I have a 3D matrix and want to calculate the gradient for this matrix. The only step which I did is defining image gradient for a 2D image as follow:
sx = [-1 0 1; -2 0 2; -1 0 1];
sy = [1 2 1; 0 0 0; -1 -2 -1];
Gx = conv2(input, sx, 'same');
Gy = conv2(input, sy, 'same');
grdmg = sqrt(Gx.^2+Gy.^2);
grddr = atan2(Gy,Gx);
I am wondering that how I can extend this code to 3D? I know that in 3D matrix I should use convn, but do not know that how I should change the sx, sy and make sz.
If you want to calculate a 3D gradient, you would have to make your kernel 3D as well. Not only are you checking for changes in the ith slice of your matrix, but you also need to check the (i-1)th slice and the (i+1)th slice as well.
As a minor comment, sy should be the transpose of sx, but you don't have that here. Make sure you change this before you do anything.
Also, with the added dimension, you have three more possible kernels because you need to check for changes horizontally in each slice, vertically in each slice, or temporally. As such, you would simply make sz either by replicating sx along the third dimension three times, or replicating sy along the third dimension three times or making a temporal kernel where the first slice is all 1, the second slice is all zero and the third slice is all -1. Therefore, calling all of these kernels szx, szy and szz, you would create them as:
szx = cat(3, sx, sx, sx);
szy = cat(3, sy, sy, sy);
szz = cat(3, ones(3,3), zeros(3,3), -ones(3,3));
To compute the magnitude, you would simply find the magnitude of all of these components together and use convn as you have suggested:
Gx = convn(input, szx, 'same');
Gy = convn(input, szy, 'same');
Gz = convn(input, szz, 'same');
grdmg = sqrt(Gx.^2 + Gy.^2 + Gz.^2);
As for the angle, this is very ambiguous because you now have three components, and if you want to find the angle, you would need to have two vectors in 3D space and find the angle in between those two vectors. For 2D this is well defined as you have already created 2 vectors, and so the angle is simply the arctan between the vertical and horizontal components. Because you have three components, and there are two to choose from, there are three possible angles you can compute, each with respect to either the x, y or z axis.
I don't have an answer for you here in this case, but this particular example from Math StackExchange may give you further insight on how to calculate the angle using the 3D gradient. The important point is that you need to know which axis you are measuring the angle with respect to.
https://math.stackexchange.com/questions/569019/how-do-you-get-3d-gradient-direction-and-magnitude
Good luck!
Related
I am carrying out 3D imaging alignment using (as an overview) the following coding scheme:
ref = imref3d(size(img(:,:,:,1)),[0 32] ,[0 32] ,[0 20] );
[optimizer, metric] = imregconfig( 'monomodal');
tform=imregtform(img(:,:,:,2), ref, img(:,:,:,1), ref, 'rigid', optimizer, metric);
transform_mat=tform.T
For clarification, the variable img is a time series of 3D images (where each 3D image is a stack of 2D cross sections). At different time points, the object I am imaging can move. Thus, the purpose of the above code is to generate a geometric transformation that optimally aligns the second time point of the 3D volume ( img(:,:,:,2) ) to a reference image, which is the first time point of the 3D volume ( img(:,:,:,1)). The geometric transformation that is finally chosen by the optimization algorithm is output to the object tform. tform contains a 4x4 matrix as one of its properties (T); it is this 4x4 matrix that encodes the translation and rotation information. As you can see from my above code, I stored this 4x4 matrix in the variable transform_mat.
After reading through a lot of scattered mathworks documentation, I determined that this transform_mat variable (representing an affine rigid body transformation matrix) is in a "post-multiplied" form, which, as I understand it, just means that it is the transposed version of what one would traditionally see in a linear algebra text book.
For my purposes, I am interested in extracting specific rotation information from transform_mat. Here is what I have done so far:
rot_postmultiply=transform_mat(1:3,1:3); %extracting the elements that encode rotation-related info
rot_premultiply=rot_postmultply'; %transposing
Just to quickly interject, I created the premultiplied version of the rotation matrix because I believe that many of the functions that act on rotation matrices assume that it is in its premultiplied form.
