How can I fit data to a piecewise function, where the breakpoint of the function is also a parameter to be optimised?
58 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Rahul
am 29 Sep. 2025 um 20:45
Bearbeitet: Matt J
vor etwa 3 Stunden
I have data with x and y values. This data should conform to a function: an assymmetric parabola. Here, the parameters that define the shape of the parabola should be different on either side of the maximum point of the parabola i.e. the breakpoint is where the maximum value of y occurs.
I was hoping to use 'fit' and to define an anonymous function for my data. But I'm not able to work out how to define an anonymous, piecewise function, especially where the breakpoint is one of the parameters to be determined by the fitting procedure, as it is not immediately clear from the data itself where the maximum value of y should occur.
Any help would be appreciated.
0 Kommentare
Akzeptierte Antwort
Matt J
vor etwa 20 Stunden
Bearbeitet: Matt J
vor etwa 13 Stunden
Once you've chosen the coefficients of the first parabola [a1,b1,c1], the breakpoint is determined from,
d=-b1/(2*a1)
Only the leading coefficient of the second parabola is a free parameter:
F=@(x) asymParabola(-2,1,0,-0.6,x);
fplot(F,[-10,10]);axis padded %example plot
ft = fittype(@(a1,b1,c1,a2, x) asymParabola(a1,b1,c1,a2, x) )
function y=asymParabola(a1,b1,c1,a2, x)
d=-b1/(2*a1);
b2=-d*2*a2;
c2=polyval([a1,b1,c1],d)-polyval([a2,b2,0],d);
left=(x<=d);
y=x;
y(left)=polyval([a1,b1,c1],x(left));
y(~left)=polyval([a2,b2,c2],x(~left));
end
4 Kommentare
Torsten
vor etwa 4 Stunden
Bearbeitet: Torsten
vor etwa 4 Stunden
If the given x,y values have no noise/errors, then the maximum y-value and the maximum of the parabola are one and the same.
Why ? Both parabola can intersect below their respective maxima, and nonetheless the point of intersection can be the maximum y-value of the piecewise function.
But it seems you interpreted the question correctly.
Weitere Antworten (3)
Walter Roberson
am 29 Sep. 2025 um 21:20
(a1*x.^2 + b1*x + c1) .* (x <= d) + (a2*x.^2 + b2*x + c2) .* (x > d)
Note that for this to work, the coefficients must be constrained to be finite
2 Kommentare
Paul
vor etwa 23 Stunden
Sounds like both sides of the function should have the same value at x = d, at least that's how interpret the question. If so, then I think the function would look something like
(a1*(x-d).^2 + b1*(x-d) + c) .* (x <= d) + (a2*(x-d).^2 + b2*(x-d) + c) .* (x > d)
Catalytic
vor etwa 10 Stunden
Bearbeitet: Catalytic
vor etwa 10 Stunden
You can also parametrize the model function directly in terms of the break point coordinates (xbreak, ybreak) and two curvature parameters -
F= @(a1,a2,xbreak,ybreak, x) modelFun(a1,a2,xbreak,ybreak, x);
xbreak=3; ybreak=5;
fplot( @(x) F(-2,-0.6,xbreak,ybreak,x), [1,5]);
xline(xbreak,'--')
fType = fittype(F);
function y=modelFun(a1,a2,xbreak,ybreak, x)
X=x-xbreak;
LHS=(X<=0);
RHS=~LHS;
y=X.^2;
y(LHS)=a1.*y(LHS) + ybreak;
y(RHS)=a2.*y(RHS) + ybreak;
end
0 Kommentare
Matt J
vor etwa 3 Stunden
Bearbeitet: Matt J
vor etwa 3 Stunden
Why ? Both parabola can intersect below their respective maxima, and nonetheless the point of intersection can be the maximum y-value of the piecewise function.
F=@(x) asymParabola(-2,1,0,-6,5,-20 ,x);
fplot(F,[-10,-1]);axis padded %example plot
function y=asymParabola(a1,b1,c1, a2, s, rightSlope, x)
%Requirements: a1<0, a2<0, s>=0, m<=0
d=-b1/(2*a1)-s;
c2=polyval([a1,b1,c1],d);
left=(x<=d);
right=~left;
xright=x(right);
y=x;
y(left)=polyval([a1,b1,c1],x(left));
y(right)=a2*(xright-d).^2 + rightSlope*(xright-d) +c2;
end
0 Kommentare
Siehe auch
Kategorien
Mehr zu Get Started with Curve Fitting Toolbox finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!