I'm thinking of pixelize + threshold, but I need a more formalized way of finding the darkest spots on the photo, with tweakable parameters.
Something I can think of is the parts with lowest spatial frequency and least luminous pixels together, i.e. the spots with darkest pixels spread over a largest area (on average) should win but small black dots should be considered too, with some sort of adaptive weighing.
desired result; left image transformed to right image:
luminance bitmap transformed into contrast 'kernels'

UPD: The required result is, in fact, a metric. Which should be a ratio of locally cumulated contrast over a known area.
Local points of negligibly small size should not count, no matter how contrast they are; bigger areas with mediocre contrast features should win [because their cumulative luminosity power is higher]. The closest thing I can think of is a 2D probability distribution function; kernelized, smoothed out, spatially-oriented.

  • $\begingroup$ it's not quite clear what you're looking for. Maybe adding a picture and clearly marking in it what these "spots" are would help. $\endgroup$ Oct 20, 2020 at 20:29
  • $\begingroup$ @MarcusMüller i need result similar to the image in the description, but strictly parameterized and comparable; what i do not need: edge detection / median luminosity calculation; what i do need: finding and quantifying areas with highest contrast $\endgroup$
    – ivan866
    Oct 20, 2020 at 20:52
  • $\begingroup$ I'm confused: your question says you want the biggest contiguous dark area. That's something with a very low contrast. Could you maybe explain what kind of contrast you mean? $\endgroup$ Oct 20, 2020 at 21:04
  • $\begingroup$ @MarcusMüller i need areas with highest contrast over a largest area possible ratio; i.e. contrast of 50% over 100 pixels radius counts equally same as contrast of 100% over 3 pixels wide area; i need contrast 'kernels', as i call them; perhaps some spectral methods will do $\endgroup$
    – ivan866
    Oct 20, 2020 at 21:09
  • 1
    $\begingroup$ hence the "kind of"; please, really, add this to your question's text, it's crucial! $\endgroup$ Oct 20, 2020 at 21:24

1 Answer 1


Pass a sliding window of eg 100x100 pixels over the image stepped 1 pixel at a time in each dimension. Note the maximum and the minimum pixel value. The local contrast is now the local_max/local_min.

A more robust method might be to use a tapered window (eg Gaussian) and a sliding histogram, using eg the 10-percentile and 90-percentile to avoid salt and pepper noise corrupting results.

If (as stated in your update) you are interested in contrast of features larger than a pixel, downsample your image first.


  • $\begingroup$ so, basically a convolution $\endgroup$
    – ivan866
    Oct 21, 2020 at 6:17
  • $\begingroup$ Yeah a convolution except that you are tracking the min/max of the delayline rather than the sum. $\endgroup$
    – Knut Inge
    Oct 21, 2020 at 14:10

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