7

I am trying to understand the FindPeaks function in V10. I have two questions. Here is some measured data with noise. 

data = {{182.6`, 0.08622910543312512`}, {182.7`, 
0.08537299262452944`}, {182.79999999999998`, 
0.08509477035006817`}, {182.9`, 0.0845805258347419`}, {183.`, 
0.0833228764826812`}, {183.1`, 0.0839126306989993`}, {183.2`, 
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0.0830234147404793`}, {183.5`, 0.08134478053584769`}, {183.6`, 
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0.07955082017792586`}, {184.`, 0.07920192550765708`}, {184.1`, 
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0.07586422613054548`}, {185.39999999999998`, 
0.07503402555148043`}, {185.5`, 0.07535492258842982`}, {185.6`, 
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0.07475705268743123`}, {185.9`, 0.07411938622704159`}, {186.`, 
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0.07336815610833888`}, {186.5`, 0.07363242236825435`}, {186.6`, 
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0.0733269167127279`}, {186.89999999999998`, 
0.0731251716924564`}, {187.`, 0.07260278352658113`}, {187.1`, 
0.07306968191644458`}, {187.2`, 
0.07290770361093614`}, {187.29999999999998`, 
0.07283251650826833`}, {187.4`, 0.07283442908533787`}, {187.5`, 
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0.07245508553439435`}, {188.2`, 0.07184010069755509`}, {188.3`, 
0.07221620673218568`}, {188.39999999999998`, 
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0.07250651648005164`}, {188.79999999999998`, 
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0.07529888198343905`}, {192.89999999999998`, 
0.07570648686665757`}, {193.`, 
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0.031440349336531115`}, {198.2`, 
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0.03241545932012303`}, {198.5`, 
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0.03273789660624955`}, {198.7`, 0.032659053442336675`}, {198.8`, 
0.032791614417334115`}, {198.89999999999998`, 
0.032825069131018285`}, {199.`, 
0.03351436361528482`}, {199.09999999999997`, 
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0.035955794122767386`}, {200.8`, 0.03571660745453507`}, {200.9`, 
0.035964735338650945`}, {201.`, 0.03617643844690327`}, {201.1`, 
0.03564424939619923`}, {201.2`, 
0.036812938339660095`}, {201.29999999999998`, 
0.036710308908121075`}};

If I drop the x values I can get FindPeaks to work well

ListPlot[{data[[All, 2]], FindPeaks[data[[All, 2]], 5]}]

Mathematica graphics

This does a good job of finding the local maximum at about 100. Question: how does the second parameter work? Is the scale of the Gaussian blurr just the number of points? I loose my maximum if I set the scale to 19 and have too many peaks if it is 2.

Second question. How do I get my x axis values back? Below is a terrible hack. Is there a better method?

a =.; b =.;
a = FindPeaks[data[[All, 2]], 5];
b = Transpose[{data[[All, 1]][[a[[All, 1]]]], a[[All, 2]]}];
ListPlot[{data, b}]

Mathematica graphics

A good reference not using FindPeaks can be found here.

Thanks

Hugh
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2 Answers2

3

Regarding your second question:

Second question. How do I get my x axis values back? Below is a terrible hack. Is there a better method?

I think it is best to use TimeSeries:

ts = TimeSeries[data];
ListPlot[{ts, FindPeaks[ts, 0.1]}]

In order to select a parameter for FindPeaks it might be better to use these plot commands:

ts = TimeSeries[data];
Show[{ListLinePlot[ts, PlotStyle -> Gray], 
  ListPlot[FindPeaks[ts, 2], PlotStyle -> {Red, PointSize -> 0.004}]}, ImageSize -> 800]

enter image description here

Anton Antonov
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  • Thanks for pointing out TimeSeries. This is an old question and I think it predates TimeSeries. It looks like it could be useful I must find out more about it. – Hugh Sep 25 '15 at 21:20
  • @Hugh Sure! I also have a solution of not using FindPeaks. Are you interested in using FindPeaks or not? Do you have some criteria with which you are/were able to judge which set of peaks is good or not? – Anton Antonov Sep 25 '15 at 21:34
  • The trouble with this data is that I want just one peak and too much smoothing looses the peak all together. This is just one interval in a long time series and I can't afford to tune each peak individually. – Hugh Sep 25 '15 at 21:41
  • @Hugh I applied Quantile Regression to this problem and since you posted this link: https://mathematica.stackexchange.com/questions/23828/finding-local-minima-maxima-in-noisy-data , I went ahead and implemented a solution for that problem and posted an answer. Would that or similar solution be of interested to you? – Anton Antonov Sep 27 '15 at 19:53
2

In a comment, Bob Hanlon offers a simpler solution to your second problem (getting the x-axis values back), namely:

b = data[[First /@ a]]

Although you could of course convert to a TimeSeries[] and work on that rather than the 1D data.

Now, with regards to the first question, you can investigate the behaviour of the scale parameter by Gaussian-filtering the data yourself, which is all FindPeaks[] is doing:

ListLinePlot[
    Transpose[{data[[All, 1]], GaussianFilter[data[[All, 2]], {2*#, #}]}] & /@ {5, 20, 50}]

You can see how the maxima disappears as the scale parameter increases, and see which peaks survive the Gaussian blur as described in the documentation. You can also specify a third "sharpness" parameter if necessary.

enter image description here

dr.blochwave
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    Thank you. This is an old post but one I am still trying to understand. I now see that there is a special Gaussian kernel which FindPeaks uses. I had previously been thinking that I had to do the filtering myself prior to using FindPeaks – Hugh Sep 25 '15 at 21:30
  • @Hugh glad I can help - if you look in the documentation for FindPeaks you'll see it uses an example similar to mine abou the GaussianFilter step. – dr.blochwave Sep 25 '15 at 21:40
  • I suspect I want the reverse of sharpness. I only want one peak in the interval. Too much smoothing and there is no peak too little and there are several. Furthermore this is just one part of a much longer time history which I can't look at manually. – Hugh Sep 25 '15 at 21:54
  • @Hugh does the 4th option of all peaks above a threshold help at all? FindPeaks[data,σ,s,t] finds only peaks with values greater than t, and you can also set independent scales for sharpness and threshold. – dr.blochwave Sep 25 '15 at 21:57
  • I have a huge range in my data with peaks in valleys and small valleys between some peaks and rising and falling trends so a threshold is not going to work. I am looking at vibration spectra. – Hugh Sep 25 '15 at 22:02
  • @Hugh is this question of any use? http://mathematica.stackexchange.com/questions/5575/how-to-find-all-the-local-minima-maxima-in-a-range – dr.blochwave Sep 25 '15 at 22:04
  • No, I have noise so I need some smoothing. That's a nice post though. – Hugh Sep 25 '15 at 22:06
  • @Hugh I guess you could smooth before using the answers in the link. But automatically determining parameters given huge variations in the data range may well be beyond FindPeaks[]! – dr.blochwave Sep 25 '15 at 22:08