Numpy efficient matrix self-multiplication (gram matrix)

Question

I want to multiply B = A @ A.T in numpy. Obviously, the answer would be a symmetric matrix (i.e. B[i, j] == B[j, i]).

However, it is not cl

Answer 1

As noted in @PaulPanzer's link, dot can detect this case. Here's the timing proof:

In [355]: A = np.random.rand(1000,1000)
In [356]: timeit A.dot(A.T)
57.4 ms ± 960 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [357]: B = A.T.copy()
In [358]: timeit A.dot(B)
98.6 ms ± 805 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Numpy dot too clever about symmetric multiplications

Answer 2

You can always use sklearns's pairwise_distances

Usage:

from sklearn.metrics.pairwise import pairwise_distances
gram = pairwise_distance(x, metric=metric)

Where metric is a callable or a string defining one of their implemented metrics (full list in the link above)

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But, I wrote this for myself a while back so I can share what I did:

import numpy as np

def computeGram(elements, dist):
    n    = len(elements)
    gram = np.zeros([n, n])
    for i in range(n):
        for j in range(i + 1):
            gram[i, j] = dist(elements[i], elements[j])

    upTriIdxs       = np.triu_indices(n)
    gram[upTriIdxs] = gram.T[upTriIdxs]

    return gram

Where dist is a callable, in your case np.inner