Several Image Processing Toolbox functions related to spatial transformations use "tform" structures. A tform structure has data and function handles needed for applying a spatial transformation in the forward or the inverse direction, and sometimes in both directions.
Contents
maketform
Function maketform has several syntaxes for constructing different types of tform structures. For example, you can construct an affine transform by providing an A matrix:
A = [2 0 0; 0 2 0; 0 0 1]; % Stretch by a factor of 2 in both directions. tform = maketform('affine', A);
tformfwd
Toolbox function tformfwd transforms points from input space to output space. For example, to apply the affine transform above to the point (u,v) = (2,3), do this:
uv = [2 3]; xy = tformfwd(tform, uv)
xy =
4 6
tforminv
Function tforminv transforms points from output space to input space. If we apply the inverse transformation to the point xy computed above, we should get back the original point.
uvp = tforminv(tform, xy)
uvp =
2 3
Introducing George
The spatial transformation functions imtransform, tformarray, tformfwd, tforminv, etc., were introduced to Image Processing Toolbox version 3 in 2001. While designing these functions, we almost always had input-space output-space diagrams on our whiteboards so we could keep our notation straight.
Because I have no drawing skills, I always drew a crude head-and-shoulders outline on these diagrams as the "image" being transformed. I started calling this figure "George." Here's what he looks like:
% http://blogs.mathworks.com/images/steve/37/george.mat load george plot(x,y), axis ij, axis equal, axis([-1 1 -1 1]), grid on
Why "George"? Well, any graduate of the Georgia Institute of Technology will recognize George's full name: George P. Burdell. If you don't know George and are curious, I'm sure Google will turn something up for you.
I like to use George to illustrate different affine transforms.
Scaling example
A1 = [2 0 0; 0 2 0; 0 0 1]; tform1 = maketform('affine', A1); uv1 = tformfwd(tform1, [x y]); subplot(1,2,1) plot(x,y), axis ij, axis equal, axis([-2 2 -2 2]), grid on, title('George') subplot(1,2,2) plot(uv1(:,1), uv1(:,2)), axis ij, axis equal, axis([-2 2 -2 2]) grid on title('Scaled by 2')
Rotation example
theta = pi/4; A2 = [cos(theta) sin(theta) 0; -sin(theta) cos(theta) 0; 0 0 1]; tform2 = maketform('affine', A2); uv2 = tformfwd(tform2, [x y]); subplot(1,2,1) plot(x,y), axis ij, axis equal, axis([-2 2 -2 2]), grid on, title('George') subplot(1,2,2) plot(uv2(:,1), uv2(:,2)), axis ij, axis equal, axis([-2 2 -2 2]) grid on title('Rotated by 45\circ')
Translation example
A3 = [1 0 0; 0 1 0; 1 -1 1]; tform3 = maketform('affine', A3); uv3 = tformfwd(tform3, [x y]); subplot(1,2,1) plot(x,y), axis ij, axis equal, axis([-2 2 -2 2]), grid on, title('George') subplot(1,2,2) plot(uv3(:,1), uv3(:,2)), axis ij, axis equal, axis([-2 2 -2 2]) grid on title('Translated')
Get
the MATLAB code
Published with MATLAB® 7.1
Greetings
I tried your code
load george
plot(x,y), axis ij, axis equal, axis([-1 1 -1 1]), grid on
and error in using plot
??? Error using ==> plot
Too many input arguments.
What could cause it
thanks
Athaur - I’d guess that there’s a problem in the way you typed the code into the command window. Perhaps you left a space between “plot” and “(x,y)”? Try putting the plot, axis, and grid calls on separate lines.
hi,
i would like to do some image transformations (namely - translation,
scaling and rotation). I have an input image that is always larger than
the output that i want to produce (e.g. input[384,256],
output[150,130]).
i combined my translation, scaling and rotation values in a 3×3 matrix
but i don’t know how to apply it now to the image. i found in help that
i should use maketform function but I want to do all three
transformation at once and i want the output image to be exactly the
size specified.
how do i do this?
you can e-mail me as well if you like.
thanks,
K.
