How to fit a data to a model without using Statistical Toolbox?

조회 수: 38 (최근 30일)
Decy
Decy 2022년 11월 6일
편집: Bruno Luong 2022년 11월 14일
For my data (30x2 table), given a model y = a1 + a2 * x^4 + a3 * x^5, I wanted to find the coefficients of my model. I was always directed to linear regression and data fitting which says it needs statistical toolbox but I don't have one. How do I make do of this?
  댓글 수: 1
Mathieu NOE
Mathieu NOE 2022년 11월 7일
hello
for polynomial fit you don't need any specific toolbox
read / look for polyfit / polyval

댓글을 달려면 로그인하십시오.

답변 (3개)

Bruno Luong
Bruno Luong 2022년 11월 14일
편집: Bruno Luong 2022년 11월 14일
Polynomial fits can be done with linear algebra, no fminsearch is required, but yeah clearly the model cannot fit data with three coefs as stated
% Same data as the Excel file
data=[-0.0834 -0.2409
-0.0826 -0.2405
-0.0775 -0.236
-0.0719 -0.2275
-0.0658 -0.2264
-0.0521 -0.2174
-0.048 -0.2135
-0.04 -0.2127
-0.0279 -0.2064
-0.0266 -0.2062
-0.022 -0.2034
-0.0197 -0.1971
-0.0193 -0.197
-0.0186 -0.1925
-0.0119 -0.1882
-0.0112 -0.1876
-0.0096 -0.166
0.0034 -0.1624
0.0065 -0.1562
0.0108 -0.1551
0.0158 -0.1539
0.0181 -0.1447
0.0312 -0.1422
0.0323 -0.1364
0.036 -0.1214
0.0667 -0.0999
0.0679 -0.0842
0.0734 -0.061
0.0877 -0.0533
0.0884 -0.0503
-0.0834 -0.2409
-0.0826 -0.2405
-0.0775 -0.236
-0.0719 -0.2275
-0.0658 -0.2264
-0.0521 -0.2174
-0.048 -0.2135
-0.04 -0.2127
-0.0279 -0.2064
-0.0266 -0.2062
-0.022 -0.2034
-0.0197 -0.1971
-0.0193 -0.197
-0.0186 -0.1925
-0.0119 -0.1882
-0.0112 -0.1876
-0.0096 -0.166
0.0034 -0.1624
0.0065 -0.1562
0.0108 -0.1551
0.0158 -0.1539
0.0181 -0.1447
0.0312 -0.1422
0.0323 -0.1364
0.036 -0.1214
0.0667 -0.0999
0.0679 -0.0842
0.0734 -0.061
0.0877 -0.0533
0.0884 -0.0503
];
x=data(:,1);
y=data(:,2);
M=x.^[0,4,5];
P=M\y;
a3=P(3);
a2=P(2);
a1=P(1);
P=[a3 a2 0 0 0 a1];
P5 = polyfit(x,y,5);
xi=linspace(min(x),max(x));
ymodel=polyval(P,xi);
yP5=polyval(P5,xi);
close all
h = plot(x,y,'og',xi,ymodel,'-b',xi,yP5,'-r');
legend(h,'data','model with three coefs','Polynomial 5th degree (6 coefs)','location','northwest')

Santosh Fatale
Santosh Fatale 2022년 11월 11일
Hi Decy,
If you have polynomial function for your model and interested to find coefficients from the data, you can use "polyfit" function as follows:
p = polyfit(x, y, n);
Here inouts 'x' and 'y' are your data points and inout 'n' represent the highest degree of the polynomial. The coefficients of polynomial are in variable 'p' and are in descending order of power of independent variable in the polynomial.
For more information about "polyfit" function follow this link:
  댓글 수: 2
the cyclist
the cyclist 2022년 11월 11일
Regarding this solution, I don't believe it will work for @Decy's case. They are not trying to fit the typical
y = a0 + a1*x + a2*x^2 * a3*x^3 + a4*x^4 + a5*x^5
They are trying to fit
y = a1 + a2 * x^4 + a3 * x^5
Note that they do not have linear, quadratic, or cubic terms.
I don't believe polyfit can do this fit.
Decy
Decy 2022년 11월 12일
Hello. yes, i tried polyfit but the error appeared was this:
Error using ()
Subscripting into a table using one subscript (as in t(i)) is not supported. Specify a row subscript and a variable
subscript, as in t(rows,vars). To select variables, use t(:,i) or for one variable t.(i). To select rows, use t(i,:).

댓글을 달려면 로그인하십시오.


Mathieu NOE
Mathieu NOE 2022년 11월 14일
hello @Decy
it seems the model choosen is not realy a good match with the data (or vice versa)
here a demo using fminsearch with your specific polynomial model
% load data
data = csvread('data.csv');
x=data(:,1);
y=data(:,2);
%polynomial fit : y = a1 + a2 * x^4 + a3 * x^5
f = @(a1,a2,a3,x) a1 + a2 * x.^4 + a3 * x.^5;
obj_fun = @(params) norm(f(params(1),params(2),params(3),x)-y);
sol = fminsearch(obj_fun, [mean(y),1,1]);
a1_sol = sol(1);
a2_sol = sol(2);
a3_sol = sol(3);
x_fit = linspace(min(x),max(x),100);
y_fit = f(a1_sol, a2_sol, a3_sol, x_fit);
Rsquared = my_Rsquared_coeff(y,f(a1_sol, a2_sol, a3_sol, x)); % correlation coefficient
figure(3);plot(x,y,'o',x_fit,y_fit,'-')
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
  댓글 수: 1
Mathieu NOE
Mathieu NOE 2022년 11월 14일
a simple demo to show that the model you want to use would be appropriate if the data would show a S curved shape , like this :
% load data
data = csvread('data.csv');
x=data(:,1);
% y=data(:,2);
% dummy y data
a1 = -0.1743;
a2 = 300;
a3 = 2e4;
y= a1 + a2 * x.^4 + a3 * x.^5 + 0.01*randn(size(x));
%polynomial fit : y = a1 + a2 * x^4 + a3 * x^5
f = @(a1,a2,a3,x) a1 + a2 * x.^4 + a3 * x.^5;
obj_fun = @(params) norm(f(params(1),params(2),params(3),x)-y);
sol = fminsearch(obj_fun, [mean(y),0,0]);
a1_sol = sol(1);
a2_sol = sol(2);
a3_sol = sol(3);
x_fit = linspace(min(x),max(x),100);
y_fit = f(a1_sol, a2_sol, a3_sol, x_fit);
Rsquared = my_Rsquared_coeff(y,f(a1_sol, a2_sol, a3_sol, x)); % correlation coefficient
figure(3);plot(x,y,'o',x_fit,y_fit,'-')
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

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Linear and Nonlinear Regression에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by