# Finding the mean of a histogram

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Denny Mannakulathil on 12 Jul 2020
Answered: Image Analyst on 12 Jul 2020
How can i find the mean of the histogram(R)?
clear;
figure(1); imshow(im); axis on;
[xP, yP] = ginput(4);
xP(5) = xP(1);
yP(5) = yP(1);
[x,y] = meshgrid(1:size(im,2),1:size(im,1));
xC = mean(xP); yC = mean(yP);
for i = 1:4
m = (yP(i+1) - yP(i)) / (xP(i+1) - xP(i));
b = yP(i) - xP(i) * m;
for j = 1:3
if(yC > xC * m + b)
else
end
end
end
figure(2);
imshow(im);
R = im(:,:,1)
R = R(R ~= 0); R = R.';
figure(3);
subplot(3,1,1)
histogram(R)
xlim([0 260])
set(get(gca,'children'),'facecolor',[1 0 0])
set(get(gca,'children'),'edgecolor',[1 0 0])

What do you mean by mean? The mean number of bins or the mean of the variables that the histogram represents? For the second you can just take the mean of R. For the first look at histogram properties and average the "values" property.

Image Analyst on 12 Jul 2020
The mean of the histogram will not be as accurate as the mean of the image since it's quantized into bins. Here's an illustration:
h = histogram(grayImage, 16) % 16 bins
grid on;
meanImageGL = mean2(grayImage)
grayLevels = h.BinEdges(1:end-1);
counts = h.Values;
meanBinnedGrayLevel = sum(grayLevels .* counts) / sum(counts)
xline(meanImageGL, 'Color', 'g', 'LineWidth', 2);
xline(meanBinnedGrayLevel, 'Color', 'r', 'LineWidth', 2);
% Or you could use the gray levels at the centers of the bins
centerBinGrayLevels = (h.BinEdges(1:end-1) + h.BinEdges(2:end)) / 2;
meanBinnedGrayLevel2 = sum(centerBinGrayLevels .* counts) / sum(counts)
You'll see
meanImageGL =
118.724487304688
meanBinnedGrayLevel =
110.93896484375
meanBinnedGrayLevel2 =
118.93896484375
So, which do you want?

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