Skip to main content

list manipulation - Finding Local Minima / Maxima in Noisy Data


I'm trying to find local minima / maxima in noisy data, consisting of data values taken at certain time intervals. Ideally, the function should take a pair of lists (one containing time values and one containing observed data values) and return the coordinates of the maxima and minima.


An example of the data is found below:


temptimelist = Range[200]/10;
tempvaluelist = Sinc[#] &@temptimelist + RandomReal[{-1, 1}, 200]*0.02;


While the questions here, here and here have a good range of answers, they don't fit with my requirements since (as far as I can see) they determine whether data is a local maximum / minimum by comparing it with the adjacent values. If I apply such an algorithm, my extrema will be identified as follows:


data with peaks identified


What I've done is to write the following code.


NoisyExtremaFinder = Function[{timeList, valueList, aroundRange},
(*NoisyExtremaFinder[] takes a pair of lists timeList_ and valueList_ and determines extrema in valueList_, returning a list containing the coordinates (as pairs of time_ and value_) of the minima as the first entry and the maxima as the second entry.

As the data is assumed to be noisy, we provide the option aroundRange_ to allow the user to determine the sensitivity of the search. Specifically, when aroundRange_=n, the function will compare each value with the preceding and subsequent n values to determine whether it is an extrema*)

extremaPosition =
Flatten@Position[

Map[#, Partition[valueList, 2*aroundRange + 1, 1, {-(1 + aroundRange), 1 + aroundRange}, {}]] - valueList, 0.] &;
(*extremaPosition[] is a custom function that determines the position of local Maxima or Minima in valueList_, with the sensitivity determined by the value of aroundRange. When aroundRange_=n, the function will compare each value with the preceding and subsequent n values to determine whether it is an extrema. You can either do extremaPosition[Max] or extremaPosition[Min] *)

extremaPoints =
Transpose@{timeList[[#]], valueList[[#]]} &@extremaPosition[#] &;
(*extremaPoints[] is a custom function that determines the coordinates of local Maxima or Minima in valueList_

Custom Functions Used: extremaPosition[]*)

{extremaPoints[Min], extremaPoints[Max]}];


Basically, instead of deciding whether a point is an extrema by comparing it only with adjacent terms, it decides whether the point is an extrema by comparing with the aroundRange preceeding and aroundRange subsequent points.


Then, by adding the following lines of code, we can see the final results:


NoisyExtremaFinder[temptimelist, tempvaluelist, 10]; (*parameter "10" chosen by estimating the distance from peak to peak*)
ListPlot[Transpose[{temptimelist, tempvaluelist}],
Epilog -> {PointSize[Medium], Red, Point[extremaPoints[Min]], Green,
Point[extremaPoints[Max]]}, ImageSize -> 700]

better peaks




Is there any better way to code to achieve the goal I have in mind of comparing a point with n preceeding and subsequent points?


I was also advised by rm-rf (in chat) to consider smoothing the noisy data first before trying to find the local minima / maxima. However, I was concerned that I could add unwanted artifacts into the data by the smoothing process. With regards to smoothing, is there an algorithm that is commonly used to filter experimental data? Filters that I have looked at include the Savitsky-Golay filter and the Low-Pass filter.



PS: My data sets have around 10,000 data points each.




Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.