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downsample data adapively/​"intellige​ntly"

조회 수: 15(최근 30일)
Andreas 2021년 9월 16일
편집: Andreas 2021년 9월 17일
I have a set of data that contains ~7800 data points (the red curve in the figure). I have reduced the number of data points with downsample() to 19 data points (blue curve). However, I would like to position the points more "intelligently". Is there a function in Matlab where the points are placed with some kind of least square minimization/adaptive spacing? The data is currently stored in a struct (data.Stress and data.Strain).
The number of data points must be possible to fix by the user (e.g. =19 in this case).
The most important part of the curve is the beginning, but I think that a general minimization would give a good enough result.
Adding a point between the first and second point and remoing one point where the curve is more or less linear would give a good-enough result. But modifying the curve manually isn't so attractive as there are quite a few curves to process.
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Mathieu NOE
Mathieu NOE 2021년 9월 17일
I would suggest you do a linear interpolation, but the new x vector should have more points in the beginning compared to the end of the curve
you can try different x axis distributions (linear decreasing or exponential) ....

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채택된 답변

Jan 2021년 9월 17일
편집: Jan 2021년 9월 17일
This is not a trivial problem. In the general case it is a global optimization problem and there can be a huge number of equivalent solutions.
The Douglas-Peuker-algorithm is a fair approach, if the input is given as discrete set of points, see: or
An alternative: Calculate the cumulative sum of the absolute values of the 2nd derivative. Now find rounded equidistant steps on this line. The higher the curvature, the more points you get:
x = linspace(0, 4*pi, 500);
y = sin(x);
n = 21; % Number of points
ddy = gradient(gradient(y, x), x); % 2nd derivative
sy = cumsum(abs(ddy));
idx = interp1(sy, 1:numel(x), linspace(sy(1), sy(end), n), 'nearest');
plot(x, y);
hold on
plot(x(idx), y(idx), 'ro');
This looks very nice even for a small number of points. To my surprise it looks such good, that this must be a standard approach. Does anybody know, how this algorithm is called?

추가 답변(1개)

Andreas 2021년 9월 17일
편집: Andreas 2021년 9월 17일
Thanks @Jan and @Mathieu NOE, I used bits and pieces from your input to this (principal) code
x = linspace(0,10,500);
y = log(x);
hold on
n = 12;
pot = 2.0;
% logarithmic
scale = x(end)/(10^pot-1.0);
red_x = (logspace(0,pot,n)-1.0)*scale;
red_y = interp1(x,y,red_x,'nearest');
red_y(1) = -3.9; %"cheating" for this example, as ln(0) -> -inf
% linear
red_x = linspace(0,x(end),n);
red_y = interp1(x,y,red_x,'nearest');
red_y(1) = -3.9;
legend('Function log(x)','logarithmic distribution','linear distribution')
It gives a better interpolation for my types of curves with logarithmic distribution. Below is an example of my "real" curves.




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