K - Use imtransform to perform the spatial transformation on your image. Look at the imtransform doc for details on the XData, YData, and Size parameters. These will give you all the control you need to get your output image in the right place, with the right size.
im using maketform to perform shearing , but while taking the inverse to get back the original image image the size of image is more. what should i do so that that shearing can be reverseable ?? Plz help
Sherwin - send me code that illustrates the problem you are having.
the example of the translation only works if the image is a vector of points (like george).
When one replaces george by the cameraman.tif image the translation does not work.
What is am easy way to translate images?
M.
Mark - yes, tformfwd and tforminv are for transforming points. Use imtransform to transform images.
Dear Sir,
The example u have given is very useful and informative.I am trying to run handwritten text but it shows error with message “Cell contents reference from a non-cell array object”.
What is the solution for that?
Thanking you
Damodar - I do not understand what you mean by “I am trying to run handwritten text.”
Hi Steve,
I have typed your code:
load george
plot(x,y), axis ij, axis equal, axis([-1 1 -1 1]), grid on
and I have obtained this answer:
??? Error using ==> load
Unable to read MAT file george.mat
File may be corrupt.
How is it possible?
(I have downloaded george.mat from your site a lot of time!)
thanks
Giancarlo
Giancarlo - please try it again. I resaved the MAT-file using v6 format. Also note that the URL has changed.
hi steve
i am trying to do some transformation(rotation tranlation scale) from 2×2 matrix(image) to 3×3 matrix (basically a plane). the idea is located a pixel from image in another coordinate system(x,y,z), from 2d to 3d
how can i do it?
thanks
David - use tformarray. It can transform multidimensional arrays along arbitrarily chosen dimensions. The number of input dimensions need not match the number of output dimensions. You’ll need to define a custom transformation using maketform.
I wanted to transform one triangle to another using these functions , the code you wrote works for the vertices of the triangle , but do you have any idea about all of the inner points of the source triangle?suppos the source triangle is a part of an image and the goal is to map it to the new frameworkk which is also a triangle.
thanks
Helga - Use imtransform to transform an image. See my collection of useful documentation links.
Sir,
I’m trying to transform an image using the following affine transform equation:
u = a0 + a1*x + a2*y;
v = a3 + a4*x + a5*y;
I define my matrix A = [a0 a1 a2; a3 a4 a5; 0 0 1];
I perform the transformation on each output image pixel using the above equations and place the input pixel values at these newly obtained locations. My range for a0 and a3 is uniformly distributed between [-2.0 2.0] and the others between [-0.04 0.04].
However my affine transformation seems to have jagged edges and I’m unable to remove these. Could you please help me on this.
Sincerely,
Pranjal
Pranjal - From your description, it sounds like you are doing your own image transformation, and that you are using something like nearest-neighbor interpolation. Try using imtransform instead, and select bilinear or bicubic interpolation.
Sir,
Thankyou very much. I have tried imtransform and it works fine.
Sincerely,
Pranjal
Hi
and I’m student in computer since
I have problem with my project….and I think you able to help me to reach the solution
My problem is how to discover the features of the hand image in multi-scale space
After detection and remove the blood and ridge from the image by using log filter
I hope to be interested
hi
I am a student in embedded systems.
I have problem in my project. I think you can suggest me so that i can proceed to other stages of the project.
Problem: I have to register two images, the main aim is to register the object in my image with the object in the other image.For example say image1 and image2 show palm (one of the objects in the image) in the image.Now I like to register only the object, palm. so is it possible to register the images pixel by pixel( i.e. to register the objects pixel by pixel so that it results in shearing, translation and rotation directly on the object ) if this is possible how can this be done.
Kamal - If you have pairs of corresponding points in the two images, then you can compute a spatial registration tranform using the Image Processing Toolbox function cp2tform. You are probably looking for a method that is more automatic. There are several automatic image registration contributions on the MATLAB Central File Exchange; you could give them a try.
hi steve…
I tried your code
load george
plot(x,y), axis ij, axis equal, axis([-1 1 -1 1]), grid on
and the error was
Error using ==> load
Unable to read file george: No such file or directory.
Error in ==> imtransform at 44
load george
Is there any way to load our biomedical image for further transformations
Shalini - You have to download the file george.mat. The URL is given in this post. You can read image files using imread.
