Why is the horizontal axis 0 to 1 in a histogram for a rgb image?

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Chanille 2023-5-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1973564-why-is-the-horizontal-axis-0-to-1-in-a-histogram-for-a-rgb-image

评论： Image Analyst 2023-5-25

采纳的回答： Image Analyst

在 MATLAB Online 中打开

for my code I am trying to determine intensitiy of the connect components and plot histograms:

for k = 1:numZones

...

zoneMask = zoneMask .* im2double(edgesClean) .* DD;

% Calculate connected components in this zone

cc = bwconncomp(zoneMask);

% Calculate region properties for each connected component in this zone

statscc = regionprops(cc, 'Area', 'Centroid', 'Eccentricity', 'Perimeter', 'MajorAxisLength', 'MinorAxisLength', 'Circularity', 'MaxFeretProperties', 'MinFeretProperties');
% Calculate features for this zone
numObjects = cc.NumObjects;
if numObjects > 0
    profileCounts(k) = numObjects;
end
% Create a histogram of the intensities
figure;
imhist(intensities);
grid on;

But sometimes the values on x axis are strange (i.e. not 0 to 256 for a rgb image I have) why is this? is my code errousnous? please help me to fix so that i have accurately determined histogram of connected components intensity.

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Answer 1

Image Analyst 2023-5-25

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https://ww2.mathworks.cn/matlabcentral/answers/1973564-why-is-the-horizontal-axis-0-to-1-in-a-histogram-for-a-rgb-image#answer_1244604

Please attach one of your images and code to read it in. Evidently your image got converted to floating point. And there is no reason that your image should span the full dynamic range unless you forced it to or it was naturally like that for some reason. Usually images don't span the full dynamic range.

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Chanille 2023-5-25

在 MATLAB Online 中打开

for k = 1:numZones
    % Create a binary mask for this zone
    if k == 1
        % Create a mask for the first zone based on distance from center
        zoneMask = sqrt((X - centerc(1)).^2 + (Y - centerc(2)).^2) < radii(1);
    else
        % Create a mask for this zone based on distance from center
        zoneMask = (sqrt((X - centerc(1)).^2 + (Y - centerc(2)).^2) > radii(k-1)) & (sqrt((X - centerc(1)).^2 + (Y - centerc(2)).^2) < radii(k));
    end
    
    areaOfEachZone(k) = sum(zoneMask, 'all') * pixelSize^2;
    zoneMask = zoneMask .* im2double(edgesClean) .* DD;
    
    % Calculate connected components in this zone
    cc = bwconncomp(zoneMask);
    
    % Calculate region properties for each connected component in this zone
    statscc = regionprops(cc, 'Area', 'Centroid', 'Eccentricity', 'Perimeter', 'MajorAxisLength', 'MinorAxisLength', 'Circularity', 'MaxFeretProperties', 'MinFeretProperties');
    
    % Calculate number of connected components for this zone
    numObjects = cc.NumObjects;
    
    % Calculate the histogram for the connected components in each zone
 
    % % Calculate the intensity values of the connected components (mitochondria) in this zone
    intensities = cellfun(@(x) mean(mean(imageData(x))), cc.PixelIdxList);
    % Create a new figure for the histogram
    figure;
    histogram(intensities, 'BinEdges', 0:1:256);
    grid on;
    flippedZone = numZones - k + 1;
    title(['Histogram of Mitochondria Intensity in Zone ' num2str(flippedZone)], 'FontSize', fontSize);
    xlabel('Intensity');
    ylabel('Frequency');

Chanille 2023-5-25

You answered my question, @Image Analyst its just because of the conditions of my code its not obvious how to set up MeanIntensity hence why i made the complicated cellfun equation as a substitute. If I may I would like to ask if despite it being complicated does the cellfun work for the purposes intended?

Also from what you're saying it sounds like imhist() is better but since my cellfun is a floating point the horizontal axis will reflect this 0 to 1 not 0 to 256.

Image Analyst 2023-5-25

I don't know. I just always use regionprops because it's so much easier and it lets you get lots of other measurements at the same time. You can compare your two intensity arrays just as well as I can.

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Why is the horizontal axis 0 to 1 in a histogram for a rgb image?

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