Find circles with specific mean and standard deviation

I want to find circular areas that are slightly darker than the background and have a higher standard deviation of grayscales. Generalizing, I know the shape I’m looking for and want to use the grayscale distribution as a discrimination metric.

I can hand select areas (e.g. in ImageJ) and measure mean grayscale and std dev and see it’s about 10% darker than the background and the std dev is about twice as high, here’s an example:

Right now I’m doing this in Python with numpy arrays but it’s pretty inefficient. Couple questions:

  1. Does this type of pattern matching have a proper name already? Cursory googling didn’t give me any hits
  2. Any recommendations on an efficient implementation? I’ll try it in C++ by hand but it seems like I could benefit from an optimized numerical library as I’m shifting the template over the image

Thanks,
Andrew

welcome.

I’d call that “texture analysis”.

you could run a grayscale dilation and erosion of a given kernel and radius, then take the difference of those, and that will be the range of values for a local neighborhood.

this doesn’t cope with salt-and-pepper noise too well.

or you could run a sobel/laplacian, abs/square the values, then grayscale dilate (to find local maximum).

edit: I’d even suggest simply looking at the erosion, not taking difference between dilation and erosion… in this situation. the data is such that you get a better signal from just the erosion.

1 Like

To do that you have to blur image and substract blur image (mean(x)) to image (x) : mean(x) -x
Absdiff shoud do the job

this is merely grayscale erosion using a 9x9 square kernel.

decent signal I’d say.

That looks good for segmentation, could help with a coarse search algorithm. My ultimate goal is to use a kind of template matching, though, where I can find the best fit for the center and radius of the circle. My first idea was to use normalized correlation but I want to take the distribution (“texture”) of the circular areas into account, in addition to the actual grayscale.