Comments on: Spatial transformations: Defining and applying custom transforms http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/ Steve Eddins manages the Image & Geospatial development team at <a href="http://www.mathworks.com/">The MathWorks</a> and coauthored <a href="http://www.mathworks.com/support/books/book5291.html?category=-1&language=-1">Digital Image Processing Using MATLAB</a>. He writes here about image processing concepts, algorithm implementations, and MATLAB.<br><br><img> Mon, 23 Nov 2009 00:57:39 +0000 http://wordpress.org/?v=2.3.1 By: Steve http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22233 Steve Mon, 19 Oct 2009 19:27:15 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22233 Douglas—I don't know what you mean by "extract the data point from the deformation grid." Can you be more specific? Douglas—I don’t know what you mean by “extract the data point from the deformation grid.” Can you be more specific?

]]> By: douglas http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22213 douglas Sun, 11 Oct 2009 08:12:36 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22213 Dear Steve, thx for your answer. I can use the imtransform now. However, I have no idea on how to extract the data point from the deformation grid. Could you help? Dear Steve,
thx for your answer. I can use the imtransform now. However, I have no idea on how to extract the data point from the deformation grid. Could you help?

]]> By: Steve http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22117 Steve Tue, 15 Sep 2009 12:16:19 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22117 Douglas—Yes. This post shows how to do that using <tt>imtransform</tt>. Or did I misunderstand your question? Douglas—Yes. This post shows how to do that using imtransform. Or did I misunderstand your question?

]]> By: douglas http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22116 douglas Tue, 15 Sep 2009 12:00:50 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-22116 <pre> Can we apply the deformation grid to an image? </pre> Can we apply the deformation grid to an image? ]]> By: Steve http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-21193 Steve Wed, 29 Oct 2008 22:36:25 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-21193 Shiva—If you look at the various blogs posts in the <a href="http://blogs.mathworks.com/steve/category/spatial-transforms/" rel="nofollow">spatial transforms</a> category, you'll find some descriptions of how they work. The Image Processing Toolbox functions <tt>imtransform</tt> and <tt>maketform</tt> use <em>inverse mapping</em>, which is described in the posts. Whether it is linear depends on whether you are talking about the warping function, or the process of warping an image. Image warping is linear, meaning that warping the image (aF + bG) is the same as a*warp_of_F + b*warp_of_G. Shiva—If you look at the various blogs posts in the spatial transforms category, you’ll find some descriptions of how they work. The Image Processing Toolbox functions imtransform and maketform use inverse mapping, which is described in the posts.

Whether it is linear depends on whether you are talking about the warping function, or the process of warping an image. Image warping is linear, meaning that warping the image (aF + bG) is the same as a*warp_of_F + b*warp_of_G.

]]> By: Shiva http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-21105 Shiva Fri, 26 Sep 2008 02:03:31 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-21105 Hello Steve, I am working on a project where I need to simulate non-linear deformations in the images. Since these are MR images, the deformations can't be as severe as the examples given above. Currently, I am using the following code to create the deformations. a = 0.95; b = 1.05; Tform_values = a + ((b-a) * rand(1,30)); for i = 1:30 f = @(x) ((x(:,:).^Tform_values(i))); g = @(x, unused) f(x); tform3 = maketform('custom', 2, 2, [], g, []); img_stack_Tformed(:,:,i) = imtransform(img_stack_org(:,:,i),tform3,'size',size(img_stack_org(:,:,i))); end What I am not able to understand is how it the spatial transformation created using the function f given below. I am not sure whether the created transform is non-linear. Any information regarding this will be really helpful. Thanks, Shiva Hello Steve,

I am working on a project where I need to simulate non-linear deformations in the images. Since these are MR images, the deformations can’t be as severe as the examples given above. Currently, I am using the following code to create the deformations.

a = 0.95; b = 1.05;
Tform_values = a + ((b-a) * rand(1,30));
for i = 1:30
f = @(x) ((x(:,:).^Tform_values(i)));
g = @(x, unused) f(x);
tform3 = maketform(’custom’, 2, 2, [], g, []);
img_stack_Tformed(:,:,i) = imtransform(img_stack_org(:,:,i),tform3,’size’,size(img_stack_org(:,:,i)));
end

What I am not able to understand is how it the spatial transformation created using the function f given below. I am not sure whether the created transform is non-linear. Any information regarding this will be really helpful.

Thanks,

Shiva

]]> By: Steve http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-20605 Steve Wed, 16 Apr 2008 17:50:37 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-20605 Tim—You could try something like this: <pre> r = @(x) hypot(x(:,1), x(:,2)) theta = @(x) atan2(x(:,2), x(:,1)) f = @(x) [r(x) .* cos(theta(x) + r(x)/3), ... r(x) .* sin(theta(x) + r(x)/3)]; g = @(x, unused) f(x); </pre> and then form a custom tform as shown in this post. Tim—You could try something like this:

r = @(x) hypot(x(:,1), x(:,2))
theta = @(x) atan2(x(:,2), x(:,1))
f = @(x) [r(x) .* cos(theta(x) + r(x)/3), ...
    r(x) .* sin(theta(x) + r(x)/3)];
g = @(x, unused) f(x);

and then form a custom tform as shown in this post.

]]> By: Tim http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-20572 Tim Tue, 08 Apr 2008 15:29:07 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-20572 These are most helpful in implementing in MATLAB the algorithms specified in "Beyond Photography". Probably due to my beginner status in using the Image Processing Toolbox, I was unable to get the the transformation where r -> r but theta -> theta + r/3 This is on page 44 in "Beyond Photography". These are most helpful in implementing in MATLAB the algorithms specified in “Beyond Photography”. Probably due to my beginner status in using the Image Processing Toolbox, I was unable to get the the transformation where
r -> r but
theta -> theta + r/3
This is on page 44 in “Beyond Photography”.

]]> By: Faisal http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-9286 Faisal Fri, 14 Sep 2007 05:07:23 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-9286 Awsome stufff ! This is really interesting ! Thanks :) Awsome stufff ! This is really interesting ! Thanks :)

]]> By: Steve http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-3451 Steve Fri, 11 Aug 2006 20:07:39 +0000 http://blogs.mathworks.com/steve/2006/08/04/spatial-transformations-defining-and-applying-custom-transforms/#comment-3451 Matthew - Well, wizarding powers aside, I'm really more of an engineer than a scientist. So far, no ill effects are apparent. Matthew - Well, wizarding powers aside, I’m really more of an engineer than a scientist. So far, no ill effects are apparent.

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