find the histogram of the image(cameraman)without using the matlab built-in functions for histogram calculations.compare your result with the with those obtained using the functions imhist().display the input image,the two histograms-(yours and matla

if true
  % code
end

Answer 1

[Marwa Ahmeid]

%By Marwa Ahmeid
function [y]=myimhist(img)
[r,c]=size(img);
y=zeros(1,256);
for i=1:r
 for j=1:c
    z=img(i,j);
   y(z+1)=y(z+1)+1;
end
end

Answer 2

[Image Analyst]

The code you posted won't do it obviously. You need to use a for loop over all rows and all columns. The gray level will be the index of a "counts" array that you'll build up. Hint:

[rows, columns, numberOfColorChannels] = size(grayImage);
counts = zeros(1, 256);
for col = 1 : columns
    for row = 1 : rows

http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers

[Marwa Ahmeid]

plot histogram without using matlab imhist() function

[Image Analyst]

To plot, you can use bar() or plot().

[Marwa Ahmeid]

no ,I want matlab Code applied find the histogram of the image(cameraman)without using the matlab built-in functions for histogram calculations.compare your result with the with those obtained using the functions imhist().display the input image,the two histograms-(yours and matlab)

[Image Analyst]

Marwa, it looks like you figured out how to complete my code, though you changed the nice descriptive variable names I had to make it more cryptic and hard to read. This is how I would have completed it:

function [counts, grayLevels] = MyHistogram(grayImage)
[rows, columns, numberOfColorChannels] = size(grayImage);
counts = zeros(1, 256);
for col = 1 : columns
  for row = 1 : rows
    % Get the gray level.
    grayLevel = grayImage(row, col);
    % Add 1 because graylevel zero goes into index 1 and so on.
    counts(grayLevel) = counts(grayLevel) + 1;
  end
end
% Plot the histogram.
grayLevels = 0 : 255;
bar(grayLevels, counts, 'BarWidth', 1, 'FaceColor', 'b');
xlabel('Gray Level', 'FontSize', 20);
ylabel('Pixel Count', 'FontSize', 20);
title('Histogram', 'FontSize', 20);
grid on;
end

[Marwa Ahmeid]

thanks Image Analyst ,That's right... i am rookie in digital image processing course,and I want to assistance in the following days...As well as i want to electronic books where there are a lot of examples

[Saravana Kumar Balakrishnan Janaki]

your code throws this error 'Array indices must be positive integers or logical values.'

[Image Analyst]

Looks like I said to add 1 in the comment but didn't in the code. Thanks for catching that. Here is corrected code:

function [counts, grayLevels] = MyHistogram(grayImage)
[rows, columns, numberOfColorChannels] = size(grayImage);
counts = zeros(1, 256);
for col = 1 : columns
	for row = 1 : rows
		% Get the gray level.
		grayLevel = grayImage(row, col);
		% Add 1 because graylevel zero goes into index 1 and so on.
		counts(grayLevel+ 1) = counts(grayLevel+1) + 1;
	end
end
% Plot the histogram.
grayLevels = 0 : 255;
bar(grayLevels, counts, 'BarWidth', 1, 'FaceColor', 'b');
xlabel('Gray Level', 'FontSize', 20);
ylabel('Pixel Count', 'FontSize', 20);
title('Histogram', 'FontSize', 20);
grid on;
end

Answer 3

[T G]

Thank u so much Image Analyst for helping..

Answer 4

[Ankan Banerjee]

Can histogram be calculated using find() command?

[Image Analyst]

Yes, if you used a loop to find all the values that fell in a certain range and then sum them up into the appropriate bin, but why would you want to do it that more complicated way?