Effacer les filtres
Effacer les filtres

I have x and y values, but, how can I draw the shape?

3 vues (au cours des 30 derniers jours)
med-sweng
med-sweng le 19 Déc 2013
Commenté : Image Analyst le 26 Juil 2018
I have the following x and y values:
149.5000 75.5000 451.5000 279.5000
which denote (respectively):
x-min y-min x-max y-max
Having those dimensions, how can I draw them as a shape that will represent the boundary with the above values?
Thanks.
  5 commentaires
med-sweng
med-sweng le 19 Déc 2013
Modifié(e) : med-sweng le 19 Déc 2013
@José-Luis. If "convexhull()" draws a convex, yes, that would do. How can we draw it using the dimensions above? Can it also be filled with a colour? Thanks
Image Analyst
Image Analyst le 19 Déc 2013
With a convex hull, if the shape were a "E" shape, you'd get a rectangle. The convex hull is like if you stretched a rubber band around all of your vertices. It's the shape of the rubber band. All of the concave parts would get "filled in."

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Réponses (2)

José-Luis
José-Luis le 19 Déc 2013
Modifié(e) : José-Luis le 19 Déc 2013
doc convhull
nVal = 50;
xx = 149.5 + rand(nVal,1).*(451.5-149.5);
yy = 75.5 + rand(nVal,1).*(279.5-75.5);
k = convhull(xx,yy);
plot(xx(k),yy(k),'r-',xx,yy,'b+')
doc patch if you want a filled polygon.
  10 commentaires
Image Analyst
Image Analyst le 20 Déc 2013
Modifié(e) : Image Analyst le 20 Déc 2013
convhull is not mandatory. The original poster merely said that he had corners of the bounding box of some shape and he wanted to plot it. No convex hull is necessary. Moreover, even if the shape inside the bounding box was weird, say a star shape or irregular blob shape, and say you wanted to plot that, then there is no reason to call convhull. For example, look at this code that draws a soft star. No convhull is called. And med-swing didn't ask for convhull he just said he wanted to know how to plot the bounding box defined by a vector where the x and y were all in the same vector.
% Demo macro to draw a rounded star (like a splat).
%
clc;
close all;
clear all;
workspace;
% Select the inner and outer radius.
outerRadius = 44 % You can change this
innerRadius = 19 % You can change this
% Select the number of lobes around the circle.
numberOfLobes = 8; % You can change this
period = 2 * pi / numberOfLobes;
meanRadius = (outerRadius + innerRadius)/2
amplitude = (outerRadius - innerRadius)/2
t = (0:.005:1)*2*pi; % Independent parameter.
variableRadius = amplitude * cos(2*pi*t/period) + meanRadius
subplot(2,2,1);
plot(variableRadius);
ylim([0 outerRadius]);
title('VariableRadius');
period = 2*pi; % Need to change this now.
x2 = variableRadius .* cos(2*pi*t/period);
y2 = variableRadius .* sin(2*pi*t/period);
subplot(2,2,2);
plot(t, x2);
title('x2 vs. t');
subplot(2,2,3);
plot(t, y2);
title('y2 vs. t');
subplot(2,2,4);
plot(x2,y2,'b')
title('x2 vs y2');
% Maximize window.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]); % Maximize figure.
set(gcf,'name','Image Analysis Demo','numbertitle','off')
I think José-Luis may have misunderstood the question when med-sweng asked how to draw "a shape that will represent the boundary with the above values" - he had the shape already - an x min and max, and a y min and max. He was not asking how to find a different shape of the bounding shape. If he had , then a convex hull might have been a good suggestion (or a call to regionprops), but he already has the bounding box, given by that array, and just needed to know how to plot it.
Youssef  Khmou
Youssef Khmou le 20 Déc 2013
he did not specify the shape so there is an infinity, you first gave the first shape ; a rectangle, then José gave a deformed rectangle using rand. The idea is clear now .

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Image Analyst
Image Analyst le 19 Déc 2013
Try this:
m = [149.5000 75.5000 451.5000 279.5000]
x1 = m(1);
x2 = m(3);
y1 = m(2);
y2 = m(4);
x = [x1 x2 x2 x1 x1];
y = [y1 y1 y2 y2 y1];
plot(x, y, 'r', 'LineWidth', 3);
grid on;
  7 commentaires
Ramesh Bala
Ramesh Bala le 26 Juil 2018
also could you say how can one draw an ellipse with the major axis coords without knowing the minor axis values?
Image Analyst
Image Analyst le 26 Juil 2018
Not sure what this means exactly. Try reading the FAQ: https://matlab.wikia.com/wiki/FAQ#How_do_I_create_an_ellipse.3F

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