Hi, I’ve been looking to apply a transform between two images using correlated point pairs. I’ve noticed that the function cp2tform does this, but that maketform can also be used for this function when supplied with a set of input points and output points: maketform(’transformation’, U, X). What is the difference between these two functions when maketform is used in this way.
Thanks,
Dave
Dave - There’s no difference if you supply the minimum number of points needed to infer the transformation function. cp2tform is used when you have a lot of points, thus overspecifying the transformation. It computes a least-squares solution.
Hi Steve,
While trying to understand how to perform affine transformation on an image using MATLAB, I came into the following problem: image A looks OK for the first coordinate
translation (A starts at [-50 -100]). The second translation is actually an inverse process. However, image B does not start at [0.5 0.5]; it starts at [50 100], instead. Why is that? I’ve post this question in MATLAB Central, but there was no response. Thanks in advance.
———————————-
im = double(imread(’lena.tif’));
T = [1 0 0; 0 1 0; -50 -100 1] % Coordinate translation
tform = maketform(’affine’, T);
[A xdata ydata] = imtransform(im, tform, ‘bilinear’);
figure, imshow(xdata, ydata, uint8(A)), title(’A');
pixval on, axis on, axis([xdata(1) xdata(2) ydata(1) ydata(2)]);
T = [ 1 0 0; 0 1 0; 50 100 1] % Inverse translation
tform = maketform(’affine’, T);
[B xdata ydata] = imtransform(A, tform, ‘bilinear’);
figure, imshow(xdata, ydata, uint8(B)), title(’B');
pixval on, axis on, axis([xdata(1) xdata(2) ydata(1) ydata(2)]);
——————————
Yuan-Liang - In your second call to imtransform, it looks like you haven’t passed in information about the coordinate system for the image A. Therefore, imtransform assumes A is in the “default” location, with its upper left pixel centered at (1,1).
Look at the UData and VData input parameters to imtransform.
I do rotation on medical image(irm),and i want to rerotate this image rotated in the same angle,without any changes in the dimensions,but i can not do this.
Homam - What have you tried and why didn’t it work for you?
i really need this information as it is very important for me in my study
Gehad—What information?
Hi,
I am trying to write a code for translation and rotation of image. I tried following code for rotating my image but it doesn’t give me the translated image even though it shows the right xdata and ydata. I am not sure why it is happening. thanks.
I=imread(’IMoriginal.jpg’);
imagesc(I); colormap gray;
A = [1 0 0; 0 1 0; 50 50 1];
T = maketform(’affine’, A);
[B,xdata,ydata]=imtransform(I, T)
figure; imagesc(B)
I am writing a code for automatic image registration of multi frame tiff images. I created a code for manual selection of control points. My images have 5 black registration points, I can find the centroids of them but what I need is to set a limit on the pixel count. For instance I use bwlabel and I get 55-112 points depending on the intensity range I use some of the labels are on 3-5 pixel points, the registration dots are much larger. Can I set a minimum required pixel count for bwlabel??? Please Help!!! thanks
mike
Michael—Use the function bwareaopen.
thanks I found it just after I posted the comment. You are very helpful. i used Identifying Round Objects post. how can I output only the round object centroid coordinates.
Nevermind I got it. thanks
Samin—See my July 7, 2006 post on translation confusion.
Hi Steve,
I am having some problems using imtransform and tformfwd.
I have to extract a section of an image(a square region) after rotating the image(aligning its axis with the horizontal and vertical axis). When I use imtransform, the size changes and therefore I am unable to find the coordinates of a particular reference point after transformation. Plz help
Anand—Please review my various “spatial transformations” posts between January and August of 2006. (See the blog archives.) The posts cover various coordinate system issues. Or look at the documentation for imtransform, especially the information about the optional output arguments XData and YData. These arguments define the coordinate system for the output image.
I am trying to register two or more images which are sub pixel shifted, rotated, using affine transform on a common grid,but I am not able to get it is there any other method so that I can transfer all these pixels from multiple low resolution images to a single grid
Liya - Here’s a reference for an algorithm to do what you want using phase correlation.
B.S. Reddy, B.N. Chatterji, An FFT-based technique for translation, rotation and scale-invariant image registration, IEEE Transactions on Image Processing 5 (1996) 1266–1271.
