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Plotting color separation of an RGB image

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Tero
Tero le 16 Sep 2020
Commenté : Tero le 17 Sep 2020
Hi all,
I have two RGB photos of a beam of light on a wall that is separated into its colors by a prism. One can see the other picture shows more distinct and clear color separation in the horizontal direction, compared to the other, which is more smooth in nature. Question is, what would be the technique to plot this difference we see? I'm looking for a simple method to show on a single plot (or equivalent) that one is more gradual, and the other is more discrete. Any ideas?
Thanks in advance,
Tero

Réponse acceptée

Tero
Tero le 17 Sep 2020
So, I came up with a different approach, but not sure how scientific this is. I converted both images to Lab color space, and calculated the color difference (Euclidian mean) to the "white point", which I defined for both images to be around the center of the whitest region. Now we can observe a greater derivative for the image with more clear color transitions - red plot, while the blue plot keeps it more flat.
Be free to argue, but unless none, I will accept my answer :)
  2 commentaires
Walter Roberson
Walter Roberson le 17 Sep 2020
Tero
Tero le 17 Sep 2020
Got it, Thanks!

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Walter Roberson
Walter Roberson le 16 Sep 2020
(First step is to break that combined image that you posted into two images. I assume that your original source already has it as two images.)
Convert each image to HSV using rgb2hsv(), and take the mean hue of each column -- mean(HSV(:,:,1),1) .
Now plot the two mean_hue lines. The first of them shows an **increase* in Hue before the cliff, whereas the second of them shows a decrease in Hue in that section.
Quantifying the meaning of this might be difficult.
  1 commentaire
Tero
Tero le 16 Sep 2020
Thanks for answering. I followed your instructions and the result is below. Blue plot is the upper of the two photos, and red plot is the bottom one. I guess it does the job, assuming the reader is aware of hue and color wheels. Otherwise the intuitiveness can become a problem, at least on an excecutive level :)

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