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Rotate an image around a point

version 1.3 (2.28 KB) by

Performs rotation of an image around any point inside of the image.

Updated

The rotation around a point is performed in following sequence:
2) Rotation of the image around the image's center with IMROTATE.
3) Cropping of the image.

Georg Lousy

Fantastic work!

ALEXEY SHASHKOV

Zainab Un

Zainab Un (view profile)

Thank you...wonderful work

sara

sara (view profile)

Bauyrzhan Aubakir

Jan Motl

Jan Motl (view profile)

Thanks Maider for the comment.

Regarding the small angle problem, I believe that for small angles (like 0.1) and small images (like 300x200) rotateAround should give images identical with the source images (and also identical with the results from imrotate).

Only for bigger images (like 1000x1000) or bigger angles (like 1) the results should be different.

Regarding the loosing of pixels. If I rotate an image by 0 degrees, I should always get the original image (regardless of the point of rotation or image size). With round function, I am not loosing any pixel on testing machines. With floor function, I am loosing pixels.

But it is possible, that internals of imrotate can differ in rounding.

Nevertheless, the function as it is behaves correctly in MATLAB R2012b and MATLAB R2014a.

Ehsan

Ehsan (view profile)

It's Really work!
Thank you, your code solve a major problem in my project.............

Maider

Maider (view profile)

I meant to change lines 35 and 36 in rotateAround from
shiftX = round(pointX-(centerX+newX));
shiftY = round(pointY-(centerY-newY));
to
shiftX = floor(pointX-(centerX+newX));
shiftY = floor(pointY-(centerY-newY));

Maider

Maider (view profile)

Thanks for this submision. This is a very clever way to use imrotate.
But the function can be improved. For example:
1. If you use rotateAround_test and use pointX = 1; and pointY = 1; and run, the result shows that position (1,1) and (2,1) are black as they belong to the lost pixels in the rotation, that should not happen.
2. It does not work well with small angles such as 0.1. In such case rotateAround produces a rotation identical to imrotate.
Running
%% Set test parameters
imageX = 300;
imageY = 200;
pointX = 1;
pointY = 1;
angle = 0.1;

%% Create test image - black dot on white field
image = ones(imageY, imageX);
image(pointY, pointX) = 0;
rot = imrotate(image,angle,'bilinear', 'crop');

%% Rotate the image around the black dot
rotated = rotateAround(image, pointY, pointX, angle,'bilinear');

>> norm(rot - rotated)

ans =

0

I suggest to change lines 35 and 36 in rotateAround from
shiftX = round(pointX-(centerX+newX));
shiftY = round(pointY-(centerY-newY));
to
centerX = floor(imageWidth/2+1);
centerY = floor(imageHeight/2+1);

That solved both problems for me.

Gaurav Mittal

Gaurav Mittal (view profile)

Thanks, works great.

Yun Inn

Jan Motl

Jan Motl (view profile)

Thanks for suggestion. I swapped the parameters of the function call.

Juan P. Viera

Juan P. Viera (view profile)

Really useful. I think you should change the rotation line:
for

jamesyu92

jamesyu92 (view profile)

Works great for me.