How to find the value of the (maximum, minimum, average) intensity in the image on MATLAB?

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Ahmed 2024-10-7

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评论： DGM 2024-10-15

How to find the value of the (maximum, minimum, average) intensity in the image on MATLAB? solve all please

% Load the image
im = imread('peppers.png');
% convert to grayscale
gray_image = rgb2gray(im);
% display the rgb and grayscale image
figure; subplot(1,2,1); imshow(im);
subplot(1,2,2); imshow(gray_image);
[row, col] = size(gray_image);
% You should see the value for row is 384
% You should see the value for col is 512
disp(row);
disp(col);
%% Your work starts
% Q1
% Locate the position and find the value 
% of the pixel with maximum intensity
% You need to change __ to your code
max_value = max(im(:));
max_row_index = max ____ ;
max_col_index = ____;
% Use nested for loop 
____
% Q2
% Locate the position and find the value 
% of the pixel with minimum intensity
min_value = ____ ;
min_row_index = ____ ;
min_col_index = ____ ;
% Use nested for loop 
____
% Q3
% Find the average intensity
all_values = ____ ;
average_intensity = ____ ;
% Use nested for loop 
____
% Q4
% Swap the maximum intensity and the minimum intensity
____
% display the images

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DGM 2024-10-7

在 MATLAB Online 中打开

Read max() and min(). You can get the index of the first pixel matching the minimum value. There's no reason to expect that only a single pixel is at the minimum value, but the code template presumes that.

Read ind2sub(). You can convert the linear index given by max() and min() to row and column subscripts. Bear in mind that if you want to find all the minimal/maximal pixel locations, then there's probably no good reason to try to address scattered points using subscripts instead of linear indices.

Read mean(). Get the array mean. I have no idea why the template is asking you for "all values". What do they want? Why is it a predicate for the mean? Are they literally asking for a list of every element in the array?

Convert the image to float and do a simple linear rescaling. Cast back to the original class.

% an image
inpict = imread('cameraman.tif');
% the things you needed anyway
[mn mnidx] = min(inpict(:));
[mx mxidx] = max(inpict(:));
% the input and output ranges defining the scale transformation
inrange = double([mn mx]);
outrange = fliplr(inrange);
% rescale the data
scalefactor = diff(outrange)/diff(inrange);
outpict = (double(inpict) - inrange(1))*scalefactor + outrange(1);
outpict = cast(outpict,class(inpict));

There's no need for any loops anywhere. It's only wasteful and more difficult to read.

Destiny 2024-10-14

编辑：Destiny 2024-10-14

在 MATLAB Online 中打开

I also have this assignment. It's for a gen ed course for non-computer science majors and we're all confused. I believe we're supposed to use loops but unfortionately we weren't even taught the syntax. It's like asking us to say something in another spoken lanuage without giving us the vocabulary. Is there a way to do this in a loop even if it's lenghty? We were given an example code but it doens't something different than finding the max, min, adn average.

Example code here:

% % This is example code we went through during the discussion section

% % How for loop works

% https://www.mathworks.com/help/matlab/ref/for.html

% Create it image

it_image = zeros(8, 8);

row_idx = [2, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7];

col_idx = [2, 5, 2, 5, 2, 4, 5, 6, 7, 2, 5, 2, 5, 2, 5, 6, 7];

for i=1:length(row_idx)

i_idx = row_idx(i);

j_idx = col_idx(i);

it_image(i_idx, j_idx) = 255;

end

% display it image

imshow(it_image);

% swap the background and foreground

it_image = imcomplement(it_image);

imshow(it_image);

% create mirror image 1

it_image_mirror_1 = zeros(8, 8);

% nested for loop

for i=1:8

for j=1:8

it_image_value = it_image(i, j);

it_image_mirror_1(i, 9-j) = it_image_value;

end

% display mirror image 1

imshow(it_image_mirror_1);

% create mirror image 2

it_image_mirror_2 = zeros(8, 8);

% nested for loop

for i=1:8

for j=1:8

it_image_value = it_image(i, j);

it_image_mirror_2(9-i, j) = it_image_value;

end

% display mirror image 2

imshow(it_image_mirror_2);

% create mirror image 3

it_image_mirror_3 = zeros(8, 8);

% nested for loop

for i=1:8

for j=1:8

it_image_value = it_image(i, j);

it_image_mirror_3(9-i, 9-j) = it_image_value;

end

% display mirror image 3

imshow(it_image_mirror_3);

% display all four images together

figure; hold on;

subplot(2, 2, 1); imshow(it_image);

subplot(2, 2, 2); imshow(it_image_mirror_1);

subplot(2, 2, 3); imshow(it_image_mirror_2);

subplot(2, 2, 4); imshow(it_image_mirror_3);

% % How if works

% https://www.mathworks.com/help/matlab/ref/if.html

% If end

if 6 > 5

disp("Hello");

end

% If else end

if 6 < 5

disp("Hello");

else

disp("World");

end

DGM 2024-10-15

在 MATLAB Online 中打开

I still think the assignment template is baloney.

This is a start comparing finding min and mean with or without loops. If we need to know the location of the (first) minimum/maximum, there are different ways to either obtain that location expressed as either a linear index or as row/column subscripts. Choose what you feel best suits your interpretation of the assignment, and expand it from there.

