mean value in a sphere - matlab

Question

I'm trying to calculate the mean value of the pixels inside a circle. In the future this needs to be extended to 3D, but for now a 2D sollution would already help me out.
As can be seen in the image, some pixels are entirely inside the circle, but some are only partly inside the circle. The ones partly in the circle also need to contribute only partly to the mean value. The pixels are square. This will simplify the mathematics I hope.
I can calculate the distance from the pixelcorners to the central point, from this you can find the pixels enterly inside and enterly outside. The rest needs correction. But how to find this correction.
[edit] thanks to Heath Raftery the problem is solved! [/edit]
the integral of a circle with radius r
As an example: I want to know the average pixelvalue of pixels in this circle. I know it is 0.3425, since 34.25% of the circle has a value of 1 and the rest is 0.
Function to check what part of a pixel is in the circle:
function [ a ] = incirc( x,y,r )
%only handles the top right quadrant of a circle
if x<0||y<0,error('only positive x,y');end
%integral of sqrt(r^2-x^2) dx
F = #(x,r) (1/2)*(x*sqrt(r^2-x^2)+r^2*atan(x/sqrt(r^2-x^2)));
%find corner locations
x=[x-0.5,x+0.5];
y=[y-0.5,y+0.5];
d = sqrt(x.^2+y.^2); %distance to closed and furthest corner
if max(d)<r,a=1;return;end %inside circle
if min(d)>r,a=0;return;end %outside circle
%intersections with edges (r^2 = x^2+y^2)
inters = [sqrt(r^2-y(1)^2),sqrt(r^2-y(2)^2),sqrt(r^2-x(1)^2),sqrt(r^2-x(2)^2)]; %x(1) x(2) y(1) y(2)
%remove imaginary and out of range intersections
inters(imag(inters)~=0)=NaN;
inters(inters<1E-5)=NaN; %to find values that are zero
inters([~((x(1)<inters(1:2))&(inters(1:2)<x(2))),~((y(1)<inters(3:4))&(inters(3:4)<y(2)))])=NaN;
idx = find(~isnan(inters));
if numel(idx)~=2,error('need two intersections of circle with pixel');end
%check area of pixel inside circumference
if all(idx==[1,2]) %2 intersections on y-edge
a=(F(y(2),r)-F(y(1),r)) - x(1); %area
elseif all(idx==[3,4]) %2 intersections on x-edge
a=(F(x(2),r)-F(x(1),r)) - y(1); %area
elseif all(idx==[1,3]) %one intersection on y-edge one on x-edge (left&bottom)
a=(F(inters(1),r)-F(x(1),r))- (y(1)*(inters(1)-x(1)));
elseif all(idx==[2,4]) %one intersection on y-edge one on x-edge (top&right)
a=(inters(2)-x(1))+(F(x(2),r)-F(inters(2),r))-(y(1)*(x(2)-inters(2)));
else
error('geometry')
end
a=real(a);
if a<0||a>1
error('computational error');
end
end
Script to test the function
M = ones(100); %data
M(1:50,:)=0;
pos=[50.2,50];
r = 2;
%calculate what the result should be
h=50-pos(2)+0.5;
A=pi*r^2;
wedge = acos(h/r)/pi;
triangle = h*sqrt(r^2-h^2);
res=(A*wedge-triangle)/A
S=0;N=0;
for i = 1:size(M,1)
for j = 1:size(M,2)
x=abs(j-pos(1));
y=abs(i-pos(2));
n=incirc( x,y,r );
M_(i,j)=n;
S = S+M(i,j)*n;
N = N+n;
end
end
result = S/N
result = 0.3425
You can see the algorithm finds the part of the pixel in the circle.

Answer 1

The question is missing a question, but I'll assume that it's not how to calculate whether pixels are fully inside or outside the circle. That's a relatively simple task. That is, a pixel is fully inside if the furtherest corner of the pixel to the centre is less than a radius away from the centre, and a pixel is fully outside if the closest corner of the pixel to the centre is more than a radius away from the centre.
The question of what proportion of pixels on the circumference fall within the circumference is much trickier. There are two fundamental solutions:
Exact and hard.
Approximate and a bit easier.
In both cases, note the horizontal and vertical symmetry means only the top right quadrant need be considered.
Then, for (1), translate the circle centre to the origin (0, 0) and treat the circumference as the function y(x) = sqrt(r^2 - x^2). Then, the area of an overlapping pixel within the circle is the integral:
integral(y(x) - y0, from x0 to x1, with respect to x)
where y0 is the bottom coordinate of the pixel, x0 is the left coordinate and x1 is the right coordinate.
This integral can be solved exactly with a trigonometric identity and a trigonometric substitution.
For (2), just generate a set of random points within the pixel and count how many of them fall within the circumference. As the set gets larger, the proportion of points that fall within the circumference to the count of all point approaches the proportion of the pixel within the circumference.

Answer 2

You can use inpolygon, to get the indices which lie inside the circle, once you have those indices you can get your pixels and do what you want.
M = rand(100); %data
[nx,ny] = size(M) ;
[X,Y] = meshgrid(1:ny,1:nx) ;
pos=[20,20];
r = 5;
phi=linspace(0,2*pi,100);
imagesc(M);
axis image
hold on
plot(pos(1),pos(2),'rx')
xc = pos(1)+r*sin(phi) ;
yc = pos(2)+r*cos(phi) ;
plot(xc,yc,'-r');
% hold off
%% get indices which are inside the circle
idx = inpolygon(X(:),Y(:),xc,yc) ;
xi = X(idx) ; yi = Y(idx) ;
plot(xi,yi,'.r')
mypixels = M(idx) ;

Answer 3

You can also use rangesearch to get the points lying within the given radius of the circle. As below:
M = rand(100); %data
[nx,ny] = size(M) ;
[X,Y] = meshgrid(1:ny,1:nx) ;
pos=[20,20];
r = 5;
phi=linspace(0,2*pi,100);
imagesc(M);
axis image
hold on
plot(pos(1),pos(2),'rx')
xc = pos(1)+r*sin(phi) ;
yc = pos(2)+r*cos(phi) ;
plot(xc,yc,'-r');
% hold off
%% Use nearest neighbour search
idx = rangesearch([X(:),Y(:)],pos,r) ;
xi = X(idx{1}) ; yi = Y(idx{1}) ;
plot(xi,yi,'.r')
mypixels = M(idx{1}) ;