How can I fit multiple datasets with one ore more formulas with the same parameters?

I have multiple data with the same Xdata (size 1x7) and different Ydata (Y1,Y2,Y3) all with size 1x7. I have to fit this data to one formula of the form:

x=a/(((1+(sqrt((b.^2).*Ypa+(Ype.^2)))./c.^d)

The totaal Ydata is combined in one matrix and then divided in parallel and perpendicular direction for the three different angles. This matrix has of length of 1x42 Y=[Y1pa Y2pa Y3pa Y1pe Y2pe Y3pe].. I have used several methods to try to find the values of the parameters so it will fits to all the three measurement data sets, but I have problems with the dimensions of the matrix. I try to solve it the following way:

mf=@(C,Y)(C(1)./(((1+(sqrt((C(2)^2).*Y(1:21))+(Y(22:42).^2)))./C(3)).^C(4)));
R1=lsqcurvefit(mf,[280,0.25,1,0.7],Xtotaal,Ytotaal);

a=R1(1);
b=R1(2);
c=R1(3);
d=R1(4);

I have to fit this formula for the three data sets (X,Y1),(X,Y2) and (X,Y3). So I used the vectors Itotaal=[X X X] and Ytotaal=[Y1 Y2 Y3]. But when running this code following error occurs:

Index exceeds matrix dimensions.

Error in Final_fit_try>@(C,Y)(C(1)./(((1+(sqrt((C(2)^2).*Y(1:21))+(Y(22:42).^2)))./C(3)).^C(4)))

Error in lsqcurvefit (line 199)
            initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});

Error in Final_fit_try (line 23)
R1=lsqcurvefit(mf,[280,0.25,1,0.7],Xtotaal,Ytotaal);

Caused by:
    Failure in initial user-supplied objective function evaluation. LSQCURVEFIT cannot continue.

What must be the size of the vectors Xtotaal and Ytotaal? I tried different sizes but the error still occurs.

Answer 1

[Star Strider]

The problem is that your ‘x’ is (1x7) and you have configured ‘Y’ to be (1x42). The easiest way to solve this is to configure ‘x’ to be (7x1) and ‘Y’ to be (7x6), and then change your ‘mf’ function to reflect those changes.

I do not have your data, so you will have to experiment with this:

Xtotaal = x(:);
Ytotaal = [Y1pa(:) Y2pa(:) Y3pa(:) Y1pe(:) Y2pe(:) Y3pe(:)];
mf = @(C,Y)(C(1)./(((1+(sqrt((C(2)^2).*Y(:,1:3))+(Y(:,4:6).^2)))./C(3)).^C(4)));

NOTE: This is UNTESTED CODE but should work. If it does not work, attach your data here (preferably as a ‘.mat’ file). If you are not familiar with them, see the documentation for the save function and its friends for details.

[Nikky]

Thanks for your help but it still doen't work with that dimensions. The problem is that my measurement data exist of one xdata and three vectors with ydata. But my input for the formula is none of these values, but the input is the xdata divided into parallel and perpendicular direction.

I attached my matlab file. I hope it is clear from the script what I wanted my script to do.

[Star Strider]

I cannot figure out what you are doing. Your function should take your independent variable as an argument and return an array the size of our dependent variable. It does not.

I will let you sort that, and delete my Answer here in a few hours.

[Nikky]

In my matlab file Bp and Bl are my indenpendent variables and I have combined them in one matrix because I could not figure out how I can write down an anonymous funcion with two independent variables as input.

I also tried to solve the problem in an other way by splitting up the problem and use three formulas with for each formula one xdata vector as input. But I could not figure out how to fit three formulas with the same parameters to the three datasets at the same time, resulting in the same parameter values for each formula. The goal of this script is to figure out what the values are of the 4 parameters so it will fit for all the three datasets (with for all the three datasets different inputs in the formula).

So I have three datasets (B1,I1), (B2,I2), (B3,I3) and the general formula: I=a./(((1+(sqrt((b^2).*B+B.^2)))./c.^d)) and I have to determine the values for a, b, c and d and they have to be the same for the three datasets.

[Star Strider]

What are the numerical values of ‘(B1,I1), (B2,I2), (B3,I3)’? If they are all different, then ‘I’ becomes a (7x3) matrix, matching ‘B’, also a (7x3) matrix.

