histogram by not using the default imhist

I1 = imread('RDL.jpg');
raw = im2double(I1(:,:,1));
m=max(max(raw));
m1=min(min(raw));
binsize=(m-m1)/10;
for i=1:1:size(raw,1)
	     for j=1:1:size(raw,2)      
		value=raw(i,j); %read input image level    
		if value >= min(min(raw)) && value <= max(max(raw))
		    for i=1:1:10
		        if value <= i*binsize+min(min(raw))+  value > (i-1)*binsize+min(min(raw))   
			    % original histogram in pixels
		           imhist(i)=imhist(i)+1;
			    % normalized histogram pdf  
		            %InputIm_normalized_histogram(i)=InputIm_histogram(i)/resolution;   
		        end
		    end
		end
	end
end
imhist(i)

I have used the code to plot a histogram from the image below. I want the graph similar to the attached image. But I am not able to get it. Kindly let me know where have I made the mistake.

[Jan]

You forgot to mention, why you assume, that there is a mistake.

[Vaswati Biswas]

Yeah I forgot to mention. Sorry for that. I was not getting any output graph from that. But no error is also shown

[Jan]

imhist(i) is a scalar, because you have defined it as a vector. Your code does not contain a command to dispaly a graph.

Answer 1

[Jan]

Initially imhist is a command. Calling it as imhist(i) + 1 should cause an error. Do you get a corresponding message?

Use a different name and initialize your counter:

history = zeros(1, 10);

This line is meaningless:

if value >= min(min(raw)) && value <= max(max(raw))

because all values are >= the minimum and <= the maximum.

if value <= i*binsize+min(min(raw))+  value > (i-1)*binsize+min(min(raw))
%                                  ^

Here you do not want a +, but an & to get a logical AND. You have evaluated min(min(raw)) already, so it is more efficient and easier to read, if you use m1 here.

A hint: "m" as maximum and "m1" as minimum are not easy to remember. maxV and minV might be better.

A leaner version of your code:

Img  = imread('RDL.jpg');
R    = im2double(Img(:,:,1));  % The red channel
maxR = max(max(R));
minR = min(min(R));

nBin    = 10;
BinEdge = linspace(minR, maxR, nBin + 1);
BinEdge(end) = Inf;   % any value > maxR to include it in last bin
hist    = zeros(1, nBin);
for iBin = 1:nBin
    hist(iBin) = sum(BinEdge(iBin) <= R(:) & R(:) < BinEdge(iBin + 1));
end

Instead of loops over the pixels, sum() is used to count the values of R matching into each bin.

[Vaswati Biswas]

Thanks for your help but using this code also I am getting the same thing. No error is there still I am not getting any output. I want a graph like this:

[Jan]

Your code does not conatin any command to produce a graph. Which detail of the show graph is required? What about

bar(hist)

[Vaswati Biswas]

Thanks. Now I am able to get the graph but the details that I want that is missing. I am not able to get how can I get that? Maybe I am missing something. My output should have a peak at the value of highest intensity. But I am not getting that

[Image Analyst]

The histogram has it's peak at the MODE (most common) intensity, not the highest intensity. You will not have a peak at the highest intensity unless the highest intensity is also the most common intensity.

[Vaswati Biswas]

@Image Analyst Thanks for clearing my doubt. Yeah I was making a mistake.

[Vaswati Biswas]

@Jan Thanks for your help it gives me the correct result now

Answer 2

[Image Analyst]

Then why not simply use histogram() or histcounts()? Why do you want to write (buggy) code yourself?

[Vaswati Biswas]

I was using imhist but it didn'y give me the result that I wanted. For example:the peak has a value at 2.2*10^4 in y axis. So my histogram should also have a peak in that value but using imhist I am not able to get it.

[Image Analyst]

That doesn't explain why you don't use the built-in histogram functions.

[Vaswati Biswas]

I didn't try built-in histogram function earlier. But later I checked that too it works fine. Thanks for your suggestion.