Could anyone help me to fit y=1/x on my data?

조회 수: 2(최근 30일)
Elnaz P
Elnaz P 2021년 10월 4일
댓글: Walter Roberson 2021년 10월 4일
Hi,
could someone help me how I can fit y=1/x (y=constant/x) on my data with displaying equation?
clc
close all
clear all
hold on
grid on
set(gca,'fontweight','bold','fontsize',16);
xlim([0 3])
ylim([0 350])
x =[0.5,1,2];
y =[273.3333333,121.8287037,66.72297297];
e =[32.63764351,23.3035763,14.67708363];
errorbar(x,y,e ,'o b','MarkerSize',12, 'LineWidth',8);
x1 =[0.5,1,2];
y1 =[248.4821429,115.0833333,51.58333333];
e1 =[26.74351835,9.791376249,6.60976072];
errorbar(x1,y1,e1 , ' ^ b','MarkerSize',12, 'LineWidth',8);
x2 =[0.5,1,2];
y2 =[278.4895833,118.4151786,67.5];
e2 =[32.11137659,26.89490119,15.47805907];
errorbar(x2,y2,e2 , 'square b','MarkerSize',12, 'LineWidth',8);
x3 =[0.5,1,2];
y3 =[226.25,117.2678571,53.20833333];
e3 =[30.41523969,25.3971604,22.89566137];
errorbar(x3,y3,e3 , 'd b','MarkerSize',12, 'LineWidth',8);
x4 =[1,2];
y4 =[110.7954545,57.5];
e4 =[27.12006334,16.83560216];
errorbar(x4,y4,e4 , '* b','MarkerSize',12, 'LineWidth',8);
hold off

채택된 답변

Mathieu NOE
Mathieu NOE 2021년 10월 4일
hello
here you are my friend
plot
code
clc
close all
clearvars
hold on
grid on
set(gca,'fontweight','bold','fontsize',16);
xlim([0 3])
ylim([0 350])
x =[0.5,1,2];
y =[273.3333333,121.8287037,66.72297297];
e =[32.63764351,23.3035763,14.67708363];
errorbar(x,y,e ,'o b','MarkerSize',12, 'LineWidth',8);
x1 =[0.5,1,2];
y1 =[248.4821429,115.0833333,51.58333333];
e1 =[26.74351835,9.791376249,6.60976072];
errorbar(x1,y1,e1 , ' ^ b','MarkerSize',12, 'LineWidth',8);
x2 =[0.5,1,2];
y2 =[278.4895833,118.4151786,67.5];
e2 =[32.11137659,26.89490119,15.47805907];
errorbar(x2,y2,e2 , 'square b','MarkerSize',12, 'LineWidth',8);
x3 =[0.5,1,2];
y3 =[226.25,117.2678571,53.20833333];
e3 =[30.41523969,25.3971604,22.89566137];
errorbar(x3,y3,e3 , 'd b','MarkerSize',12, 'LineWidth',8);
x4 =[1,2];
y4 =[110.7954545,57.5];
e4 =[27.12006334,16.83560216];
errorbar(x4,y4,e4 , '* b','MarkerSize',12, 'LineWidth',8);
%% curve fit using fminsearch
% We would like to fit the function
% y = a + b./x
% to the data.
ym = mean([y;y1;y2;y3;[NaN y4]],'omitnan'); % create the mean y value for each x value
f = @(a,b,x) a + b./x;
obj_fun = @(params) norm(f(params(1), params(2),x)-ym);
sol = fminsearch(obj_fun, [mean(ym),-1]); % "good" initial guess => will give good fit
a_sol = sol(1);
b_sol = sol(2);
y_fit = f(a_sol, b_sol, x);
Rsquared = my_Rsquared_coeff(ym,y_fit); % correlation coefficient
% plot(x, y_fit, '-',x,ym, 'r .', 'MarkerSize', 40)
plot(x, y_fit, 'r-')
title(['Fit a + b./x to data : R² = ' num2str(Rsquared) ], 'FontSize', 20)
xlabel('x data', 'FontSize', 20)
ylabel('y data', 'FontSize', 20)
eqn = " y = "+a_sol + " + " +b_sol +" / x";
legend(eqn)
hold off
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Rsquared = my_Rsquared_coeff(data,data_fit)
% R2 correlation coefficient computation
sum_of_squares = sum((data-mean(data)).^2);% The total sum of squares
sum_of_squares_of_residuals = sum((data-data_fit).^2);% The sum of squares of residuals, also called the residual sum of squares:
Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;% definition of the coefficient of correlation is
end
  댓글 수: 2
Walter Roberson
Walter Roberson 2021년 10월 4일
You could look at the eqn variable, which describes it as a string, or you could ook at the sol variable, where sol(1) is the additive constant and sol(2) is the "c" value that x is to be divided into.

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추가 답변(1개)

Walter Roberson
Walter Roberson 2021년 10월 4일
x =[0.5,1,2];
y =[273.3333333,121.8287037,66.72297297];
C = mean(x.*y)
C = 130.6471
predy = C./x;
syms c
residue = expand(sum((c./x - y).^2))
residue = 
bestc = solve(diff(residue,c),c)
bestc = 
vpa(bestc)
ans = 
133.68702033999999825720719638325
predy2 = double(bestc)./x;
plot(x, y, ':*', x, predy, '--.b', x, predy2, '-+r')
legend({'raw', 'mean', 'calculus'})
The calculus approach provides the theoretical optimum, and is valid for any number of points. If you do not have the symbolic toolbox, it is possible to create a formula that could be applied
The C value calculate from mean() should be a cheap approximation that is not too far off.
syms X [1 3]
syms Y [1 3]
syms c
residue = sum((c./X - Y).^2);
general_solution = solve(diff(residue,c),c)
general_solution = 
Obviously the 2 factors out, and the general formula becomes
c = sum(y./x)./sum(1./x.^2)

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