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Centroid of polyarea
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How to get a centroid of polyarea?
Here is my code.
Thanks.
figure, imshow('000445.png')
hold on
xy = [];
n = 0;
but = 1;
while but == 1
[xi,yi,but] = ginput(1);
plot(xi,yi,'r.')
n = n+1;
xy(:,n) = [xi;yi];
end
t = 1:n;
ts = 1: 0.1: n;
xys = spline(t,xy,ts);
plot(xys(1,:),xys(2,:),'r-');
A = polyarea(xys(1,:),xys(2,:));
plot(xys(1,:),xys(2,:),'r-');
title (['Area = ' num2str(A)]);
axis image
hold off
0 Commenti
Risposta accettata
Chandra Kurniawan
il 9 Gen 2012
Hi,
I modified your first code becomes :
I = imread('peppers.png');
[r c o] = size(I);
imshow(I); hold on;
xy = [];
n = 0;
but = 1;
while but == 1
[xi, yi, but] = ginput(1);
plot(xi, yi, 'r.');
n = n + 1;
xy(:, n) = [xi; yi];
end
t = 1 : n;
ts = 1 : 0.1 : n;
xys = spline(t, xy, ts);
plot(xys(1,:), xys(2,:), 'r-');
A = polyarea(xys(1,:), xys(2,:));
plot(xys(1,:), xys(2,:), 'r-');
title (['Area = ' num2str(A)]);
axis image
%hold off
Then, I create my own code.
J = logical(zeros(r, c));
xcoor = floor(xys(1,:));
ycoor = floor(xys(2,:));
for x = 1 : numel(xcoor)
J(ycoor(x),xcoor(x)) = 1;
end
J = imdilate(J,strel('square',20));
J = bwmorph(J,'thin',inf);
J = imfill(J,'holes');
stat = regionprops(J,'Centroid');
plot(stat.Centroid(1),stat.Centroid(2),'go',...
'markerfacecolor','b')
And the result is :
![](https://www.mathworks.com/matlabcentral/images/broken_image.png)
2 Commenti
Sean de Wolski
il 9 Gen 2012
You could use poly2mask() instead of the dilation/skeletonization. I do not believe the method you are using would be correct at boundaries. I.e. where the strel is not fully represented on boundary of the image, the thinning operation will be shifted in and not centered since the strel was not centered at the edge.
Though regionprops/works for this, in two dimensions the formula is well defined:
http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon
Più risposte (1)
Sean de Wolski
il 9 Gen 2012
Once you know area, A, and coordinates: x, y:
As = sum(A)/2;
x_bar = (sum((x(2:end)+x(1:end-1)).*A)*1/6)/As;
y_bar = (sum((y(2:end)+y(1:end-1)).*A)*1/6)/As;
2 Commenti
THAMMISHETTI NIKESH
il 12 Nov 2012
x=[0 10 10 12 12 20 20 12 10 8 8 0 0]; y=[3 3 0 0 3 3 6 6 20 20 6 6 3]; As=polyarea(x,y); X_bar=0; Y_bar=0; h=length(x)-1; for a=1:h X_bar=(1/(6*As))*(x(a)+x(a+1))*(x(a)*y(a+1)-x(a+1)*y(a))+X_bar; Y_bar=(1/(6*As))*(y(a)+y(a+1))*(x(a)*y(a+1)-x(a+1)*y(a))+Y_bar; end
I used the above code, hope it helps you
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