Curve fitting for non-linear data
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I am trying to fit some data using lsqcurvefit in MATLAB but I am fairly new to this area.
xdata1 = [0 60 660 1250];
ydata1 = [0 18 23 31];
In the image below, the red line is the fit I want to achieve. Sadly, Polyfit does not provide suitable results.
How can I achieve this fit? Thank you in advance!
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Matt J
2018-10-11
0 个投票
I believe piece-wise linear fitting was in the scope of Bruno's free-knot spline fitting package,
更多回答（2 个）
Chaoyu Zhang
2018-10-11
编辑：Chaoyu Zhang
2018-10-15
You can use the method described below,
The target equation (3rd order or maybe higher) is
y = a*x.^3 + b*x.^2 + c*x + d;
A * p = y;
p is the parameters of the equation,
p = [a;b;c;d]
A is the matrix made of x.^3,x.^2,x,1,
A = [x(1).^3 x(1).^2 x(1) 1; ... ; x(n).^3 x(n).^2 x(n) 1]
y is the vector made of y,
y = [y(1); ... ;y(n)];
p = (A.'*A)^(-1)*A.'*y;
Now you get the parameters you need.
5 个评论
Torsten
2018-10-11
This in principle is what polyfit does, but the OP was not satisfied with the fit curve.
Image Analyst
2018-10-15
Matt, do you have a reference for the better accuracy claim? The closest I could find in the help said "If A is a square matrix, then A\B is roughly equal to inv(A)*B, but MATLAB processes A\B differently and more robustly." But I don't think that says it's more accurate.
Well, the poorer accuracy comes from the inversion of (A.'*A), since cond(A.'*A) is the square of cond(A).
>> A=rand(1000,100);
>> cond(A.'*A)
ans =
632.4462
>> cond(A)
ans =
25.1485
When solving with mldivide(), the QR decomposition is used, which avoids this inversion. With A=Q*R,
(A.'*A)^(-1)*A.'*y
=(R.'*R)^(-1)*R.'*Q.'*y
=R^(-1)*Q.'*y
So, the inversion involves only R^(-1) and cond( R )=cond(A).
Here is a test showing the increased error sensitivity of inv(A.'*A)*(A.'*y).
N=1000;
M=15;
A=vander(linspace(1,3.3,M)) + eye(M);
xt=rand(M,1);
yt=A*xt;
y=yt+randn(M,N)*1e-6;
x1=inv(A.'*A)*(A.'*y);
x2=A\y;
Error1=mean( sqrt(sum((x1-xt).^2)) )
Error2=mean( sqrt(sum((x2-xt).^2)) )
should give something like
Error1 =
5.0094
Error2 =
2.8313e-04
Image Analyst
2018-10-11
0 个投票
You cannot get that unless you put in a model curve for that shape. Otherwise functions are not going to know that it's a piecewise linear fit or some sharply kinked log function or whatever. And having more data points would help too. Then you can use fitnlm.
I'm attaching several examples for piecewise linear fit and non-linear fits.
4 个评论
Julian
2018-10-11
Image Analyst
2018-10-11
Then can you just say that the curve is linear between the first 2 points, and linear from the second point onwards? So just have two formulas that define the "curve".
Julian
2018-10-11
Alex Sha
2022-1-29
For three order of polyfit: y = p1+p2*x+p3*x^2+p4*x^3
Root of Mean Square Error (RMSE): 5.67944047963309E-15
Sum of Squared Residual: 1.2902417664678E-28
Correlation Coef. (R): 1
R-Square: 1
Parameter Best Estimate
---------- -------------
p1 3.92787554473843E-15
p2 0.340654276995852
p3 -0.000698994200659206
p4 3.57048623250019E-7
while, if taking function as: y = p1*x/(p2+x)^2-p3*x
Root of Mean Square Error (RMSE): 0
Sum of Squared Residual: 0
Correlation Coef. (R): 1
R-Square: 1
Parameter Best Estimate
---------- -------------
p1 9035.3274708592
p2 119.630504666011
p3 -0.0199834390844168
Obviously, the second function should be more reasonable than 3rd polyfit.
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