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Let's say I have a a very large list of data in the format {{n,f[n]}} so something like {{1,600},{2,700},{3,1100}...} and if this data is plotted, then it will look like a sum of Gaussian functions+a quadratic function when fitted (ie it will have multiple peaks, but it will be sitting on a quadratic curve).

My goal is to remove the background, in this case probably a quadratic function so that the peaks sit on the horizontal axis. I have thought of a way to do this, but I cannot do the list manipulations.

The method is basically look for the minimum data-value f[n] every 20 datapoints across (or bin of width 20). Now in every bin we have one 'minimum' datapoint. Then for every bin, search for all datavalues that is at least 6sigma higher than its corresponding 'minimum' datapoint(sigma is Sqrt[f[n]], and add these 'high' points to a list called background. If a 'high' point doesn't exist within the bin, add the minimum datapoint to background instead. Now we have a list of background points which are essentially points which 'hug' the bottom of the dataset, and removed the outliers which are too low

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    $\begingroup$ How is this different than the question you asked 6 hours ago that has almost exactly the same title (and it has an answer by the way)? What about that answer wasn't satisfactory? It's likely that you should just edit the previous post with whatever new details are in this question. $\endgroup$
    – march
    Oct 30, 2015 at 3:39
  • $\begingroup$ Sorry I forgot i posted it $\endgroup$ Oct 30, 2015 at 3:49
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ Oct 30, 2015 at 5:25

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