Hi Steve,
I am working on registering of a t2 wieghted mri image onto a histology image of the spinal cord.The dimensions of the histology image(which i am using as the base image for imaage registration) is , while the unregistered image is a part of the mri image which i have extracted using bwboundaries and cropim. The dimensions of the unregistered image is and is not fixed as we use cropim.Now whenever i try to register the image using the inbuilt registration tool cpselect and imtransform i do not get the desired effect.I am using a affine transformation. Plz could you guide me in the right direction ?
The dimensions of the base image is 318×442 double and that of the unregistered image is 36×38 double.
Robin—Since you didn’t say exactly what you tried to do, or what undesired effect you received, I can’t guess what might be wrong. You might want to take a look at the three “Image Registration” demos in the Image Processing Toolbox. You can find them here.
I have been working on a coordinate transformation, i was wondering if u might have complete blocks for simulink for the transformations.
Ivan—See the Video and Image Processing Blockset.
Dear Steve,
I have a video sequence of a man walking. I know how to extract the contour of this man from every frame. Now I should save these boundaries in a similar format (imagine that I want to center the shapes and pile up them). Any idea?
My problem can also be seen in this way: you have “George” and “translatedGeorge” and you want to check if they correspond to the same person, so you have to find a transformation T (if it exists) that gives George=T*translatedGeorge
Can you suggest me any approach?
Thanks for your helpfulness.
Mari—No, I don’t any suggestions for you.
Hi,I have a small doubt.I have got a binary image.I have to distort the image using the distortion function x’=0.5*x+x*x/255 and y’=0.5*y+y*y/255;and image size is 255 by 255.how to do the distortion using above function? i mean how to display image using the distorted coordinates? Somehow if i can have a function to display with the given coordinates then can we do it?and wt is that function?Thanks
continued for the above…where x,y are the original coordinates and x’,y’ are the transformed coordinates
Ram—See my blog entry for today (March 31, 2008).
Steve, Can you please provide some tutorial on how stereo maps are constructed. For example, the Mars Rovers and USC’s “Stanely” (that won the first DARPA Grand Challenge) both constructed stereo maps used to guide the vehicles. I would like to learn the algorithms, how the maps are constructed, and how they are used. Thanks, Jon
Jon—I’m sorry, but I’m not familiar with stereo map algorithms.
Thank you steve,but i have doubt that- maketform(’affine’,[.5 0 0; .5 1.5 0; 100 200 1]); how do u define this matrix[.5 0 0; .5 1.5 0; 100 200 1]? is it standard matrix or randomly chosen or how it relates to the distortion function that i have? sorry if it too silly question..many thanks..
if i have transform like x->a*x+x*x then how do we define the matrix?
Ram—The documentation for maketform includes the affine transform equation that relates input-space and output-space coordinates. As for what the terms mean, there is a tremendous amount of information available online about affine transforms. For a discussion that’s specific to both MATLAB and image processing, see section 5.11 of Digital Image Processing Using MATLAB.
Hi Steve,
I need to change x,y co ordinates of a given image to x’,y’ according to a distortion function given by
x’=a*x+((1+a)x*x)/L and y’=a*y+((1+a)y*y)/L where a is any scalar between 0 and 1 and L=256. Can you please help me out. Thanks.
Hi Steve,
Is it possible to make rotation and translation together?
This should to save calculation time in my opinion.
I tried like this:
XForm = [cos(Phi) -sin(Phi); sin(Phi) cos(Phi); TX TY];
tform = maketform(’affine’, XForm);
TX and TY - translation vectors
Phi - rotation angle
but now i don’t know what is the first operation in the maketform function, rotation and than translation or translation and than rotation?
Thanks Sladjan
Sri—See my post about defining custom transformations.
Sladjan—The equation showing how the affine transform matrix is used is in the reference page for maketform. You can take a look how your matrix gets applied with that equation to see the order of things. In general, rotation and translation are both affine transforms, and the composition of two affine transforms is also affine. You form the composition by multiplying the two affine matrices together.
Ram—x->a*x+x*x is not an affine transformation. You’ll have to do it using a custom transformation.
