Matlab edge expansion vectorization - matlab

Question

I have the following matlab code in a project of mine. image_working at this point is a logical image, the result of edge detection. The below loop expands each white point to be essentially a cross with width width (this is so a later call to imfill() will find more closed regions. The four if statements check that each point is within the original bounds.
[edge_row, edge_col] = find(image_working);
for width = 1:width_edge_widen
for i = 1:length(edge_row)
if (edge_row(i) + width <= m)
image_working(edge_row(i) + width, edge_col(i)) = 1;
end
if (edge_row(i) - width >= 1)
image_working(edge_row(i) - width, edge_col(i)) = 1;
end
if (edge_col(i) + width <= n)
image_working(edge_row(i), edge_col(i) + width) = 1;
end
if (edge_col(i) - width >= 1)
image_working(edge_row(i), edge_col(i) - width) = 1;
end
end
end
I suspect there's a good way to vectorize it and avoid the top-level loop, but I'm at a loss as to how to go about it. Simply indexing (like image_working(edge_row, edge_col)) doesn't work, since this will give give a rectangular region rather than the individual points. Linear indexing (calling inds = find(image_working)) is undesirable because it's difficult to do both vertical and horizontal shifts, although there may well be a vectorizable transform on the indices that I haven't thought of. Any advice?

Answer 1

First, a "vectorized" solution will not neseserally be the fastest here, this depends in how sparse your binary image is. Second, here are a few solutions that will be faster than your code:
First create a random binary image
im0=rand(2000)>0.999; % this means sparsity (density) ~ 1e-3
Solution 1 - for loop (for a cross of width 1, but you can change it as needed):
im=im0;
sd=size(im);
width=1;
[x y]=find(im((1+width):sd(1)-(width+1), (1+width):sd(2)-(width+1)));
x=x+width; y=y+width;
for n=1:numel(y)
im(x(n)-width:x(n)+width,y(n))=1;
im(x(n),y(n)-width:y(n)+width)=1;
end
Solution 2 - vectorized and one line (for a cross of width 1):
im=conv2(single(im0),[0 1 0; 1 1 1; 0 1 0],'same')>0;
Solution 3 - vectorized logical indexing (for a cross of width 1)
im =( im0(2:end-1,2:end-1) | im0(1:end-2,2:end-1) |...
im0(2:end-1,1:end-2) | im0(3:end ,2:end-1) |...
im0(2:end-1,3:end));
im =[zeros(1,size(im,2)); im; zeros(1,size(im,2))];
im= [zeros(size(im0,1),1) im zeros(size(im,1),1)];
You'll see that for sparse images the for loop will be faster than the other methods.
Solution 1: Elapsed time is 0.028668 seconds.
Solution 2: Elapsed time is 0.041758 seconds.
Solution 3: Elapsed time is 0.120594 seconds.
For less sparse images say (~1%) you can use the vectorized solution (Solution 2), as the for loop will very quickly become less efficient but I would check performance on your data before deciding.
Bonus edit, for fun I vectorized the cross filter in solution 2 to have arbitrary width as follows:
f=#(width) circshift(vander([1 zeros(1,2*width)]),[width -width]);
so im=conv2(single(im0),f(1),'same')>0; is equivalent to what is written in Solution 2, but now you can use any f(width_size) you want.