Fitting multiple data sets to single curve in least square sense

조회 수: 3 (최근 30일)
Abdulllah
Abdulllah 2022년 3월 11일
댓글: AndresVar 2022년 3월 11일
I am trying to fit multiple data sets i.e., x1,x2,x3 -> y1,y2,y3 to a single cuve f following this exapmle. However, it returns error that the second coulmn must be a vector. The ultimate goal is to fit to a curve such that the sum(abs(f(x)-y))<3. How can I do it in matlab starting with the following example
x1=(0:1:10)'; % Explanatory variable
x2=(0:1:10)'+1;
x3=(0:1:10)'+2;
x=[x1 x2 x3];
y = 5 + 3*x + 7*x.^2;
y = y + 2*randn((size(x)));% Add some noise to response variable
% Define function that will be used to fit data
% (F is a vector of fitting parameters)
f = @(F,x) F(1) + F(2).*x + F(3).*x.^2;
F_fitted = nlinfit(x,y,f,[1 1 1]); %Error: Requires a vector second input argument.
% Display fitted coefficients
disp(['F = ',num2str(F_fitted)])
% Plot the data and fit
figure
plot(x,y,'*',x,f(F_fitted,x),'g');
legend('data','fit')
The actuall dataset is attached.

답변 (2개)

Abolfazl Chaman Motlagh
Abolfazl Chaman Motlagh 2022년 3월 11일
the nlinfit function take a vector as second input y, because it's cost function for optimization and regression is based on scalar output. you should fitt each model separately.
x1=(0:1:10)';
x2=(0:1:10)'+1;
x3=(0:1:10)'+2;
x=[x1 x2 x3];
y = 5 + 3*x + 7*x.^2;
y = y + 2*randn((size(x)));
model = @(F,x) F(1) + F(2).*x + F(3).*x.^2;
for i=1:size(x,2)
F_fitted(i,:) = nlinfit(x(:,i),y(:,i),model,[1 1 1]);
end
for i=1:size(x,2)
subplot(1,3,i)
plot(x(:,i),y(:,i),'*',x(:,i),model(F_fitted(i,:),x(:,i)),'g')
title(['(X_' num2str(i) ',Y_' num2str(i) ')'])
end
  댓글 수: 9
Abolfazl Chaman Motlagh
Abolfazl Chaman Motlagh 2022년 3월 11일
let us review our conversation, because i think there's misunderstanding between us.
-------------------------------------------------------------------------
you have two variable x and y. both have same number of elements. column are x and y are different observations. but we know that all datasets obey same relation. which means there is a function like f(x) that describe all relations. which means . j here is index of dataset (which is column) and i is index of element within a dataset (which is row). so basically:
so basically all elements of x and y are some information we can use in finding function f.
--------------------------------------------------------------------------
if this is the problem making x and y, a column vector using x(:) and y(:) and then finding best fit on all data will do the work. (my second comment)
but here comes my objection. if column of x (and y) are describing same functionality and relation. then ploting them should illustrate this fact. but as we can see in scatter plot that i put in third comment, we can see the pairs don't show same relation for any i. this means there is no functionality describing all this relations at once.
AndresVar
AndresVar 2022년 3월 11일
@Abdulllah like Abolfazi said, your data won't fit a single f(x). The data fits a*exp(b*x) pretty well but for different values of a.
There might be another parameter that is missing that collapses the data.
Find the different coefficient by fitting each series or find them like this: simulataneous Curve fitting - (mathworks.com)

댓글을 달려면 로그인하십시오.


AndresVar
AndresVar 2022년 3월 11일
편집: AndresVar 2022년 3월 11일
x1,x2,x3 are column vectors that you combined into a matrix, so then y was evaluated to a matrix also.
nlinfit expects vectors not matrices
to fix it, combine x1 x2 x3 into 1 long column vector using ";"
x=[x1;x2;x3]
Then you should probably sort it and apply same sorting to y Sort array elements - MATLAB sort (mathworks.com)
  댓글 수: 1
Abdulllah
Abdulllah 2022년 3월 11일
If I combined x1,x2,x3 to a single vector and do the same with y, Matlab finds a very bad solution. I am looking for a better options to get a single optimium function to fit all the datasets.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Descriptive Statistics에 대해 자세히 알아보기

제품


릴리스

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by