How to fit a series of datapoints using 3 piecewise linear fit lines

3 visualizaciones (últimos 30 días)
I have the following datapoints:
xdata = [-4.09434456 -3.40119738 -2.99573227 -2.7080502 -2.48490665 -2.30258509 -2.14843441 -2.01490302 -1.89711998 -1.79175947 -1.69644929 -1.60943791 -1.5293952 -1.45528723 -1.38629436 -1.32175584 -1.26113122 -1.2039728 -1.14990558 -1.09861229 -1.04982212 -1.00330211 -0.95885035 -0.91629073 -0.87546874 -0.83624802 -0.7985077 -0.76214005 -0.72704873 -0.69314718 -0.66035736 -0.62860866 -0.597837 -0.56798404 -0.5389965 -0.51082562 -0.48342665 -0.4567584 -0.43078292 -0.40546511 -0.3807725 -0.35667494 -0.33314445 -0.31015493 -0.28768207 -0.26570317 -0.24419696 -0.22314355 -0.20252426 -0.18232156 -0.16251893 -0.14310084 -0.12405265 -0.10536052 -0.08701138 -0.06899287 -0.05129329 -0.03390155 -0.01680712 0. 0.0165293 0.03278982 0.04879016 0.06453852 0.08004271 0.09531018 0.11034806 0.12516314 0.13976194 0.15415068 0.16833531 0.18232156 0.19611488 0.20972053 0.22314355 0.23638878 0.24946086 0.26236426 0.27510329 0.28768207 0.30010459 0.31237469 0.32449605 0.33647224 0.34830669 0.36000273 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1.5040774 1.50777426 1.5114575 1.51512723 1.51878354 1.52242654 1.5260563 1.52967294 1.53327655 1.53686722 1.54044504 1.54401011 1.54756251 1.55110234 1.55462968 1.55814462 1.56164725 1.56513765 1.56861592 1.57208213 1.57553636 1.5789787 1.58240924 1.58582805 1.58923521 1.59263079 1.59601489 1.59938758 1.60274892 1.60609901 1.60943791 1.6127657 1.61608246 1.61938824 1.62268314 1.62596721 1.62924054 1.63250319 1.63575522 1.63899671 1.64222774 1.64544835 1.64865863 1.65185863 1.65504842 1.65822808 1.66139765 1.66455721 1.66770682 1.67084654 1.67397643 1.67709656 1.68020698 1.68330776 1.68639895 1.68948062 1.69255282 1.69561561 1.69866905 1.70171319 1.70474809 1.70777381 1.71079041 1.71379793 1.71679643 1.71978597 1.7227666 1.72573837 1.72870133 1.73165555 1.73460106 1.73753792 1.74046617 1.74338588 1.7462971 1.74919985 1.75209421 1.75498022 1.75785792 1.76072736 1.76358859 1.76644166 1.76928661 1.77212349 1.77495235 1.77777323 1.78058617 1.78339122 1.78618842 1.78897783 1.79175947 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which would give a spectrum (as shown). I would like to fit it using 3 piecewise linear lines and also find the first 'break-point' (just after the first zero slope line). Can anyone help me with that?
Thanks

Respuesta aceptada

Bruno Luong
Bruno Luong el 29 de Dic. de 2020
Editada: Bruno Luong el 29 de Dic. de 2020
Use this FEX
load data.mat % attached above
ydata(end) = []; % your xdata/ydata do not match in length !
slope0 = struct('p', 1, 'x', [min(xdata) max(xdata)], 'v', [0 0]);
% Download BSFK function here
% https://www.mathworks.com/matlabcentral/fileexchange/25872-free-knot-spline-approximation
pp = BSFK(xdata,ydata, 2, 3, [], struct('KnotRemoval', 'none', 'pntcon', slope0));
% Check
figure
xi = linspace(min(xdata),max(xdata));
yi = ppval(pp,xi)
plot(xdata,ydata,'b',xi,yi,'r');
for xb=pp.breaks
xline(xb);
end
grid on
  12 comentarios
Bruno Luong
Bruno Luong el 23 de En. de 2021
Editada: Bruno Luong el 23 de En. de 2021
Sorry but all that is the conclusion of your own eye and/or the data is too noisy at the left.
BSFK simply minimizes based on the least square; Meaning that the break it finds so as minimizes the l2 norm of the residual. To me the second figure fits better than then first figure, frankly.
If you enter the data less noisy I bet that BSFK gives the "right" break point.
If you want it as the first figure then just the do it manually, instead of the computer algorithm which is totally objective even if yo don't like it.

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