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    <p>How to optimize this will depend on some properties of your data, but here are some ideas:</p> <ol> <li><p>Have you tried profiling the code? Using one of the <a href="http://docs.python.org/2/library/profile.html" rel="nofollow">Python profilers</a> can give you some useful information about what's taking the most time. Often, a piece of code you've just written will have one biggest bottleneck, and it's not always obvious which piece it is; profiling lets you figure that out and attack the main bottleneck first.</p></li> <li><p>Do you know what typical values of <code>i</code> are? If you have some idea, you can speed things up by starting with <code>i</code> greater than 0 (as @vhallac noted), or by increasing <code>i</code> by larger amounts — if you often see big values for <code>i</code>, increase <code>i</code> by 2 or 3 at a time; if the distribution of <code>i</code>s has a long tail, try doubling it each time; etc.</p></li> <li><p>Do you need all the data when doing the least squares regression? If that function call is the bottleneck, you may be able to speed it up by using only some of the data in the range. Suppose, for instance, that at a particular point, you need <code>i</code> to be 200 to see a large enough (above-noise) change in the data. But you may not need all 400 points to get a good estimate of the slope — just using 10 or 20 points, evenly spaced in the <code>start:end</code> range, may be sufficient, and might speed up the code a lot.</p></li> </ol>
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    1. COThanks for your suggestions. From profiling I learned that the regression takes mere seconds, by far most of the execution time is taken up by the nested loop. The value for `i` ranges between 3 and 1500 typically. My main problem is that the data behaves a bit like an exp(-x) function, so starting with a steep slope, followed by strong curvature finishing with a very gentle slope. But I'll give it a try using a different increment of `i`
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