From this 3x3 rotation matrix, I want to extract the rotations (in radians) about the STATIC x-axis, y-axis, and z-axis of the reference image. My initial attempt at carrying this out is as follows:
eul = rotm2eul(rot_premultiply);
The rotm2eul function provides me the 3 Euler Angles associated with this 3x3 rotation matrix. eul is a 1x3 vector and, according to documentation, the "default order for Euler angle rotations is 'ZYX' ". However, I am not certain that Euler Angles are actually describing the information I want to extract (i.e. rotations about the static x, y, z axes of the reference image).
I do not have a strong linear algebra/geometric transformation background, but my understanding of Euler Angles is that each rotation that takes place changes the coordinate system. For example, after the rotation about the Z axis (the first value in the eul vector), we now have new X and Y axes (call them X' and Y'). Then the "Y-axis rotation" (the second value in eul) is actually the rotation about Y'...not Y. Repeat this argument for the final "X-rotation" (which would really be about an X'' axis).
If anyone could offer some insight about how to proceed (and if my concerns about Euler Angles are correct), I would greatly appreciate it!
Also, sorry if the word "static" is the incorrect terminology. Hopefully, I have provided sufficient context so that no confusion arises.
If I understand you correctly, all you need is rotm2axang that converts a rotation given as an orthonormal rotation matrix, rotm, to the corresponding axis-angle representation, axang. The input rotation matrix must be in the premultiply form for rotations.
The output of rot2axang is an n-by-4 matrix of n axis-angle rotations. The first three elements of every row specify the rotation axis, and the last element defines the rotation angle (in radians).
If you dont have access to the robotics system toolbox that has that function, consider this alternative instead.
EDIT:
Rotation of a point in 3D space by an angle alpha about an arbitrary axis defined by a line between two points (x1,y1,z1) and (x2,y2,z2) can be achieved by the following steps:
translation so that the rotation axis passes through the origin
rotation around x axis so that the rotation axis lies in the xz plane (or projection on the xz plane)
rotation around y axis so that the rotation axis lies along the z axis (or projection on the z axis)
rotation by alpha around the z axis
the inverse of step 3
the inverse of step 2
the inverse of step 1
Are you asking about the angles in steps 2,3?
Otherwise, are you already in the origin for the axis around which you rotate?
If indeed you need the angles in step 2,3 then:
step #1 is done via a the translation matrix T:
T = [ 1 0 0 -x1
0 1 0 -y1
0 0 1 -z1
0 0 0 1 ];
Step #2 is done via the following:
Given that v = (a,b,c) is the unit vector along the rotation axis, then d = sqrt(b^2 + c^2) as the length of the projection onto the yz plane. Then rotate by the angle between the projection of rotation axis in the yz plane and the z axis given by the dot product of the z component of the unit vector v and its yz projection. This angle (beta) is determine by: cos(beta) = dot([0,0,c],[0,b,c]) /(c*d) = c/d , sin(beta) = cross( [0,0,c],[0,b,c]) /(c*d) = b/d
The rotation matrix Rx is therefor:
Rx = [ 1 0 0 0
0 c/d -b/d 0
0 b/d c/d 0
0 0 0 1 ] ;
Step 3 we rotate around the y axis so that the rotation axis lies along the positive z axis. Using dot and cross product relationships the cosine of the angle is d, the sine of the angle is a. The rotation matrix about the y axis Ry is
Ry = [ d 0 -a 0
0 1 0 0
a 0 d 0
0 0 0 1];
Last, we rotate around z simply by the angle alpha
Rz = [ cos(alpha) sin(alpha) 0 0
-sin(alpha) cos(alpha) 0 0
0 0 1 0
0 0 0 1 ] ;
I think the answers to this question should help you out: to decompose a 3D rotation into Cartesian components.
I am attempting to use the inverse mapping approach to multiply an inv(3x3) matrix to an image(540x420) to rotate it! I cannot use any toolbox functions and am attempting to do a double FOR loop. I am confused about how this works if anyone could help me out it would be greatly appreciated!
NOTE: A is the image! I have a bilinear interpretation code after this that will fix the image accordingly. I am just a having a hard time wrapping my head around this (3x3) * (540x420) matrix.
