lsqcurvefit with 2 points trouble

Question

Mindy 2023-5-2

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1956654-lsqcurvefit-with-2-points-trouble

评论： Mathieu NOE 2023-5-3

采纳的回答： Mathieu NOE

在 MATLAB Online 中打开

Hello Matlab Community,

I need help with a little code ...

I want to do this :

, obtain by this code :

xdata2 = [0.9 1.5 13.8 19.8 24.1 28.2 35.2 60.3 74.6 81.3];
ydata2 = [455.2 428.6 124.1 67.3 43.2 28.1 13.1 -0.4 -1.3 -1.5];
fun = @(x,xdata2)x(1)-x(2)*xdata2(1:2).^2;
x0 = [100,-1];
x = lsqcurvefit(fun,x0,xdata2,ydata2(1:2))
times = linspace(xdata2(1),xdata2(end));
figure
hold on
plot(xdata2,ydata2,'ko')
plot(times,x(1)-x(2)*times.^2,'b--')
ylim([-10 500])
xlim([0 100])
legend('Data','Fitted curve')
title('Data and Fitted Curve')

with my own data but I obtain this :

with the following code:

xdata2 = 2e-5:2e-5:18e-5;
ydata2 = [0.997 0.991 0.983 0.974 0.964 0.954 0.944 0.935 0.926];
fun = @(x,xdata2)x(1)-x(2)*xdata2(1:2).^2;
x0 = [100,-1];
x = lsqcurvefit(fun,x0,xdata2,ydata2(1:2))
times = linspace(xdata2(1),xdata2(end));
figure
hold on
plot(xdata2,ydata2,'ko')
plot(times,x(1)-x(2)*times.^2,'b--')
legend('Data','Fitted curve')
title('Data and Fitted Curve')

. I expect a line which passes by at least the 2 first points... Do you know why please?

Thanks in advance.

Regards,

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Mathieu NOE 2023-5-2

hello

I am not sure to understandwhat shape / model you want to fit

from your function description, looks like you want to fit an inverted parabola ? in the first picture the legend speaks about an exponential fit , so what do you want ?

Matt J 2023-5-2

It is strange that you don't just use polyfit. There are no bounds being applied in your fitting process.

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Answer 1

Mathieu NOE 2023-5-2

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在 MATLAB Online 中打开

maybe you want a parabola , you can use polyfit to obtain the best fit :

x = 2e-5:2e-5:18e-5;

y = [0.997 0.991 0.983 0.974 0.964 0.954 0.944 0.935 0.926];

% y = a - b*x²

% Fit a polynomial p of degree "degree" to the (x,y) data:

degree = 2;

p = polyfit(x,y,degree);

% Evaluate the fitted polynomial p and plot:

f = polyval(p,x);

eqn = poly_equation(flip(p)); % polynomial equation (string)

Rsquared = my_Rsquared_coeff(y,f); % correlation coefficient

figure(3);plot(x,y,'o',x,f,'-')

legend('data',eqn)

title(['Data fit - R squared = ' num2str(Rsquared)]);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function Rsquared = my_Rsquared_coeff(data,data_fit)

% R2 correlation coefficient computation

% The total sum of squares

sum_of_squares = sum((data-mean(data)).^2);

% The sum of squares of residuals, also called the residual sum of squares:

sum_of_squares_of_residuals = sum((data-data_fit).^2);

