Image intensity as a function of pattern strength

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Hi All
Attached are images of square wave gratings, a "barcode" and random noise. Please let us assume for the discussion, that all of the images have the same number of pixels and the same average luminance.
When we look at it, we can see stright away that the square wave gratings is most "patterns", the barcode has less initsirc order and the noise image is simply a noise.
I have tried several functions, but they keep showing that the noise image has the most clusters, and I also could not find a way to know how strong a pattern is in an image.
Many thanks
Ron
  2 Comments
Walter Roberson
Walter Roberson on 12 Sep 2021
Will the lines always be vertical? If so that is information that can be used.
Are the patterns perfect, or could there be occasional missing pixels ? I notice what you posted is JPEG, which is prone to aliasing artifacts never every vertical line.
Ron Meidan
Ron Meidan on 13 Sep 2021
Hi Walter
The bars are just used as an example. As you mentioned the term "perfect", this is exactly what I was asking about. With our eyes, it is plain to see a perfect pattern, and I wondering if MATLAB could also tell how "perfect" an image is.
Many thanks
Ron

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Accepted Answer

Image Analyst
Image Analyst on 12 Sep 2021
There are several attributes for each image. You can measure all of them and then decide which of them corresponds to the "strength" of the pattern. For example you can measure the mean and standard deviation. You can measure the number of black blobs. You can measure the area and solidity of each blob, etc. For example for the 3 with different size bars in them, are they all the same strength or not? If not what determines strength?
Perhaps this will help:
  3 Comments
Image Analyst
Image Analyst on 13 Sep 2021
Again there are lots of ways to describe it - lots of properties/attributes. If you look at color, the top row and bottom row are identical. If you look at shape or y position they are not. And not all of the shapes in the top row are identical. They differ in their x value across the image. You cannot compare things in advance without saying what to compare. However you can measure everything you can think of and take the standard deviations of the properties. So for the top row, the SD of the shape, area, color etc. will be zero but the x values are different so there will be a non-zero stdev for x location. Then if you have two images you can compare the two sets of SDs to see if one set has SDs on average that are lower than the other set.

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