**You are now following this question**

- You will see updates in your followed content feed.
- You may receive emails, depending on your communication preferences.

# Compact way to calculate the centroid of a boundary of a set of points

118 views (last 30 days)

Show older comments

### Accepted Answer

Jan
on 30 Sep 2022

Edited: Jan
on 30 Sep 2022

The centroid of the boundary is the mean value of the coordinates:

x = rand(40, 1).^2; % More points on the left

y = rand(40, 1).^2; % More points on the bottom

k = boundary(x, y);

c = mean([x(k), y(k)], 1); % Center of points of boundary

plot(x, y, 'r.');

hold('on');

plot(x(k), y(k), 'b');

plot(c(1), c(2), '*g');

You see, that this is not the center of mass, but the centroid of points. To get the center of mass:

[CoMx, CoMy] = centroid(polyshape(x(k), y(k))); % Center of Mass

plot(CoMx, CoMy, '*k');

% or:

[cx, cy] = CenterOfMass(x(k), y(k))

function [cx, cy] = CenterOfMass(x, y)

% This fails, if lines of the polygon intersect.

x = x(:);

y = y(:);

x_ = circshift(x, 1);

y_ = circshift(y, 1);

A = x .* y_ - x_ .* y;

As = sum(A) * 3;

cx = sum((x_ + x) .* A) / As;

cy = sum((y_ + y) .* A) / As;

end

##### 16 Comments

Bruno Luong
on 30 Sep 2022

"The centroid is the mean value of the coordinates"

I'm not sure if I have the same definition. I would say it is as the mass is ditributed on the boundary, so it can be calculated as the weigthed mean of the center of the segments. The weight is the length of the segment.

But as OP doesn't make any description, I'll hesitate to give that solution.

Sim
on 30 Sep 2022

Edited: Sim
on 30 Sep 2022

Thanks both @Jan and @Bruno Luong! I was thinking at the "centroid as the mean value of the coordinates", but there might be other definitions, I do not know about other ones (@Bruno Luong?)...

About the center of mass of the points, I did not think about it.... :-)

...But then, which of the two centers would make more sense / would be more appropriate to characterize the points?

William Rose
on 30 Sep 2022

@Sim,

@Jan has done a very clever thing. He creates an array of points that roughly covers the unit square, so the centroid should be near (.5,.5). The points are distributed in a parabolic way so there are more points near the origin than you'd expect for a simple 2D uniform distribution. This is smart because it creates a data set in which the mean of all the points (green *) is noticeably different from the centroid (black *).

Here is a non-random set of points that also illustrates the difference: a square plate with a skinny notch, and 7 points at the base of the notch:

x = [0 0 1 1 .03 .027 .023 .02 .023 .027 .03 1 1 ]';

y = [0 1 1 .51 .51 .508 .502 .50 .498 .492 .49 .49 0 ]';

pgon=polyshape(x,y);

c = mean([x, y], 1);

[COMx, COMy] = centroid(pgon);

plot(x, y, 'r.');

hold('on');

plot(pgon)

plot(c(1), c(2), '*g');

plot(COMx, COMy, '*k');

@Bruno Luong also has a smart question, which had not occurred to me: Is this surface a filled-in polygon, or is all its mass on the edge?

Bruno Luong
on 30 Sep 2022

Edited: Bruno Luong
on 30 Sep 2022

My proposition is assumed a unifrorm Mass of the edge (boundary) which gives the result that is more or less invariant to the density of the sample points of the boundary.

x = [0 0 1 1 .03 .027 .023 .02 .023 .027 .03 1 1 ]';

y = [0 1 1 .51 .51 .508 .502 .50 .498 .492 .49 .49 0 ]';

pgon=polyshape(x,y);

%% Point centroid

c = mean([x, y], 1);

%% Region centroid

[COMx, COMy] = centroid(pgon);

%% Boundary centroid

xyw = [x([1:end 1])';

y([1:end 1])'];

w = sqrt(sum(diff(xyw,2).^2,1));

xym = (xyw(:,1:end-1)+xyw(:,2:end))/2;

bdrc = sum(xym.*w,2) ./ sum(w,2);

plot(x, y, 'c.');

hold('on');

plot(pgon)

h1=plot(c(1), c(2), '*g');

h2=plot(COMx, COMy, '*k');

h3=plot(bdrc(1), bdrc(2), 'or');

legend([h1,h2,h3], 'points', 'region', 'boundary')

William Rose
on 30 Sep 2022

Sim
on 1 Oct 2022

@William Rose Sorry, I forgot to reply to "Is this surface a filled-in polygon, or is all its mass on the edge?"

I have a bunch of points, and I create a boundary to wrap my points. Therefore, I would say that most of the mass (given by my points) lies inside the boundary and not on the boundary / edge.

:-)

Sim
on 1 Oct 2022

Edited: Sim
on 1 Oct 2022

I have tried to make a summary of all your answers (and I got a bit confused about the meaning of the Bruno Luong's "region centroid" and "boundary centroid"):

x = rand(40, 1).^2; % More points on the left

y = rand(40, 1).^2; % More points on the bottom

k = boundary(x, y); % Get the boundary that wraps the points

pgon=polyshape(x,y);

figure('position',[100 100 600 600],'Color',[1 1 1],'Visible','on');

hold on

% Show the points

plot(x, y, 'r.');

% Show the boundary

plot(x(k), y(k), 'b');

% Center 1

% From Jan: "center of points of boundary" or "centroid of the boundary"

c = [];

c = mean([x(k), y(k)], 1);

h1 = plot(c(1),c(2),'linewidth',2,'MarkerSize',15,'Marker','.','MarkerEdgeColor','g');

