How can I compute the Area and the Centroid of the following shape?

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M
M 2022-9-11
评论： Star Strider 2024-11-4，12:04
Ran in:
How can I compute the Area and the Centroid of the following shape?
I have used the following code to construct it:
xA = 0;
xB = 1;
xf = 1
xf = 1
x = linspace(0, xf, xf*1e4 + 1);
a = 9.2; %
c = 0.5; % center of function
Y = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
plot(x, Y, 'linewidth', 1.5), grid on, xlim ([0 1.1])
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Star Strider
Star Strider 2022-9-11
Ran in:
The polyshape approach is the easiest way to go on this.
However it is possible to calculate this using the trapz function and the centroid definition —
xA = 0;
xB = 1;
xf = 1
xf = 1
x = linspace(0, xf, xf*1e4 + 1);
a = 9.2; %
c = 0.5; % center of function
Y = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
cx = trapz(x,x.*Y) ./ trapz(x,Y)
cx = 0.7138
cy = trapz(Y,x.*Y) ./ trapz(Y,x)
cy = 0.3934
p = polyshape(x,Y);
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
[xc,yc] = centroid(p)
xc = 0.7176
yc = 0.3974
figure
plot(x, Y, 'linewidth', 1.5), grid on, xlim ([0 1.1])
The values from the two approaches are not exactly the same, however they are reasonably close.
.
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Star Strider
Star Strider 2022-9-12
Ran in:
As always, my pleasure!
That is esentially the approach I used, with trapz.
The polyshape approach may be more accurate, however the difference is slight —
xA = 0;
xB = 1;
xf = 1;
x = linspace(0, xf, xf*1e4 + 1);
a = 9.2; %
c = 0.5; % center of function
Y = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
cx = trapz(x,x.*Y) ./ trapz(x,Y)
cx = 0.7138
cy = trapz(Y,x.*Y) ./ trapz(Y,x)
cy = 0.3934
p = polyshape(x,Y);
Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.
[xc,yc] = centroid(p)
xc = 0.7176
yc = 0.3974
x_accuracy = abs(xc-cx)/mean([xc cx])
x_accuracy = 0.0053
y_accuracy = abs(yc-cy)/mean([yc cy])
y_accuracy = 0.0100
figure
plot(x, Y, 'linewidth', 1.5, 'DisplayName','Data'), grid on, xlim ([0 1.1])
hold on
plot(cx,cy,'r+', 'DisplayName','trapz')
plot(xc,yc,'rx', 'DisplayName','polyshape+centroid')
hold off
legend('Location','best')
The relative deviation of the individual values from the mean of each set is about 0.5% for the x-coordinate and about 1% for the y-coordinate. I am not certain how accurate they can be.
.
Star Strider
Star Strider 2024-11-4，12:04
I used the expressiion described in By integral formula in the Centroid article iin Wikipedia.
Later, I also wrote an anonymous function for it:
ctrd = @(x,y) [trapz(x, x.*y) / trapz(x, y); trapz(y, x.*y) / trapz(y, x)]; % Returns: [Centroid x-Coordinate; Centroid y-Coordinate]
The (x,y) arguments are the ‘x’ and ‘y’ vectors describing the outline of the region.
.

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更多回答（4 个）

Bruno Luong
Bruno Luong 2022-9-12
编辑：Bruno Luong 2022-9-12
Ran in:
xA = 0;
xB = 1;
a = 9.2;
c = 0.5;
sfun = @(x)sigmf(x, [a c]);
Area = integral(@(x)sfun(x),xA,xB);
xc = integral(@(x)sfun(x).*x,xA,xB)/Area
xc = 0.7138
yc = integral(@(x)sfun(x).^2,xA,xB)/(2*Area)
yc = 0.3935
Sam Chak
Sam Chak 2022-9-12
Ran in:
Hi @M
If your intention is to find the defuzzified output value for membership function mf at the interval in x using the centroid method, then you can try this method:
xA = 0;
xB = 1;
x = xA:0.0001:xB;
mf = sigmf(x, [9.2 0.5]);
plot(x, mf), grid on, xlim([-0.2 1.2]), ylim([0 1.2]), xlabel('\it{x}'), ylabel('\mu(\it{x})')
xc = defuzz(x, mf, 'centroid')
xc = 0.7138
xline(xc, '--', sprintf('%.4f', xc), 'LabelVerticalAlignment', 'middle');
Note that there should be no significant difference between
mf = sigmf(x, [9.2 0.5]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
and
mf = sigmf(x, [9.2 0.5]);
Image Analyst
Image Analyst 2022-9-11
centroid
Centroid of polyshape
collapse all in page
Syntax
[x,y] = centroid(polyin)
[x,y] = centroid(polyin,I)
Description
example
[x,y] = centroid(polyin) returns the x-coordinates and the y-coordinates of the centroid of a polyshape.
example
[x,y] = centroid(polyin,I) returns the coordinates of the centroid of the Ith boundary of polyin.
This syntax is only supported when polyin is a scalar polyshape object.
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Image Analyst
Image Analyst 2022-9-12
You didn't look up the help for polyshape, did you? It's just as trivial as @Star Strider showed:
p = polyshape(x,Y);
[x,y] = centroid(p)

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Kwanele
Kwanele 2024-11-2，23:11
% Define the functions
f1 = @(x) 0.5 * (80.94 - 0.25 * exp(2 * x.^2));
f2 = @(x) 8.997 - 0.5 * exp(x.^2);
% Define the limits of integration
a = 0;
b = 1.7;
% Use MATLAB's integral function
numerator = integral(f1, a, b);
denominator = integral(f2, a, b);
% Calculate the centroid
centroid = numerator / denominator;
% Display the results
fprintf('Numerator: %.4f\n', numerator);
fprintf('Denominator: %.4f\n', denominator);
fprintf('Centroid: %.4f\n', centroid);

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