Hi Steve,
I want to transform a image one position into correct rectangular position.AT what algorithm use it pls reply to me
Krishnamoorthy—Your question is pretty vague. You might find the imtransform, maketform, and cp2tform functions to be helpful. Also, look at the image registration demos on the Image Processing Toolbox demos page.
Hi,
I have an image,named ‘I’, that I’d like transform from pixel coordiantes to UTM coordianates. For do this, I know some points of image called GPC(ground control points). For example the pixel (x=50,y=50) correspond with UTM coordinates 4900000 370000.
First of all, I use cp2tform in this way:
geo_aug07 = cp2tform(XY,xy,’projective’);
where XY is pixel matrix end xy is UTM matrix. XY=[ 50 50, … …,… …] xy=[4900000 370000,… ….,…. ….] geo_aug07 is a structure.
Then, I use imtransform in this way:
[I_proj3,XX,YY] = imtransform(I,geo_aug07);
where I is input image, I_proj3 is output image. using imshow (XX,YY,I_proj3) I have an output image according to projective transformation. In this image I can see each pixel with the respective coordinate UTM.
The geometry end the size are different between I end I_proj3. size(I_proj3)>size(I).
Now my problem is to obtain from output image an image like the input image(with the same size) where each pixel is matched with UTM coordinate obtained form output image. For do this I think to use an inverse function of imtransform, but I don’t know how works.
Does exist an inverse Matlab function of imtransform?
Thanks
Andrea—Assuming your spatial transformation has an inverse, you would form a tform struct with the inverse transformation and then use imtransform to apply it.
I formed an inverse tform struct with fliptform and then I apply imtransform. The result is not correct. why? is it possibile that does not exist an inverse trasformation of imtransform?
Andrea—What’s not correct about it?
Hi Steve,
When doing an image transformation with some predefined coefficients, I end up with some of the output pixel coordinates going in to the negative field. I am using a relation like this to map the pixels: I2(x,y,:)=I1(u,v,:)and I get the error for negative indexes.
1-How can I deal with the negative indexes?
2-What do you think is the best way to interpolate the pixels?
Best,
Leonardo
Leonardo—I think the best way to handle questions 1 and 2 is to use imtransform. But if you want to implement the spatial image transform yourself, you have a variety of choices. There is no “best” answer to either question. For input-space coordinates outside the defined image boundary, for example, you could assume the image pixels are zero, or you could use the nearest image pixel on the boundary. For interpolation, you have many choices. Bilinear interpolation is probably most common, although bicubic interpolation is now the default in many applications, including the Image Processing Toolbox. And there are circumstances in which nearest neighbor would be best.
Hi Steve,
when using tformfwd to find corresponding points in the new space, the resulting co-ordinates from tformfwd do not match the transformed image display (using for instance datacursor to select a couple of points). As an example, I’m working on the following piece of code:
tTransf = eye(3);
NormMat = [0.6851 -0.3149; -0.3149 0.6851];
tTransf(1:2,1:2)= NormMat;
tform=maketform(’affine’,tTransf);
new_im = imtransform(im,tform);
The problems occur when I call:
new_pt = tformfwd(tform, pt);
The co-ordinates given by new_pt do not match the diplayed point!!!I should add that in my work, I have not control over NormMat, it is computed iteratively. But I have noted that tformfwd work fine when the values in NormMat are all positive.
Any suggestion please ?
Omar—You didn’t say how you are displaying the image, so I can only guess at the problem. The function imtransform will by default compute the result only within the bounding box of the transformed image. This has the implicit effect of removing translations. Try getting the using the xdata, ydata as additional output arguments from imtransform and then passing them to imshow. For more info, see my posts on translation confusion and on controlling the input- and output-space grids.
Hi Steve,
I have an image which i have captured at an angle. Basically a perspective distorted image.
How can i use maketform to bring it back to a rectangular shape.
I tried imtransform with some parameters and managed to get it to turn a little. But i get the same result when i use imresize to some specific size.
Kindly help
Thanks
Ash
Ash—Try using cp2tform and cpselect.
Hi Steve,
Tried using cp2tform and cpselect. But i dont have a base image. I need to just remove the projective distortion. I’ll try again.
Thanks
Ash
Hi Steve,
Many thanks for your reply.
