Nathaniel Smith wrote: > 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. >> >> -- Sorry, I was a bit confused there. Actually, c = np.dot(a, b) was just what I needed.
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