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# histograms of crossing count

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esser maalaoui on 14 Jan 2013
i have 2 histograms of crossing count of image (horizontal and vertical), crossing count is the number of times the pixel value changes from 0 to 1 allong vertical or horizontal scan line. the task here is to devide this histogram into five bins with equal width and use five gaussian-chaped weight windows to get the final values please help me it is urgent
Walter Roberson on 3 Feb 2013

Image Analyst on 14 Jan 2013
You can do this without loops using conv2(), sum(), and histc():
binaryImage = randi(2, 5, 10)-1 % Generate sample data.
% Get size so we can get edges for histograms.
[rows columns] = size(binaryImage);
% Find differences between element and prior one.
diffImageVertical = conv2(binaryImage, [1;-1], 'valid')> 0
diffImageHorizontal = conv2(binaryImage, [1,-1], 'valid') > 0
% Count number of rising edges.
edges = 0:1:(rows-1);
countsV = histc(sum(diffImageVertical, 1), edges)
edges = 0:1:(columns-1);
countsH = histc(sum(diffImageHorizontal, 2), edges)
##### 2 CommentsShowHide 1 older comment
Image Analyst on 14 Jan 2013
It probably would, the larger the image the more you'd benefit. conv2 is highly optimized.

Amith Kamath on 14 Jan 2013
Thanks for the interesting question! I'm guessing you're trying something like this. The
I = (rand(500,500) >= 0.5);
%imshow(I)
hChanges = zeros(499,1);
vChanges = zeros(499,1);
for i = 1:499
for j = 1:499
if(I(i,j+1) ~= I(i,j))
vChanges(i) = vChanges(i) + 1;
end
if(I(i+1,j) ~= I(i,j))
hChanges(j) = hChanges(j) + 1;
end
end
end
hHist = hist(hChanges,5); % 5 bins.
vHist = hist(vChanges,5);
hHist and vHist should contain the 5 coefficients you're looking for. I'm not really sure what Gaussian shaped weight windows means, but I'm sure you can fit a gaussian on this data using normfit and the like!