It is not the display that I’m after. I just used it to check my results.
What I’m trying to do, is to track a point, say Pt, in the original image, which is transformed into NPt in the transformed space. Using the function tformfwd(tform, Pt) alone has given me some correct results and in cases, completely wrong results. So I’ve written this little piece of code (as an example):
im = imread(’pout.tif’);
mat = [-0.4218 0.7922; 0.9157 0.7922];
tTransf = eye(3);
tTransf(1:2,1:2) = mat;
tform=maketform(’affine’,tTransf);
[new_im xdata ydata] = imtransform(im,tform);
figure(1); imshow(im);
figure(2); imshow(new_im);
Pt = [116 93];
NPt = tformfwd(tform, Pt);
NPt = NPt + [1 1] - [xdata(1) ydata(1)];
Pt here is the left eye of the kid. That code seems to be working fine when the elements of mat are smaller than one.
As an example where I’ve got wrong results, I simply change mat to mat = [3 -1; 2 -2] and run the code again. That code no longer produces correct results.
What do you reckon?
Regards
Omar
Omar—I don’t see anything wrong. I tried the following code:
mat = [3 -1; 2 -2]; tTransf = eye(3); tTransf(1:2,1:2) = mat; tform = maketform('affine', tTransf); im = imread('pout.tif'); [new_im xdata ydata] = imtransform(im, tform); Pt = [116 93]; NPt = tformfwd(tform, Pt); imshow(new_im, 'XData', xdata, 'YData', ydata); hold on axis on plot(NPt(1), NPt(2), 'y*')And here’s the result:
Looks like the point is being plotted in the correct location.
Hi Steve,
The display is correct as you have demonstrated.
But I want to access the pixel new_im(NPt) (precisely, new_im (NPt(2), NPt(1) ), for some processing operations in its neighbourhood, also in the context of my work the location of NPt has to be as precise as possible, otherwise my algorithm wouldn’t converge.
In the example above, NPt = tformfwd(tform, Pt) returns (534 -302) which is out of the range of new_im. To solve the problem, in my first example I used the formula NPt = NPt + [1 1] - [xdata(1) ydata(1)]; as follows:
[new_im xdata ydata] = imtransform(im,tform);
Pt = [116 93];
NPt = tformfwd(tform, Pt);
NPt = NPt + [1 1] - [xdata(1) ydata(1)];
which is working fine and removes the translation effect. However, for mat = [3 -1; 2 -2] that trick does not work. I’ve also noticed that in the second example when mat = [3 -1; 2 -2], xdata is equal to [50 1302.5] and ydata is equal to [-822 -3], leading to an imgae of size 820 x 1254, where the transformed new_im is of size 583×923.
I wonder if there is a scaling factor (in addition to translation) I have to take care of ?
Having said that, my question is about how to access new_im at NPt, when the computed NPt is out of range.
Regards
Omar
Omar—The computed NPt is not out of range. It is exactly where your own defined transformation took it. To determine how to convert NPt into image pixel coordinates, you need to use both [xdata(1) ydata(1)] and [xdata(2) ydata(2)], because there is potentially a scale factor applied by imtransform in order to keep the output image size from growing too much. You can control this directly by using the XYScale parameter when you call imtransform.
Hello,
I have a random displacement field with vectors corresponding to each pixel in an image and I want to ask if it’s possible to apply that field directly to an image using imtransform? I understand that I might have to create a tform using ‘custom’ option. Do you know if such a thing is possible?
Thank you,
A
Amalia—Yes, that is possible and can be fun. See my post about defining and applying custom transformations.
Steve,
Function findbounds returns slightly different results form those of XDATA and YDATA in imtransform. The relative difference is up to %0.25.
Malik—Once the bounds are determined in output space, you have to establish a grid of output-space pixel locations at which to compute output image pixels. Since the grid is regularly spaced, the border pixels might not line up exactly with the output of findbounds. That means that XData and YData have to be adjusted to match the output pixel grid. See the subfunction recompute_output_bounds inside imtransform, as well as the code that calls it.
Steve,
I need to use a image transform in which the output would be a bigger resized image not only by expanding pixels forward out the original one but also expanding them in the inner direction in case of holes. Being more understandable, the code would generate a ring with greater perimeter and smaller hole than a ring used as an input image.
