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    <h2>Most recent version:</h2> <pre><code>def zipij7(m=mem, ms=mems): cpy = sorted(ms + [m]) loc = cpy.index(m) return zip(ms[:(loc)], repeat(m)) + zip(repeat(m), ms[(loc):]) </code></pre> <p>Benches slightly faster for me than truppo's, slower by 30% than Michal's. (Looking into that now)</p> <hr> <p>I may have found my answer (for now). It seems I forgot about making a list comp version for `zipij()``:</p> <pre><code>def zipij1(m=mem, ms=mems, f=ij): return [f(i, m) for i in ms] </code></pre> <p>It still relies on my silly <code>ij()</code> helper function, so it doesn't win the award for brevity, certainly, but timings have improved:</p> <pre><code># 10000 1.27s # 50000 6.74s </code></pre> <p>So it is now my current "winner", and also does not need to generate more than one list, or use a lot of function calls, other than the <code>ij()</code> helper, so I believe it would also be the most efficient.</p> <p>However, I think this could still be improved... I think that making N <code>ij()</code> function calls (where N is the length of the resultant list) is not needed:</p> <ul> <li>Find at what index <code>mem</code> would fit into <code>mems</code> when ordered</li> <li>Split mems at that index into two parts</li> <li>Do <code>zip(part1, repeat(mem))</code></li> <li>Add <code>zip(repeat(mem), part2)</code> to it</li> </ul> <p>It'd basically be an improvement on <code>zipij4()</code>, and this avoids N extra function calls, but I am not sure of the speed/memory benefits over the cost of brevity. I will maybe add that version to this answer if I figure it out.</p>
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    1. POEfficient reordering of coordinate pairs (2-tuples) in a list of pairs in Python
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    1. COExactly how do you measure? You latest solution is slower than the inlined version I suggested. Even with an already sorted list and low "m" it is still slower. With a large "m" and a shuffled list it is 10 times slower.
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    2. COMy latest solution? Definitely not.. Both Michal and I are getting twice as fast as the previous best solutions. Check his answer. I am timing differently than he is, but we are still getting the same results. Our latest implementations differ slightly, though. Mine is slightly faster still.
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    3. COI've updated my answer now with the exact benchmark I'm running
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