How to segment a curve into linear and nonlinear pieces

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Ahmed Zohaib
Ahmed Zohaib el 19 de Jul. de 2023
Comentada: Bruno Luong el 19 de Jul. de 2023
I have a dataset which looks linear at the beginning (up to ~3.5 along x-axis) and then it shows and upward climb. I want to locate an ideal point uptill where the data is linear at the beginning and fit a line. Thanks a lot. The data is attached (1st column is X and 2nd column is Y).
  2 comentarios
Bruno Luong
Bruno Luong el 19 de Jul. de 2023
linear about 3.5 ? When I zoom in it is not clear where the linear starts. Not 3.5, not 2, may be on the interval (0,1)
Ahmed Zohaib
Ahmed Zohaib el 19 de Jul. de 2023
Thanks for the feedback. Okay, how do I locate the exact point where it loses linearity in between [0 1]?

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Respuesta aceptada

Chunru
Chunru el 19 de Jul. de 2023
load(websave("DatasetXY.mat", "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1437433/DatasetXY.mat"));
whos
Name Size Bytes Class Attributes XY 52651x2 842416 double cmdout 1x33 66 char
[~, ia] = unique(XY(:, 1)); % remove the duplicated points
XY = XY(ia, :);
plot(XY(:,1), XY(:,2));
gg = gradient(gradient(XY(:, 2), XY(:, 1)), XY(:, 1)); % 2nd derivative
gg = movmedian(gg, 20); % to remove some spikes
hold on;
whos
Name Size Bytes Class Attributes XY 818x2 13088 double cmdout 1x33 66 char gg 818x1 6544 double ia 818x1 6544 double
% ignore some initial points
ip = 20; % 20 points
gg((ip+1):end)
ans = 798×1
0 0 0 0 0 0 0 0 0 0
i = find(abs(gg((ip+1):end))>20, 1, 'first'); % 2000 as a threshod (adjust it)
i = ip+i
i = 301
plot(XY(i,1), XY(i, 2), 'ro');
%xlim([2 3])

Más respuestas (1)

Bruno Luong
Bruno Luong el 19 de Jul. de 2023
Editada: Bruno Luong el 19 de Jul. de 2023
Ths produces kind of "arbitrary" choice of the breakpoint, consider fitting the global curve with few line segments
load(websave("DatasetXY.mat", "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1437433/DatasetXY.mat"));
X=XY(:,1);
Y=XY(:,2);
% FEX https://uk.mathworks.com/matlabcentral/fileexchange/25872-free-knot-spline-approximation
% it takes a dozain of seconds here
pp = BSFK(X,Y,2,[],[],struct('Animation',true))
xbreaks=pp.breaks(2)
ybreaks=ppval(pp,xbreaks)
plot(xbreaks,ybreaks,'or','MarkerSize',15)
  1 comentario
Bruno Luong
Bruno Luong el 19 de Jul. de 2023
If I crop the data up to X <= 4 it finds the break at
xbreaks=2.3085
As I said it's a mmoving target. The data is more or less exponetial everywhere. It is just a question of scaling, totally subjective.
load(websave("DatasetXY.mat", "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1437433/DatasetXY.mat"));
X=XY(:,1);
Y=XY(:,2);
b = X < 4
% FEX https://uk.mathworks.com/matlabcentral/fileexchange/25872-free-knot-spline-approximation
% it takes a dozain of seconds here
pp = BSFK(X(b),Y(b),2,[],[],struct('Animation',true))
xbreaks=pp.breaks(2)
ybreaks=ppval(pp,xbreaks)
plot(xbreaks,ybreaks,'or','MarkerSize',15)

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