With those matrices, it should be straightforward to estimate your parameters.

[Nikky]

Sorry the formula actually is:

I=a./(((1+(sqrt((b^2).*Bp+Bl.^2)))./c.^d))

Bp (size 1x7) and Bl (size 1x7) are the perpendicular and parallel components of each xdata, B. So the problem is that I have two inputs in the formula, both calculated from the measured xdata. That's what made this problem not straight forward for me.

[Star Strider]

Repeating myself: ‘What are the numerical values of ‘(B1,I1), (B2,I2), (B3,I3)’? If they are all different, then ‘I’ becomes a (7x3) matrix, matching ‘B’, also a (7x3) matrix.’

I’m now lost.

How do ‘Bp’ and ‘BI’ relate to ‘(B1,I1), (B2,I2), (B3,I3)’?

[Nikky]

I have measured for three different angles (for each angle one dataset). Bp=B*sin(theta) Bl=B*cos(theta) with theta=0,45,90 degrees.

See the numerical values of B1=B2=B3=B, I1, I2, I3, Bp(1,2,3), Bl(1,2,3) and the formula I have to use to fit the data in the matlab file.

[Star Strider]

This runs, and makes sense in the context of what I understand about your data and the function you want to fit to it:

I_0=[192 135 114 90 79 68 64.5];                                        %%ydata 0 degrees
I_45=[196 136 113 89 78 66 58];                                         %%ydata 45 degrees
I_90=[209 152 130 108 95 82 76];                                        %%ydata 90 degrees

Bp1=[0.11840*sind(0) 0.23788*sind(0) 0.3151*sind(0) 0.4335*sind(0) 0.5105*sind(0) 0.6250*sind(0) 0.6968*sind(0)];           
Bp2=[0.11840*sind(45) 0.23788*sind(45) 0.3151*sind(45) 0.4335*sind(45) 0.5105*sind(45) 0.6250*sind(45) 0.6968*sind(45)];
Bp3=[0.11840*sind(90) 0.23788*sind(90) 0.3151*sind(90) 0.4335*sind(90) 0.5105*sind(90) 0.6250*sind(90) 0.6968*sind(90)];   %%=B

Bl1=[0.11840*cosd(0) 0.23788*cosd(0) 0.3151*cosd(0) 0.4335*cosd(0) 0.5105*cosd(0) 0.6250*cosd(0) 0.6968*cosd(0)];           %%=B
Bl2=[0.11840*cosd(45) 0.23788*cosd(45) 0.3151*cosd(45) 0.4335*cosd(45) 0.5105*cosd(45) 0.6250*cosd(45) 0.6968*cosd(45)];
Bl3=[0.11840*cosd(90) 0.23788*cosd(90) 0.3151*cosd(90) 0.4335*cosd(90) 0.5105*cosd(90) 0.6250*cosd(90) 0.6968*cosd(90)];

Itotaal = [I_0(:)  I_45(:)  I_90(:)];
Btotaal = [Bp1(:)  Bp2(:)  Bp3(:)  Bl1(:)  Bl2(:)  Bl3(:)];

mf = @(C,Y)(C(1)./(((1+(sqrt((C(2)^2).*Y(:,1:3))+(Y(:,4:6).^2)))./C(3)).^C(4)));

Q1 = mf([280,0.25,1,0.7], Btotaal);

R1=lsqcurvefit(mf,[280,0.25,1,0.7],Btotaal,Itotaal);

I will let you determine if it produces the correct results.

If you have the Global Optimization Toolbox, you might want to see if the patternsearch function or others can more quickly estimate the correct parameters.

[Nikky]

Thank you! This is the way to produce the results. Unfortunately the fitting results are not very good (see the fitting results in the matlab file).

[Star Strider]

My pleasure!

The important thing is that my code works!

I cannot control the success of the fit. That is dependent on the initial parameter estimates and your data.

That is the reason I suggested a more robust parameter estimation approach, the easiest being patternsearch. Your function quite likely has many local minima in its hyperplane, and the one it finds first depends on where it starts. You want it to find the global minimum, that should provide the best fit. You can guess randomly at different initial parameter estimates, or use patternsearch, or their friends, such as ga and GlobalSearch to estimate the parameters.

If my Answer solves your problem, please Accept it!