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    <p>The first improvement would be to simplify the three cases in the main loop: Rather than iterating while some of the sequence has elements, iterate while <em>both</em> sequences have elements. When leaving the loop, one of them will be empty, we don't know which, but we don't care: We append them at the end of the result.</p> <pre><code>def msort2(x): if len(x) &lt; 2: return x result = [] # moved! mid = int(len(x) / 2) y = msort2(x[:mid]) z = msort2(x[mid:]) while (len(y) &gt; 0) and (len(z) &gt; 0): if y[0] &gt; z[0]: result.append(z[0]) z.pop(0) else: result.append(y[0]) y.pop(0) result += y result += z return result </code></pre> <p>The second optimization is to avoid <code>pop</code>ping the elements. Rather, have two indices:</p> <pre><code>def msort3(x): result = [] if len(x) &lt; 2: return x mid = int(len(x) / 2) y = msort3(x[:mid]) z = msort3(x[mid:]) i = 0 j = 0 while i &lt; len(y) and j &lt; len(z): if y[i] &gt; z[j]: result.append(z[j]) j += 1 else: result.append(y[i]) i += 1 result += y[i:] result += z[j:] return result </code></pre> <p>A final improvement consists in using a non recursive algorithm to sort short sequences. In this case I use the built-in <code>sorted</code> function and use it when the size of the input is less than 20:</p> <pre><code>def msort4(x): result = [] if len(x) &lt; 20: return sorted(x) mid = int(len(x) / 2) y = msort4(x[:mid]) z = msort4(x[mid:]) i = 0 j = 0 while i &lt; len(y) and j &lt; len(z): if y[i] &gt; z[j]: result.append(z[j]) j += 1 else: result.append(y[i]) i += 1 result += y[i:] result += z[j:] return result </code></pre> <p>My measurements to sort a random list of 100000 integers are 2.46 seconds for the original version, 2.33 for msort2, 0.60 for msort3 and 0.40 for msort4. For reference, sorting all the list with <code>sorted</code> takes 0.03 seconds.</p>
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