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1

You should determine your problem more obviously.


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I think this part in article is starting point of the confusion : See how the $I(x+u, y+v)$ changed into a totally different form $I(x,y)+ uI_x + vI_y)$? The author should have write a bit more detailed notation as: $I(x,y)+ uI_x(x,y) + vI_y(x,y)$ i.e. pointing out that these are element-wise spatial derivatives (like Sobel) not matrix derivatives ( ...


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The approach I took was using MATLAB's functions, either regionprops() or bwconncomp() and bwdist(). The idea is to give a grade for each pixel which is a part of bad pixels object. The grade is the radius of the circle bounding the object the pixel resides in. One way to calculate the radius of the bounding circle is the MajorAxisLength property in the ...


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Erode the image using a structuring element which is the size/shape of the maximum allowable "bad region". Then dilate using the same structuring element. This will remove the bad-but-good-enough regions. From there you can work on characterizing/measuring what's left. Example given below using Matlab. % Create a blank image. Mimg = 1000; Nimg = 1000; img = ...


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Median filter? Use it to «thin out» bad pixels that appear locally sparse (for some definition of «local»). Then count the remaining bad pixels. Or convolve with a large-ish 2d kernel (eg flat rectangular window) and decimate to get a number for «how many bad pixels are there inside each eg 16x16 window». Then accumulate the score for each block in a ...


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I've not tried this, but I'm wondering if something like a simple low pass filtering, followed by a column sum will do. No time to explain code right now, but this seems to do something like what you want. Will return in a few hours and explain what it's doing. R Code Below library(imager) N <- 128 noise <- array(runif(N*N*1*1),c(N,N,1,1)) #5x5 ...


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Of course you could. But I cannot find a useful manner. Because of duality, product or convolution in the image domain can be re-expressed as a convolution or product in the frequency domain. Also because if you apply it locally, components of a frequency transformation generally produce a value akin to intensity. In the extreme and trivial case, a one-...


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