Finding the mean of a histogram
127 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
How can i find the mean of the histogram(R)?
clear;
im = imread('folders01.jpg');
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
mask = im(:,:,j);
if(yC > xC * m + b)
mask(y<x*m+b) = 0;
else
mask(y>x*m+b) = 0;
end
im(:,:,j) = mask;
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])
0 comentarios
Respuestas (2)
Thiago Henrique Gomes Lobato
el 12 de Jul. de 2020
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.
0 comentarios
Image Analyst
el 12 de Jul. de 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:
grayImage = imread('cameraman.tif');
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?
0 comentarios
Ver también
Categorías
Más información sobre Histograms en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!