How can I fit multiple lines through a common y-intercept?

4 vues (au cours des 30 derniers jours)
Michael McDonald
Michael McDonald le 22 Oct 2020
I would like to linearly fit N sets of data to a functional form yi = mi * x + b, such that the lines have individual slopes mi but are required to share a single y-intercept b.
I've included some example code below for a simple case of two vectors y1 and y2 that perfectly lie on lines that would otherwise intercept the y-axis at y=1 and y=2. In this case with identical numbers of points, all perfectly on a line already, I'm pretty sure the "correct" answer would solve to a shared intercept on the black cross at y=1.5. For this simple case I could brute force it by looping through a bunch of y-intercepts, fitting to all of them, and minimizing the residuals, but is there a more elegant linear algebra way to solve this?
Thanks!
%% Initialization
x = [1:5];
m1 = 1;
b1 = 1;
y1 = m1*x + b1;
m2 = 2;
b2 = 2;
y2 = m2*x + b2;
%% Fitting
M1B1 = polyfit(x,y1,1);
M2B2 = polyfit(x,y2,1);
%% Plotting
% figure()
plot(x,y1,'bo',x,y2,'ro',[0 x],[b1 y1],'b--',[0 x],[b2 y2],'r--',0,(b1+b2)/2,'k+')
xlim([-1 5])
ylim([0 10])
title({'What''s the best red and blue line','to fit the data and share a y-intercept?'})

Réponse acceptée

Matt J
Matt J le 22 Oct 2020
Modifié(e) : Matt J le 22 Oct 2020
p=[x(:).^[1,0] ,0*x(:); 0*x(:),x(:).^[0,1]] \[y1(:);y2(:)];
m1=p(1)
b=p(2)
m2=p(3)
  2 commentaires
Matt J
Matt J le 22 Oct 2020
Modifié(e) : Matt J le 22 Oct 2020
For N lines,
Y=vertcat(y1(:),y2(:),...,yN(:));
X=kron(speye(N),x(:));
X(:,N+1)=1;
MB=X\Y;
m=MB(1:N); %vector of N slopes
b=MB(N+1);
Michael McDonald
Michael McDonald le 23 Oct 2020
Thanks Matt! That solves the problem very elegantly indeed. I pasted some code and a picture illustrating the effect, and will mark the question as solved.
%% Initialization
x = [-3:3];
noise = 0.5;
b0 = 1
m1 = 1;
y1 = m1*x + b0 + noise*(rand(size(x))-.5);
m2 = 2;
y2 = m2*x + b0 + noise*(rand(size(x))-.5);
%% Fitting
% Individual fits -- not satisfactory, won't share y-intercept
fit1 = polyfit(x,y1,1)
fitm1 = fit1(1); fitb1 = fit1(2);
fit2 = polyfit(x,y2,1)
fitm2 = fit2(1); fitb2 = fit2(2);
% Matt J's addition: forces a shared y-intercept
p=[x(:).^[1,0] ,0*x(:); 0*x(:),x(:).^[0,1]] \[y1(:);y2(:)];
bestm1=p(1)
b=p(2)
bestm2=p(3)
%% Plotting
figure()
subplot(1,3,1)
plot(x,y1,'bo',x,y2,'ro',x,fitm1*x + fitb1,'b--',x,fitm2*x + fitb2,'r--',0,b0,'k*')
title({'Here are simple individual fits','to these two datasets'})
subplot(1,3,2)
plot(x,y1,'bo',x,y2,'ro',x,fitm1*x + fitb1,'b--',x,fitm2*x + fitb2,'r--',0,b0,'k*')
xlim([-.1 .1])
ylim([b0-.25 b0+.25])
title({'Zoom in: We''d like them','to share a y-intercept'})
subplot(1,3,3)
plot(x,y1,'bo',x,y2,'ro',x,bestm1*x+b,'b-',x,bestm2*x+b,'r-',0,b0,'k*',0,b,'ko')
xlim([-.1 .1])
ylim([b0-.25 b0+.25])
title('Much better!')

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Get Started with Curve Fitting Toolbox dans Help Center et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by