% Center 2

% From Jan: "center of mass"

CoMx = []; CoMy = [];

[CoMx, CoMy] = centroid(polyshape(x(k), y(k)));

h2 = plot(CoMx, CoMy,'linewidth',2,'MarkerSize',15,'Marker','x','MarkerEdgeColor','k');

% Center 3

% From Jan: "center of mass", called "function [cx, cy] = CenterOfMass(x, y)"

x_ = []; y_ = []; A = []; As = []; cx = []; cy = [];

x = x(k);

y = y(k);

x_ = circshift(x, 1);

y_ = circshift(y, 1);

A = x .* y_ - x_ .* y;

As = sum(A) * 3;

cx = sum((x_ + x) .* A) / As;

cy = sum((y_ + y) .* A) / As;

h3 = plot(cx, cy,'linewidth',2,'MarkerSize',15,'Marker','d','MarkerEdgeColor','r');

% Center 4

% From Bruno Luong: point centroid

% From William Rose: no name

c = [];

c = mean([x, y], 1);

h4 = plot(c(1),c(2),'linewidth',2,'MarkerSize',15,'Marker','^','MarkerEdgeColor','m');

% Center 5

% From Bruno Luong: region centroid

% From William Rose: no name

COMx = []; COMy = [];

[COMx, COMy] = centroid(pgon);

h5 = plot(COMx, COMy,'linewidth',2,'MarkerSize',15,'Marker','>','MarkerEdgeColor','b');

% Center 6

% From Bruno Luong: boundary centroid

xyw = []; w = []; xym = []; bdrc = [];

xyw = [x([1:end 1])';

y([1:end 1])'];

w = sqrt(sum(diff(xyw,2).^2,1));

xym = (xyw(:,1:end-1)+xyw(:,2:end))/2;

bdrc = sum(xym.*w,2) ./ sum(w,2);

h6 = plot(bdrc(1), bdrc(2),'linewidth',2,'MarkerSize',15,'Marker','o','MarkerEdgeColor','y');

% plot(pgon)

legend([h1,h2,h3,h4,h5,h6],...

'Jan: center of points of boundary',...

'Jan: center of mass',...

'Jan: center of mass (CenterOfMass)',...

'Bruno Luong: point centroid AND William Rose: no name',...

'Bruno Luong: region centroid AND William Rose: no name',...

'Bruno Luong: boundary centroid', ...

'fontsize',15);%,'Location','northeastoutside')

hold off

set([h1,h2,h3,h4,h5,h6],'linestyle','none')

This is the result, with 4 types of centers....

From this plot we can see that:

- the Bruno Luong's "point centroid" is the Jan's "center of points of boundary" (even though the arguments of the mean are slightly different, i.e. "[x, y]" for Bruno Luong and "[x(k), y(k)]" for Jan).. right?

% From Bruno Luong: point centroid

c = mean([x, y], 1);

% From Jan: "center of points of boundary" or "centroid of the boundary"

c = mean([x(k), y(k)], 1);

- the Jan's "center of mass" is the Jan's "center of mass (with the function CenterOfMass)"

Therefore,

- I would take the Bruno Luong's "point centroid" and the Jan's "center of points of boundary" definitions as definition for the "centroid of a boundary of a set of points" as in my initial question (even though the arguments of the mean are slightly different, i.e. "[x, y]" for Bruno Luong and "[x(k), y(k)]" for Jan)
- I would also consider the Jan's definition of "center of mass".
- Still I do not understand what the Bruno Luong's "region centroid" and "boundary centroid" mean (?).

William Rose
on 2 Oct 2022

@Sim,

The point centroid is the average of the locations of all the points.

The boundary centroid is the centroid of a wire of uniform density that runs along the boundary.

The region centroid is the centroid of a plate of uniform density, with the specified boundary.

If you place a lot of boundary points close together in a straight line, the point centroid will be pulled in that direction, but the boundary and region centroids won't be affected, because the wire location and the plate edge are still the same, whether it is just a few points in a line, or many.

If you add a long thin hairpin fold to the wire, the boundary centroid will be pulled in that direction, but the plate centroid will not be affected much, because the "cutout" made by the hairpin has a negligible area.

Sim
on 23 Nov 2023

Edited: Sim
on 23 Nov 2023

% [cx, cy] = CenterOfMass(x(k), y(k))

function [cx, cy] = CenterOfMass(x, y)

% This fails, if lines of the polygon intersect.

x = x(:);

y = y(:);

x_ = circshift(x, 1);

y_ = circshift(y, 1);

A = x .* y_ - x_ .* y;

As = sum(A) * 3;

cx = sum((x_ + x) .* A) / As;

cy = sum((y_ + y) .* A) / As;

end

performs the same as in

However, it looks like you divide by 3, while, in this paper, the author divides by 6.

Did you use the same formulas or something else? Or am I missing something? I do not understand.. :-) :-) :-)

Bruno Luong
on 23 Nov 2023

Edited: Bruno Luong
on 24 Nov 2023

@Sim The area of the polygonal (A in the book) is sum(A)/2 in Jan code.

So sum(A)*3 in Jan's code is equal to 6*area. They are the same. (EDIT typo)

### More Answers (0)

### See Also

### Categories

### Community Treasure Hunt

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

Start Hunting!**An Error Occurred**

Unable to complete the action because of changes made to the page. Reload the page to see its updated state.

Select a Web Site

Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: .

You can also select a web site from the following list

How to Get Best Site Performance

Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.

Americas

- América Latina (Español)
- Canada (English)
- United States (English)

Europe

- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)

- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom (English)

Asia Pacific

- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)