Plotting mean of the columns of image on that image

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Yawar Rehman 2014-9-1

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评论： Yawar Rehman 2014-9-3

I am struggling to plot the mean of the columns of the gray scale image on the same image. Mean of the columns of an image can be found by using Mean command in Matlab, but i don't have any idea how to plot it in image coordinates. Any idea or help is appreciated! Thank you :)

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Yawar Rehman 2014-9-3

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I think i have been using the wrong term "plotting" to express myself.

mean function of Matlab returns with the mean of the columns of an image. for finding that where does that column's mean value lie, i guess we have to go through the column pixel search. when there is an exact match or relatively nearer match that pixel coordinate is stored.

finally, mean of the stored pixel coordinates may result in a straight line as i have shown.

so far i have been using this code to do this.

load clown
M = mean(X);
figure, imshow(X,[]); 
hold on;
down=zeros(1,numel(M));
 for columnIdx = 1:numel(M)
    meanValue = M(columnIdx);
    % find locations of gray-scale values equal to the mean        
    idx = find(X(:, columnIdx) == meanValue);
    % if not exact match find relative near alues ~ to the mean
    if (isempty(idx) == 1)
        [~, idx] = min(dist(X(:,columnIdx),meanValue));
    end
    % select one value & make a vector for plotting
    down(1,columnIdx)=idx(1);
end
% plot
plot([1, numel(M)], [mean(down), mean(down)], 'r-')
hold off;

it will result in the following output,

actually what i am interested in are the 'y' pixel coordinates so that i can divide the image into 2 parts. superimposing plots cannot do this i guess. please correct me if my intuition is wrong or if possible can you please suggest a simpler way to do this. thanks :)

Guillaume 2014-9-3

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Well, you seem to have answered your own question there. Doesn't the algorithm you wrote do what you want?

The only thing I would change is the inside of your for loop to:

 meanValue = M(columnIdx);
 [~, idx] = min(abs(X(:, columnIdx) - meanValue);
 down(1, columnIdx) = idx;

Yawar Rehman 2014-9-3

i am not sure it is the correct answer though it is doing what i want! + i am curious if someone could suggest me a way for avoiding for loop

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Answer 1

Guillaume 2014-9-3

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To do what you want without a loop:

[~, down] = min(abs(bsxfun(@minus, X, mean(X))));

bsxfun just subtract the column mean from every value in X, and then you take the index of the minimum of the absolute value of that difference.

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Yawar Rehman 2014-9-3

kudos ... thank you guillaume

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Answer 2

Michael Haderlein 2014-9-2

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Actually, I believe he just gave exactly contradicting answers on Guillaume's question. As I get it, this is what the result should look like:

I did it with the following code, however, I have no idea why a more straight-forward way didn't work (maybe someone can tell me?):

img=imread('grayscaleexample.bmp');
figure, imshow(img)
ha=axes('color','none','position',get(gca,'position'));
tf=figure; 
hp=plot(1:size(img,2),mean(img)); 
set(hp,'parent',ha)
set(ha,'xlim',[1,size(img,2)])
delete(tf)

So the idea is to create another axes above the ones showing the image and plot the mean of the grayscale there. For some reason (please help) I have to draw the plot somewhere else first and then copy it to the axes where it's shown finally. When I plot directly in the axes on top of the ones with the image, the image axes will be deleted - I have no idea why.

What I mean is the following:

>> figure, imshow(img)
>> axes('position',get(gca,'position'))
>> get(gcf,'children')
ans =
     1.0e+03 *
      1.4420
      1.4370
>> plot(1:size(img,2),mean(img));
>> get(gcf,'children')
ans =
     1.4370e+03

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Michael Haderlein 2014-9-2

Yes, they create new axes if no axes are there and they also reset part of the axes properties in case there are some already. However, in this example, the old axes really were deleted as you see in my example. It has nothing to do with the background.

Yawar Rehman 2014-9-3

thank for the answer haderlein. i have commented on the question which i asked. can you please review it

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Answer 3

Guillaume 2014-9-2

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编辑：Guillaume 2014-9-2

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If you do indeed want to plot the mean on the graph (i.e. the other option I offered) then:

 imshow(img);
 hold on
 plot(1:size(img, 2), mean(img), 'r');

I'm not sure what you mean by having two portions of the image though.

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Michael Haderlein 2014-9-2

oops, you're totally right with that, of course plotting in the same axes is possible. This axes deleting behavior confused me so much that I didn't think about this anymore ;-)

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