% definition of the coefficient of correlation is

Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function eqn = poly_equation(a_hat)

eqn = " y = "+a_hat(1);

for i = 2:(length(a_hat))

if sign(a_hat(i))>0

str = " + ";

else

str = " ";

end

if i == 2

eqn = eqn+str+a_hat(i)+"*x";

else

eqn = eqn+str+a_hat(i)+"*x^"+(i-1)+" ";

end

eqn = eqn+" ";

end

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Mathieu NOE 2023-5-2

在 MATLAB Online 中打开

FYI, maybe this decaying exponential model (for the first data set)

x = [0.9 1.5 13.8 19.8 24.1 28.2 35.2 60.3 74.6 81.3];

y = [455.2 428.6 124.1 67.3 43.2 28.1 13.1 -0.4 -1.3 -1.5];

f = @(a,b,c,x) a.*exp(-(x./b).^c);

obj_fun = @(params) norm(f(params(1), params(2), params(3),x)-y);

sol = fminsearch(obj_fun, [y(1),1,1]);

a = sol(1)

a = 495.4161

b = sol(2)

b = 10.0348

c = sol(3)

c = 1.0200

yfit = f(a, b,c, x);

Rsquared = my_Rsquared_coeff(y,yfit); % correlation coefficient

figure;

plot(x, y, '+', 'MarkerSize', 10, 'LineWidth', 2)

hold on

plot(x, yfit, '-');grid on

xlabel('time t');

ylabel('P(t)');

title(['Data fit - R squared = ' num2str(Rsquared)]);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function Rsquared = my_Rsquared_coeff(data,data_fit)

% R2 correlation coefficient computation

% The total sum of squares

sum_of_squares = sum((data-mean(data)).^2);

% The sum of squares of residuals, also called the residual sum of squares:

sum_of_squares_of_residuals = sum((data-data_fit).^2);

% definition of the coefficient of correlation is

Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;

end

Mindy 2023-5-3

Thank you so much for your help.

Sorry I wasn't enough clear but yes, I want to use an osculating parabola on my first two points from my time series. I forget to update the legend ("fitted exp" -> "fitted curve"). Thanks for sharing the one with the exp. fitting also. It will be usefull for my next step!

Mathieu NOE 2023-5-3

My pleasure !

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Answer 2

Matt J 2023-5-2

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1956654-lsqcurvefit-with-2-points-trouble#answer_1227719

编辑：Matt J 2023-5-2

在 MATLAB Online 中打开

Change the units of your xdata to something more natural:

xdata2 = (2e-5:2e-5:18e-5)*1e5;

and then,

ydata2 = [0.997 0.991 0.983 0.974 0.964 0.954 0.944 0.935 0.926];

fun = @(x,xdata2)x(1)-x(2)*xdata2(1:2).^2;

x0 = [100,-1];

x = lsqcurvefit(fun,x0,xdata2,ydata2(1:2))

Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.

x = 1×2

0.9990 0.0005

times = linspace(xdata2(1),xdata2(end));

figure

hold on

plot(xdata2,ydata2,'ko')

plot(times,x(1)-x(2)*times.^2,'b--')

legend('Data','Fitted curve')

title('Data and Fitted Curve')

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Matt J 2023-5-2

编辑：Matt J 2023-5-2

在 MATLAB Online 中打开

You can, of course, change xdata2 back to their original units by multiplying x(2) by 1e10. Don't know why you would prefer the original units, however.

Alternatively, you can introduce a scaling coefficient in your model equation:

xdata2 = (2e-5:2e-5:18e-5);

ydata2 = [0.997 0.991 0.983 0.974 0.964 0.954 0.944 0.935 0.926];

fun = @(x,xdata2)x(1)-1e10*x(2)*xdata2.^2;

x0 = [100,-1];

x = lsqcurvefit(fun,x0,xdata2(1:2),ydata2(1:2))

Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.

x = 1×2

0.9990 0.0005

times = linspace(xdata2(1),xdata2(end));

figure

hold on

plot(xdata2,ydata2,'ko', times,fun(x,times),'b--')

legend('Data','Fitted curve')

title('Data and Fitted Curve')

hold off

Mindy 2023-5-3

Thank you so much for your help. xdata is expressed in seconds. These are my 10 first values of my dataset (xdata = times, ydata = velocity from cross-correlation function). In my case, acquisitions of this frequency level are useful for calculating the size of turbulence.

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lsqcurvefit with 2 points trouble

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