histogram by not using the default imhist

Question

Vaswati Biswas 2021-3-13

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/771748-histogram-by-not-using-the-default-imhist

编辑： Vaswati Biswas 2021-3-15

Capture.JPG

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.

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Vaswati Biswas 2021-3-13

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

Jan 2021-3-13

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

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Answer 1

Jan 2021-3-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/771748-histogram-by-not-using-the-default-imhist#answer_646978

编辑：Jan 2021-3-13

在 MATLAB Online 中打开

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