Using imrotate without losing quality

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Milad 2018-3-12

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评论： DGM 2024-7-5

I'm in a situation that I need to be able to randomly rotate (0-359 angle degrees) different images. I'm currently using imrotate() but considering that the images are small and the quality is important, I noticed that the images lose their qualities, specially in curve lines. I can still rotate my images 0, 90, 180 and 270 degrees without losing any qualities but I wanted to know if there is any way to rotate an image with more degrees without losing the quality.

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Shafaq 2024-7-4

编辑：Image Analyst 2024-7-5

在 MATLAB Online 中打开

I want to rotate matrix as well as image and want to show that output is inavriant. Imrotate is affecting invariance so what should I do?

% Generate a random 50x30 matrix A and a 50x1 vector b
A = rand(8, 8);
b = rand(8, 1);
% A=[1 2;3 4];
% b=[1 2]';
% Least squares solution for the original problem
x = (A' * A) \ (A' * b);
theta=90;
theta= deg2rad(theta);
% Rotation matrix for 90 degrees
R = [cos(theta), -sin(theta); sin(theta), cos(theta)];
% Extend the rotation matrix to apply to the whole problem
R_ext = blkdiag(kron(eye(1), R), eye(0)); % Adjust size accordingly
% Rotate the matrix and vector
A_rot = R_ext * A;
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To operate on each element of the matrix individually, use TIMES (.*) for elementwise multiplication.
b_rot= R_ext * b;
theta= rad2deg(theta);
A_rot=imrotate(A_rot,theta,'crop');
b_rot=imrotate(b_rot,theta,'crop');
%  Display the original and rotated matrices using imagesc
% Least squares solution for the rotated problem
x_rot = (A_rot' * A_rot) \ (A_rot' * b_rot);
% Display results
fprintf('Original least squares solution (first 5 elements):\n');
disp(x(1:end));
fprintf('Rotated least squares solution (first 5 elements):\n');
disp(x_rot(1:end));
% Check invariance
invariance_check = norm(x )- norm(x_rot);
fprintf('Invariance check (should be close to zero): %f\n', invariance_check);

Image Analyst 2024-7-5

Not sure what all that means. An image IS a matrix. And what do you mean by invariant? Invariant to what?

Why don't you start a new discussion thread and explain it?

If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:

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DGM 2024-7-5

@Shafaq has already asked this earlier here:

https://www.mathworks.com/matlabcentral/answers/2134431-i-want-to-rotate-matrix-as-well-as-image-and-want-to-show-that-output-is-invariant-imrotate

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Answer 1

Milad 2018-3-13

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https://ww2.mathworks.cn/matlabcentral/answers/387742-using-imrotate-without-losing-quality#answer_309799

(Thanks to guys for answering and helping me with this question. To sum things up I've decided to write an answer myself.)

It seems that rotation always reduces image quality unless we allow the rotated image to be larger than the original or only use multiples of 90 degrees.

The best option in imrotate() seem to be 'bicubic'. Even without resizing the image it gives a pretty good result.

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DGM 2024-7-5

编辑：DGM 2024-7-5

在 MATLAB Online 中打开

tinycb.png

This is appropriate, but there are conceivable reasons why someone might overlook this fact or otherwise be confused to hear that it's true.

Consider the following rotation of a small test pattern

% a tiny 12x9 checkerboard

A = imread('tinycb.png'); % uint8

% rotate by 90 using two interpolation options

theta = 90;

B = imrotate(A,theta); % output is 9x12

C = imrotate(A,theta,'bilinear');

% upscale to prevent the forum display from

% doing interpolation on the tiny images

B = imresize(B,20,'nearest');

C = imresize(C,20,'nearest');

% display them side by side

montage({B,C},'border',[10 10],'backgroundcolor','m');

What if the angle were off by a small amount?

% the same tiny image

A = imread('tinycb.png');

% angle is off by the tiniest amount

theta = 90 + eps(90) % looks like 90, right?

theta = 90.0000

% rotate as before

B = imrotate(A,theta); % now the output is 10x13

C = imrotate(A,theta,'bilinear');

% upscale as before

B = imresize(B,20,'nearest');

C = imresize(C,20,'nearest');

% display them side by side

montage({B,C},'border',[10 10],'backgroundcolor','m');

% how far off was that?

format long

theta

theta =

90.000000000000014

Depending on how the angles are being obtained, it's possible to run into cases where one only assumes they're hitting 90, but aren't actually. The cause might be rounding error, or it might be as gross as an off-by-one error -- for example:

th = linspace(0,360,1001); % this will hit [0 90 180 270 360]
nnz(mod(th,90) == 0)
ans = 
     5

versus

th = linspace(0,360,1000); % this only hits 0 and 360
nnz(mod(th,90) == 0)
ans = 
     2

FWIW, in the case of 90d multiples, imrotate() internally just calls rot90(), so the interpolation options are actually circumvented.

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Answer 2

Image Analyst 2018-3-12

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https://ww2.mathworks.cn/matlabcentral/answers/387742-using-imrotate-without-losing-quality#answer_309694

See if using the ‘nearest’ option of imresize gives you something you like better.

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Milad 2018-3-13

Of course I've seen your solution! Your answers are the most respected. But as I said, I don't want to use imresize(). And if you meant imrotate(), I've tried its methods too. Which still didn't give good results. I'll try the 'nearest' option of imrotate() once again to see if anything changes.

Image Analyst 2018-3-13

Sorry - meant imrotate(). Let me know if that's any better. Otherwise (if not) your only option is to upsample your image like they already suggested.

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Answer 3

Guillaume 2018-3-12

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/387742-using-imrotate-without-losing-quality#answer_309666

在 MATLAB Online 中打开

As per Walter's comments, the only way you can perform arbitrary rotations without loss of quality if by increasing the number of pixels in the image, so:

supersampled = imresize(yourimage, 4);  %use whatever scale you which > 1. imresize uses bicubic by default which is the best.
rotatedimage = imrotate(supersampled, someangle, 'bicubic');

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Image Analyst 2018-3-13

Did you even TRY or even see my solution (below)? If you just want the image rotated but want sharp edges, not blurry edges, then it might work. Of course you will get jaggies for certain angles, but the edges will be sharper.

Guillaume 2018-3-13

imresize is used to upsample the image. A feature that was occupying a single pixel now occupy more. Since we use 'bicubic interpolation' you also get antialising. You now have more resolution for your feature, so you loose less quality when you rotate it.

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Using imrotate without losing quality

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