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copied!<p>The following code gives a nice increase in performance that depends upon how dense the numbers are. Using a set of 1000 random numbers, sampled uniformly between 0 and 100, it runs about 30 times faster than your implementation.</p> <pre><code>pos_1_start = 0 for i in range(np.size(vector1)): for j in range(pos1_start, np.size(vector2)): if np.abs(vector1[i] - vector2[j]) < .02: results1 += [(vector1[i], vector2[j], i, j)] else: if vector2[j] < vector1[i]: pos1_start += 1 else: break </code></pre> <p>The timing:</p> <pre><code>time new method: 0.112464904785 time old method: 3.59720897675 </code></pre> <p>Which is produced by the following script:</p> <pre><code>import random import numpy as np import time # initialize the vectors to be compared vector1 = [random.uniform(0, 40) for i in range(1000)] vector2 = [random.uniform(0, 40) for i in range(1000)] vector1.sort() vector2.sort() # the arrays that will contain the results for the first method results1 = [] # the arrays that will contain the results for the second method results2 = [] pos1_start = 0 t_start = time.time() for i in range(np.size(vector1)): for j in range(pos1_start, np.size(vector2)): if np.abs(vector1[i] - vector2[j]) < .02: results1 += [(vector1[i], vector2[j], i, j)] else: if vector2[j] < vector1[i]: pos1_start += 1 else: break t1 = time.time() - t_start print "time new method:", t1 t = time.time() for lv1 in range(np.size(vector1)): for lv2 in range(np.size(vector2)): if np.abs(vector1[lv1]-vector2[lv2])<.02: results2 += [(vector1[lv1], vector2[lv2], lv1, lv2)] t2 = time.time() - t_start print "time old method:", t2 # sort the results results1.sort() results2.sort() print np.allclose(results1, results2) </code></pre>

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