drawing a best fit graph

6 visualizaciones (últimos 30 días)
emeka onwochei
emeka onwochei el 2 de Mzo. de 2021
Comentada: KSSV el 4 de Mzo. de 2021
is there a way to draw a best fit graph from a cut off circle, i have the [x y] coordinates for the first black points. now i want to use these coordinates to draw the exact shape on a graph. i tried polyfit but it doesnt give me the required shape
  1 comentario
KALYAN ACHARJYA
KALYAN ACHARJYA el 2 de Mzo. de 2021
Can you provide data or code to produce the plot shown?

Iniciar sesión para comentar.

Respuestas (1)

KSSV
KSSV el 2 de Mzo. de 2021
I = imread('image.png') ;
I = rgb2gray(I) ;
[y,x] = find(~I) ;
% Get center
idx = boundary(x,y) ;
x = x(idx) ;
y = y(idx) ;
imshow(I)
hold on
plot(x,y,'r')
  9 comentarios
emeka onwochei
emeka onwochei el 3 de Mzo. de 2021
this is the code i used to get the center, it works for other test images but for this. the output does not give me a tangent across the blue line. does your code pasted earlier have a way of getting the center of the ellipse?
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [xc,yc,R,a] = circfit(x,y)
%
% [xc yx R] = circfit(x,y)
%
% fits a circle in x,y plane in a more accurate
% (less prone to ill condition )
% procedure than circfit2 but using more memory
% x,y are column vector where (x(i),y(i)) is a measured point
%
% result is center point (yc,xc) and radius R
% an optional output is the vector of coeficient a
% describing the circle's equation
%
% x^2+y^2+a(1)*x+a(2)*y+a(3)=0
%
% By: Izhak bucher 25/oct /1991,
x=x(:); y=y(:);
a=[x y ones(size(x))]\[-(x.^2+y.^2)];
xc = -.5*a(1);
yc = -.5*a(2);
R = sqrt((a(1)^2+a(2)^2)/4-a(3));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[xfit,yfit,Rfit] = circfit(x,y);
plot(xfit,yfit,'*r','MarkerSize',10);
plinex=[xfit min(x)];
pliney=[yfit max(y)];
coefficients=polyfit(plinex,pliney,1);
slope=coefficients(1)
[x1p, index1] = min(plinex)
[x2p, index2] = max(plinex)
% Get y at those endpoints
y1p = pliney(index1);
y2p = pliney(index2);
% Compute the slope
perpSlope = -1 / slope
% Equation of line going through the left most x (at x1p) is
% (y - y1p) = perpSlope * (x - x1p)
y1 = perpSlope * (x1p - plinex(index1)) + y1p
y2 = perpSlope * (x2p - plinex(index1)) + y1p
line([x1p, x2p], [y1, y2], 'Color', 'r', 'LineWidth', 0.5)
KSSV
KSSV el 4 de Mzo. de 2021
I would suggest you this:
I = imread('image.png') ;
I = rgb2gray(I) ;
[y,x] = find(~I) ;
% Get center
idx = boundary(x,y) ;
x = x(idx) ;
y = y(idx) ;
%% Get circle radius and centre ;
% select three points randomly on circle
idx = [10 100 150] ;
pt1 = [x(idx(1)) y(idx(1))] ;
pt2 = [x(idx(2)) y(idx(2))] ;
pt3 = [x(idx(3)) y(idx(3))] ;
[C, R] = calcCircle(pt1, pt2, pt3) ;
%% Draw circle
th = linspace(0,2*pi) ;
xc = C(1)+R*cos(th) ;
yc = C(2)+R*sin(th) ;
imshow(I)
hold on
plot(xc,yc,'r')

Iniciar sesión para comentar.

Categorías

Más información sobre Graph and Network Algorithms en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by