Tomy—What is the question?
Sorry, I’ve submitted the comment before finish it.
The transformation would be similar to the output given by the imopen function but, for the application I am developing I need to keep as much as pixel resolution as possible. I’ve tried with different functions but noone match the results I’m looking for. could you give me any suggestion about the topic?
Thanks in advance.
Tomy—I don’t see any connection between your image transform description and imopen. Regarding the pixel resolution, you might want to look at my post about controlling the input and output grids with imtransform.
Steve,
Probably I was not clear enought in the explanation of my question, so let me show you in a simple way the transformation I am looking for:
Suposse I have a squared ring as shown next,
The function I am trying to design should out a result similar to the next figure,
The main problem I have is that I don’t know how to remap the image in order to obtain those results.
I though in determining the euclidean distance from the image pixels to the background and try to expand the evaluated pixels towards the direcction where such distance is minimum but I dont know how to expand/contract the pixels according with the object boundary location in a coherent way.
Maybe there is no relation with the imopen funtcion I mentioned in the last question but I’ve shown by using the imdilate the results are quite similar as I need, however, the output image is blurred.
Thanks
I’m really sorry but the squared rings have not been shown as I wanted.
The first ring had a 44u of outer perimeter and 16u of inner perimeter,the second one is 52u of outer perimeter and 12u of inner. This is just an example for explaining what the function should do.
Tomy—I can’t help you with your remapping function definition, but once you get that worked out, take a look at my post on custom spatial transformations.
Dear Steve:
I had wrote the following codes:
{
xscale = 5;
yscale = 5;
rotation = 30;
oldimg = imread(’house.bmp’);
theta = rotation*pi/180;
As = [xscale 0; 0 yscale];
Ath = [cos(theta) -sin(theta);sin(theta) cos(theta)];
A = Ath*As;
T = eye(3,3);
T(1:2,1:2)=A;
tform=maketform(’affine’,T);
newimg = imtransform(oldimg,tform);
size(newimg)
figure,imshow(newimg);
}
My problem is as follows:
As xscale>=2 and yscale>=2, the size of the transfomed image is not the one I expected. In the above example, the size of the origin image is 256X256, however, the new image size is 513X513.
imtransform has logic in it that automatic adjusts the size (in X-Y space) of the output pixels in order to keep the dimensions of the output image from growing too large.
To prevent this automatic (and unexpected) adjustment of the output pixel size, specify the ‘XYScale’ parameter to be 1, like this:
when i follow ur code for rotation i got errors in matlab mentioned at the end ?
My code is
for th=-90:90; x(th+181)=cos((th*pi)/180)*(8-(0.33*((th*pi)/180))*((th*pi)/180)); y(th+181)=sin((th*pi)/180)*(8-(0.33*((th*pi)/180))*((th*pi)/180)); end plot(x,y), axis ij, axis equal, axis([-15 15 -15 15]), grid on theta = pi/4; A2 = [cos(theta) sin(theta) 0; -sin(theta) cos(theta) 0; 0 0 1]; tform2 = maketform('affine', A2); uv = tformfwd(tform2, [x y]) %% after the execution of %%this line and the error appears to be in MATLAB are ??? Error using ==> tform>checkCoordinates at 141 Function TFORMFWD: SIZE(U) is inconsistent with T.ndims_in. Error in ==> tform at 42 [U, D] = checkCoordinates( f, P, A{1} ); Error in ==> tformfwd at 68 varargout = tform('fwd', nargout, varargin{:});Any suggestions ?
Qureshi—You are calling tformfwd incorrectly. It is expecting its second input to have two columns, but you are passing it [x y], which is a 542-element row vector. You’ve got a bug in your indexing offset (181 is wrong), and you are computing x and y as row vectors instead of column vectors. And there’s no need for that for-loop. Consider doing something like this:
Hi steve
Is it to possible to shift the subpixels in an image ? if possible how ?. Does matlab provide any algorithm for subpixels shifting?
Regards
Qur—Do you mean subpixel translation? Then yes, you can do that with imtransform and an affine transformation. Be sure to look at this post before you try it, though.