Histogram and median filters...AGAIN

Previously I applied a filter to a histogram and attained a slope. I am still waiting for some insight into what the slope might mean in a previous question that I asked. Now I do the same thing (I strongly believe) and now I am getting a horizontal line straight across, can someone tell me what I've possibly done wrong? Or if there's something I'm missing? Thanks, Neo

[per isakson]

You should at least provide a link to the previous question. Don't expect us to search.

[Neo]

http://www.mathworks.com/matlabcentral/answers/168553-histogram-processing-and-median-filters Sorry about that.

Answer 1

[Image Analyst]

Running a median filter on a histogram can smooth out spurious noise spikes and let you get a smoother histogram that's easier to find thresholds on.

The slope might indicates how the number of gray levels changes. If it changes a lot (high slope) then it could indicate that that is a good place for a threshold. Of course a flat slope could also indicate that, like where you threshold in the valley centered between two humps.

You really need to provide some context to get a good answer.

[Neo]

Context: I ran a code called Variational Mode Decomposition (can google and find it on Matlab) and I set it to generate a histogram of the image as well as a domain of a part of the image. I set the code to output a domain with a two modes (it could output many if you wanted to that would range across the whole image) and then I applied the median filter to the histogram with this code k = 1:size(v,3); y = medfilt1(k,3); plot (k,y); I choose k because in the code it has k=1:size(v,3)

      figure('Name', ['Masked Mode #' num2str(k)]);
      subplot(131);
      imagesc(A(:,:,k));
      colormap gray;
      axis equal;
      axis off;

which I felt was the input as "x" to generate the filter on the histogram. When I run the filter, I got a slope at first (very steep) and I thought that it would correspond to counts of the pixels over bins which as you clarified corresponds more specifically to the number of gray level changes? Hope not too overwhelming.

Note: generating domains of an image is what the code does where "x" corresponded to y = medfilt1(x,n) as the template or "original'

[Image Analyst]

Sorry, I didn't understand that. What is the "domain of a part of the image"? And how could a chunk of code be "the input as x"? And how does that chunk of code you call "x" generate a filter? And how does running this filter generate slopes? I'm just going to bow out because I have no idea how to help you.

[Neo]

Okay, thanks for your input thus far. But one last question you might be able to answer, what would constitute as the x in the median filter code?

[Image Analyst]

You can run a median filter on the histogram like this

[pixelCounts, grayLevels] = imhist(grayImage);
smoothHist = medfilt1(pixelCounts, 5);

[Neo]

Thank you. How can a histogram have noise spikes though? I thought a histogram just counted the number of pixels within a certain range of values?

[Image Analyst]

It can. Try this code:

grayImage = uint8(randi(255, 100, 300));
[pixelCounts, grayLevels] = imhist(grayImage);
bar(grayLevels, pixelCounts, 'BarWidth', 1);

Do you see any spikes in there? Of course you do - it's not perfectly smooth is it?

[Neo]

Ahhhhh. I see. But are those noise spikes or just how the histogram is presenting the data?

[Image Analyst]

Statistically it's unlikely that every single possible gray level would have the exact same number of pixels. If you roll a die 24 times, you're not going to get every value of 1-6 exactly 4 times are you? So that natural statistical variation (call it noise if you want) is real and the histogram accurately portrays it (in this case though there are cases where it won't be accurate if you have more bins than pixels you're displaying it on and your bar width is not 1.0).

[Neo]

Wow! You're so amazing! Thanks! I have a better understanding. -Until next time