Centroid of polyarea

Question

Jaejin Hwang 2012-1-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/25658-centroid-of-polyarea

采纳的回答： Chandra Kurniawan

在 MATLAB Online 中打开

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

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Answer 1

Chandra Kurniawan 2012-1-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/25658-centroid-of-polyarea#answer_33614

在 MATLAB Online 中打开

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 :

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Sean de Wolski 2012-1-9

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

Chandra Kurniawan 2012-1-9

Hi, Sean

thanks for the suggestion

I will use it in the future.

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Answer 2

Sean de Wolski 2012-1-9

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/25658-centroid-of-polyarea#answer_33612

在 MATLAB Online 中打开

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;

Source

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Jaejin Hwang 2012-1-9

That's interesting. But If I add it, I couldn't see the centroid point. What did I make mistake? Thanks.

% spline function

I= imread('000445.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-');

% calculating area

A = polyarea(xys(1,:),xys(2,:));

plot(xys(1,:),xys(2,:),'r-');

title (['Area = ' num2str(A)]);

axis image

% calculating centroid

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;

plot(x_bar,y_bar,'b*');

THAMMISHETTI NIKESH 2012-11-12

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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