% this is a single-channel uint8 image
inpict = imread('cameraman.tif');
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% calculate image stats without loops
% i'm only doing min and mean
[mn0 idxmn0] = min(inpict(:)); % still uint8
av0 = mean(inpict(:)); % uint8-scale float
% what if you wanted the row/col subscripts of the min value
% instead of the linear index?
[rowmn0 colmn0] = ind2sub(size(inpict),idxmn0);
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% alternatively, you could use a single loop
% both of these outputs are uint8-scale float
mn1 = Inf;
idxmn1 = 1;
av1 = 0;
for k = 1:numel(inpict)
    % look for first minimum
    if inpict(k) < mn1
        mn1 = inpict(k);
        idxmn1 = k;
    end
    % accumulate for mean
    av1 = av1 + double(inpict(k));
end
av1 = av1/numel(inpict); % mean from sum
% same as before
[rowmn1 colmn1] = ind2sub(size(inpict),idxmn1);
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% alternatively, you could use two loops
% to address the array using subscripts
mn2 = Inf;
rowmn2 = 1;
colmn2 = 1;
av2 = 0;
for c = 1:size(inpict,2)
    for r = 1:size(inpict,1)
        % same concept as before
        if inpict(r,c) < mn2
            mn2 = inpict(r,c);
            rowmn2 = r;
            colmn2 = c;
        end
        % same as before
        av2 = av2 + double(inpict(r,c));
    end
end
av2 = av2/numel(inpict); % same as before

Given your example, I see two interpretations to the final question. "Swap the maximum intensity and the minimum intensity". I gave one example for the literal interpretation of swapping the array extrema, but maybe that's not intended.

Alternatively, if they wanted you to do something akin to imcomplement(), that's a different story. For the simple case of a uint8 image, it might simplify to merely

% the naive assumption is a great way to cause problems
outpict = 255 - inpict; % flip 0 and 255

... but not all images are uint8. The scale of values in an image is expected to correspond to its numeric class, so invert the image based on its class. You could conditionally check the class and do a similar linear transformation of values arithmetically, but this is the simpler way.

% invert an image by swapping the nominal values 
% associated with black and white in the given array class.
% e.g.  [0 1] for float classes
%       [0 255] for uint8
%       [0 65535] for uint16
%       [-32768 32767] for int16
% and so on.
if isnumeric(inpict)
    inclass = class(inpict);
    switch inclass(1)
        case {'u','i'}
            outpict = bitcmp(inpict); % this works for all integer classes
        otherwise
            outpict = 1 - inpict; % this works for unit-scale float
    end
elseif islogical(inpict)
    outpict = ~inpict; % this works for logical class
else
    error('expected INPICT to be numeric or logical')
end

It's important then, that images are properly-scaled for their class. Just as it's problematic to universally presume the wrong pivot values, if we're assuming pivot values based on the numeric class, then the scale of the image values needs to match the class.

In fact, the example you were given includes an utterly common error which creates an improperly-scaled image for which imcomplement() then silently fails.

it_image = zeros(8, 8); % this is a float-class (should be unit-scale)
row_idx = [2, 2, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7];
col_idx = [2, 5, 2, 5, 2, 4, 5, 6, 7, 2, 5, 2, 5, 2, 5, 6, 7];
for i=1:length(row_idx)
    i_idx = row_idx(i);
    j_idx = col_idx(i);
    it_image(i_idx, j_idx) = 255; % but we're shoving uint8-scale values into it
end
% this should either be uint-scale (0 to 1) float
% or uint8-scale (0 to 255) uint8
% instead it's uint8-scale float
it_image
it_image = 8×8
     0     0     0     0     0     0     0     0
     0   255     0     0   255     0     0     0
     0   255     0     0   255     0     0     0
     0   255     0   255   255   255   255     0
     0   255     0     0   255     0     0     0
     0   255     0     0   255     0     0     0
     0   255     0     0   255   255   255     0
     0     0     0     0     0     0     0     0
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% so when we invert it, the values are messed up even more
% imshow simply truncates the negative values, so you don't notice it
it_image = imcomplement(it_image)
it_image = 8×8
     1     1     1     1     1     1     1     1
     1  -254     1     1  -254     1     1     1
     1  -254     1     1  -254     1     1     1
     1  -254     1  -254  -254  -254  -254     1
     1  -254     1     1  -254     1     1     1
     1  -254     1     1  -254     1     1     1
     1  -254     1     1  -254  -254  -254     1
     1     1     1     1     1     1     1     1
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

... but maybe that's getting off-topic.

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Answer 1

Zinea 2024-10-7

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Hi Ahmed,

This is a homework question and hence I won't be providing the exact answers. Below are some hints to help you reach the correct solution:

Q1: A function that finds the maximum value in an array is to be used. Then, a function that returns indices of elements that meet a condition is to be used. Finally, convert linear indices to row and column indices.

Q2: The same steps are to be followed as in Q1, but for the minimum value.

Q3: A function that calculates the mean of an array is to be used.

Q4: The indices obtained from Q1 and Q2 are to be used to exchange the values at those positions in the image array.

Happy learning!

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