dot(a, b). Did you mean c = \sum_j \sum_i (a[i] * b[i, j])? > This seems to be one way: > > import numpy as np > a = np.arange (3) > b = np.arange (12).reshape (3,4) > c = np.dot (a, b).sum() > > but np.dot returns a vector, which then needs further reduction. Don't know > if > there's a better way. > > -- > -- Those who don't understand recursion are doomed to repeat it > > _______________________________________________ > NumPy-Discussion mailing list > [email protected] > http://mail.scipy.org/mailman/listinfo/numpy-discussion -- Nathaniel J. Smith Postdoctoral researcher - Informatics - University of Edinburgh http://vorpus.org _______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
