In terms of histogram computation, the frequency-to-intensity computation is correct, although there is a slight error ... more on that later. Also, I would personally avoid using loops here. See my little note at the end of this post.

However, there are three problems with your code:

Problem # 1 - Histogram is not initialized correctly

histogram_values

should contain your histogram, but you are initializing the histogram with a vector 0:255

. Each intensity value must start at 0 , so you really need to do this:

histogram_values = zeros(256,1);

      

        
        
        
      

    

Problem # 2 - A small error in the loop for

Your intensity ranges from 0 to 255, but MATLAB starts indexing at 1. If you ever get intensities equal to 0, you will get an out-of-bounds error. So the right thing to do is to take p

and add it at 1 to start indexing at 1. However, one complication I have to point out is that if you have a use case uint8

, adding 1 to intensity 255 will just saturate the value up to 255. It won't go to 256 .... so it's also reasonable that you roll something like this double

to ensure you reach 256.

Thus:

histogram_values = zeros(256,1);
for i = 1:rows
  for j = 1:cols
    p = double(openImage(i,j)) + 1;
    histogram_values(p) = histogram_values(p) + 1;

  end
end

      

        
        
        
      

    

Problem # 3 - Not Calling histogram

Right

You have to override the behavior histogram

and include the edges. Basically, do the following:

histogram(histogram_values, 0:255);

      

        
        
        
      

    

The second vector indicates where we should place the columns on the axis x

.

Small note

You can perform a complete histogram calculation without any loops for

. You can try this with a combination of bsxfun

, permute

, reshape

and two sum

:

mat = bsxfun(@eq, permute(0:255, [1 3 2]), im);
h = reshape(sum(sum(mat, 2), 1), 256, 1);

      

        
        
        
      

    

If you want a more detailed explanation of how this code works under the hood, see this conversation between kkuilla and me: http://chat.stackoverflow.com/rooms/81987/conversation/explanation-of-computing-an-images -histogram-vectorized

However, the gist of it is as shown below.

The first line of code creates a 1 column 3-D vector that ranges from 0 to 255 by permute

, and then using bsxfun

with the eq

(equal) function , we broadcast to get us a 3D matrix where each slice is the same size as and a grayscale image, and gives us locations that are equal to the intensity of interest. Specifically, the first snippet tells you where the elements are 0, the second slice tells you where the elements are 1 until the last snippet where it tells you where the elements are 255.

For the second line of code, once we have calculated this 3D matrix, we will calculate the two sums - first by summing each row independently, and then by summing each column of that intermediate result. We then get the total per slice, which tells us how many values ​​there were for each intensity. It is therefore a 3D vector and so we reshape

will go back to a single 1D vector to complete the calculation.

To display a histogram I would use bar

with a flag histc

. Here's a reproducible example if we use an image cameraman.tif

:

%// Read in grayscale image
openImage = imread('cameraman.tif');
[rows,cols] = size(openImage); 

%// Your code corrected
histogram_values = zeros(256,1);
for i = 1:rows
  for j = 1:cols
    p = double(openImage(i,j)) + 1;
    histogram_values(p) = histogram_values(p) + 1;    
  end
end

%// Show histogram
bar(0:255, histogram_values, 'histc');

      

        
        
        
      

    

We get the following: