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    <p>Instead of summing the dot product of each pair, which requires <code>n * m</code> operations, you can sum all of the vectors in each list and just do the final dot product, which will only take <code>n + m</code> operations.</p> <p>Before:</p> <pre><code>def calc_slow(L1, L2): result = 0 for n, m in itertools.product(L1, L2): result += np.dot(n, m) return result </code></pre> <p>After:</p> <pre><code>def calc_fast(L1, L2): L1_sums = np.zeros(len(L1[0])) L2_sums = np.zeros(len(L2[0])) for vec in L1: L1_sums += vec for vec in L2: L2_sums += vec return np.dot(L1_sums, L2_sums) </code></pre> <p>EDIT: Even faster version, taking advantage of numpy:</p> <pre><code>def calc_superfast(L1, L2): return np.dot(np.array(L1).sum(0), np.array(L2).sum(0)) </code></pre> <p>The output is the same:</p> <pre><code>print X[0], Y[0], calc_slow(X[0], Y[0]) print X[0], Y[0], calc_fast(X[0], Y[0]) </code></pre> <p>prints:</p> <pre><code>[[1, 2, 3], [2, 4, 5], [2, 3, 4]] [[12, 14, 15], [12, 13, 14]] 711 [[1, 2, 3], [2, 4, 5], [2, 3, 4]] [[12, 14, 15], [12, 13, 14]] 711.0 </code></pre> <p>Timing it shows that there is significant improvement. Timing code:</p> <pre><code>import random import time def rand_vector(size=3): return [random.randint(1, 100) for _ in xrange(3)] def rand_list(length=200): return [rand_vector() for _ in xrange(length)] print "Generating lists..." L1 = rand_list(200) L2 = rand_list(200) print "Running slow..." s = time.time() print calc_slow(L1, L2) print "Slow for (%d, %d) took %.2fs" % (len(L1), len(L2), time.time() - s) print "Running fast..." s = time.time() print calc_fast(L1, L2) print "Fast for (%d, %d) took %.2fs" % (len(L1), len(L2), time.time() - s) </code></pre> <p>Sample outputs:</p> <pre><code>Generating lists... Running slow... 75715569 Slow for (100, 100) took 1.48s Running fast... 75715569.0 Fast for (100, 100) took 0.03s Generating lists... Running slow... 309169971 Slow for (200, 200) took 5.29s Running fast... 309169971.0 Fast for (200, 200) took 0.04s Running fast... 3.05185703539e+12 Fast for (20000, 20000) took 1.94s </code></pre> <p>The operation for two lists of 20000 elements each only took ~2 seconds, whereas I predict it would take several hours with the old method.</p> <hr> <p>The reason this works is that you can group the operations together. Assuming you have the following lists:</p> <pre><code>L1 = [a, b, c], [d, e, f], [g, h, i] L2 = [u, v, w], [x, y, z] </code></pre> <p>Then summing the dot product of each pair would look like this:</p> <pre><code>a*u + b*v + c*w + a*x + b*y + c*z + d*u + e*v + f*w + d*x + e*y + f*z + g*u + h*v + i*w + g*x + h*y + i*z </code></pre> <p>You can factor out the <code>u</code>, <code>v</code>, <code>w</code>, <code>x</code>, <code>y</code>, and <code>z</code> elements:</p> <pre><code>u*(a + d + g) + v*(b + e + h) + w*(c + f + i) + x*(a + d + g) + y*(b + e + h) + z*(c + f + i) </code></pre> <p>Then you can further factor out those sums:</p> <pre><code>(u + x)*(a + d + g) + (v + y)*(b + e + h) + (w + z)*(c + f + i) </code></pre> <p>Which is just the dot product of the summed vectors from each initial list.</p>
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