Efficiently multiply elements of each row together

Question

Given a ndarray of size (n, 3) with n around 1000, how to multiply together all elements for each row, fast? The (inelegant) second solution below

Answer 1

np.prod accepts an axis argument:

np.prod(a, axis=1)

With axis=1, the column-wise product is computed for each row.

Sanity check

assert np.array_equal(np.prod(a, axis=1), prod1(a))

Performance

17.6 µs ± 146 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

(1000x speedup)