dividing a circle into equal n parts and then generate random point inside each part

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Hassan 2013-10-19

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https://ww2.mathworks.cn/matlabcentral/answers/90748-dividing-a-circle-into-equal-n-parts-and-then-generate-random-point-inside-each-part

评论： Walter Roberson 2018-5-24

I am trying to divide a circle (given the radius and the center)into n parts and then generate random points inside each part . If anyone has a sample code or can help me with this, thanks in advance.

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the cyclist 2013-10-19

Is this a school assignment? What have you tried so far?

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Azzi Abdelmalek 2013-10-19

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编辑：Azzi Abdelmalek 2013-10-19

在 MATLAB Online 中打开

x0=2;
y0=1;
r=1;
teta=-pi:0.01:pi;
x=r*cos(teta)+x0
y=r*sin(teta)+y0
plot(x,y)
hold on
scatter(x0,y0,'or')
axis square 
%----------------------------------------
% divide your circle to n sectors
n=8
tet=linspace(-pi,pi,n+1)
xi=r*cos(tet)+x0
yi=r*sin(tet)+y0
for k=1:numel(xi)
plot([x0 xi(k)],[y0 yi(k)])
hold on
end

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a.t. 2018-5-24

The points should be equally spread across the area (=a sector of the circle)

Walter Roberson 2018-5-24

Probably the easiest way to do equal spacing by area is to convert to Cartesian coordnates, and find grid points that are within the boundaries of the sector.

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Image Analyst 2013-10-19

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https://ww2.mathworks.cn/matlabcentral/answers/90748-dividing-a-circle-into-equal-n-parts-and-then-generate-random-point-inside-each-part#answer_100228

Start with the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_set_of_random_locations_within_a_circle.3F

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Hassan 2013-10-19

how can i generate one random point in each part (each part contains only one point)

Image Analyst 2013-10-19

编辑：Image Analyst 2013-10-19

在 MATLAB Online 中打开

Like I said, use the FAQ. Just modify it to use the angles you want. Below is does it for a sector going between pi/4 and 3pi/4:

% Create a random set of coordinates in a circle.
% First define parameters that define the number of points and the circle.
n = 5000;
R = 20;
x0 = 50; % Center of the circle in the x direction.
y0 = 90; % Center of the circle in the y direction.
% Now create the set of points.
% For a full circle, use 0 and 2*pi.
%angle1 = 0;
%angle2 = 2*pi;
% For a sector, use partial angles.
angle1 = pi/4;
angle2 = 3*pi/4;
t = (angle2 - angle1) * rand(n,1) + angle1;
r = R*sqrt(rand(n,1));
x = x0 + r.*cos(t);
y = y0 + r.*sin(t);
% Now display our random set of points in a figure.
plot(x,y, '.', 'MarkerSize', 5)
axis square;
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
fontSize = 30;
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
title('Random Locations Within a Circle', 'FontSize', fontSize);