Count black pixels within a circle [closed] - matlab

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I have one vector of radii and second vector of hundreds of [X,Y] coordinates. For each possible coordinate-radius pair I have count all black pixels within a circle (whose center is placed in the coordinate) on an input binary image.
What is the fastest way to do it? My only idea is to iterate through every pixel of an image, check the circle equation and then the pixel color but it doesn't seem much optimized for several hundred such operations.

Answer 1

Matlab is great for working with images thanks to the matrix syntax. It does also work with indices so most time you can avoid "iterating through pixels" (although sometimes you'll still have to).
Instead of checking all the pixels within each circle, and having to detect how many pixels were counted twice, another approach is to create a mask, the same size of you image. Blank this mask for each of your circles (so overlapping pixels are only 'blanked' once), then apply the mask on your original picture and count the remaining illuminated pixels.
For an example, I have to take some sample data, the image:
load trees
BW = im2bw(X,map,0.4);
imshow(BW)
And 20 random point/circle coordinates (you can change the number of points and the min/max radii easily):
%// still building sample data
s = size(BW) ;
npt = 20 ; Rmin=5 ; Rmax=20 ; %// problem constants
x = randi([1 s(2)] ,npt,1); %// random X coordinates
y = randi([1 s(1)] ,npt,1); %// Y
r = randi([Rmin Rmax],npt,1); %// radius size between 5 to 20 pixels.
Then we build your custom mask :
%// create empty mask with enough overlap for the circles on the border of the image
mask = false( s+2*Rmax ) ;
%// prepare grid for a sub-mask of pixels, as wide as the maximum circle
xmask = -Rmax:Rmax ;
[xg,yg] = ndgrid(xmask,xmask) ;
rg = sqrt( (xg.^2+yg.^2) ) ; %// radius of each pixel in the subgrid
for ip=1:npt
mrow = xmask+Rmax+y(ip) ; %// calc coordinates of subgrid on original mask
mcol = xmask+Rmax+x(ip) ; %// calc coordinates of subgrid on original mask
cmask = rg <= r(ip) ; %// calculate submask for this radius
mask(mrow,mcol) = mask(mrow,mcol) | cmask ; %// apply the sub-mask at the x,y coordinates
end
%// crop back the mask to image original size (=remove border overlap)
mask = mask(Rmax+1:end-Rmax,Rmax+1:end-Rmax) ;
imshow(mask)
Then you just apply the mask and count :
%% // Now apply on original image
BWm = ~BW & mask ; %// note the ~ (not) operator because you want the "black" pixels
nb_black_pixels = sum(sum(BWm)) ;
imshow(BWm)
nb_black_pixels =
5283

Answer 2

Here is one implementation:
Advantages:
No loops, meshgrid/ndgrid. Instead used faster bsxfun and pdist2
Dots are counted only once, even when the circles overlap.
Variable radius used (radius of all circle is not same)
Code:
%// creating a binary image with little black dots
A = randi(600,256);
imbw = A ~= 1;
%// Your binary image with black dots
imshow(imbw);
%// getting the index of black dots
[dotY, dotX] = find(~imbw);
nCoords = 10; %// required number of circles
%// generating its random coordinates as it is unknown here
Coords = randi(size(A,1),nCoords,2);
%// calculating the distance from each coordinate with every black dots
out = pdist2(Coords,[dotX, dotY]).'; %//'
%// Getting only the black dots within the radius
%// using 'any' avoids calculating same dot twice
radius = randi([10,25],1,size(Coords,1));
pixelMask = any(bsxfun(#lt, out, radius),2);
nPixels = sum(pixelMask);
%// visualizing the results by plotting
hold on
scatter(dotX(pixelMask),dotY(pixelMask));
viscircles([Coords(:,1),Coords(:,2)],radius.'); %//'
hold off
Output:
>> nPixels
nPixels =
19