[xold,yold,k] = size(A);
xnew = xold;
ynew = yold;
Rot = [cosd(angle) -sind(angle) 0 ; sin(angle) cos(angle) 0; 0 0 1];
InverseRot = double(inv(Rot));
E = zeros(xnew, ynew,k);
E = double(E);
A = double(A);
for i=1:ynew
for j=1:xnew
E(i,:) = %This is where I'm confused
end
end
You're not multiplying the inverse rotation matrix by the image, you're multiplying it by the homogeneous coordinates of a point in the new image to get the corresponding homogeneous coordinates in the original image.
newpt = [xnew ynew 1].';
origpt = InverseRot*newpt;
xold = origpt(1);
yold = origpt(2);
Notes:
Your rotation matrix mixes the degree and radian versions of the trig functions. That will mess up your results.
Rotations without translations are about the origin. Your image is not centered at the origin. You'll need to do something about that.
The inverse of a rotation matrix is the same as its transpose. If you fix the note above, you'll end up with an affine transformation matrix rather than a straight rotation matrix. This also has an easier way to calculate the inverse. If the affine transformation matrix is [R | t; 0 | 1], the inverse is [RT | -RT*t; 0 | 1] where RT is the transpose of R.
The coordinates in the original image that you get after transformation are not likely to be integers, so you won't be able to simply assign the value of a pixel in the old image to a pixel in the new image. As you said, you'll have to find some way to interpolate the value and assign that to the point in the new image.
Given scatter data, or a matrix, I would like to generate a nice plot such as the one shown below, with all 3 histograms and a colored matrix. I'm specifically interested in the diagonal histogram, which ideally, would correspond to the diagonals of a matrix:
Source figure: www.med.upenn.edu/mulab/jpst.html
The existing command scatterhist is not that powerful to generate this type of graph. Any ideas?
Thanks!
EDIT:
Following #Cris Luengo's hints, I came up with the following code which does some first work at the inclined histogram: WORK IN PROGRESS (HELP WELCOME)!!
b = [0 1 2 3 4 5 6 7 8 9 10];
h = [0.33477 0.40166 0.20134 0.053451 0.008112 0.000643 2.7e-05 0 0 0 0];
wid = 0.25; bb = sort([b-wid b-wid b+wid b+wid]);
kk = [zeros(numel(h),1) h(:) h(:) zeros(numel(h),1)];
kk = reshape(kk',[1,numel(kk)]);
pp=patch(bb,kk,'b');axis([-.5 5 0 .5])
set(gca,'CameraUpVector',[-1,.08,0]);axis square
EDIT 2: Using rotation
phi = pi/4;
R = [cos(phi),-sin(phi);sin(phi),cos(phi)];
rr = [bb' kk'] * R;
bb = rr(:,1); kk = rr(:,2);
patch(bb,kk,'b'); axis([-.5 3 -4 .5])
Here is a recipe to plot the diagonal histogram, if you can do that I’m sure you can figure out the rest too.
Compute the histogram, the bin counts are h, the bin centers are b.
Build a coordinate matrix, attaching the coordinates of a point on the x-axis at the left and right ends of the histogram:
coords = [b(:),h(:)];
coords = [coord;b(end),0;b(1),0];
Using patch you can now plot the histogram as follows:
patch(coords(1,:),coords(2,:));
To plot a rotated histogram you can simply multiply the coords matrix with a rotation matrix, before using patch:
phi = pi/4;
R = [cos(phi),-sin(phi);sin(phi),cos(phi)];
coords = R * coords;
You might need to shift the plot to place it at the right location w.r.t. the other elements.
I recommend that you place all these graphic elements in the same axes object; you can set the axes’ visibility to 'off' so that it works only as a canvas for the other elements.
It will be a bit of work to get everything placed as in the plot you show, but none of it is difficult. Use the low-level image, line,patch and text to place those types of elements, don’t try to use the higher-level plotting functions such as plot, they don’t provide any benefits over the low-level ones in this case.
I have two images which one of them is the Original image and the second one is Transformed image.
I have to find out how many degrees Transformed image was rotated using 3x3 transformation matrix. Plus, I need to find how far translated from origin.
Both images are grayscaled and held in matrix variables. Their sizes are same [350 500].
I have found a few lecture notes like this.
Lecture notes say that I should use the following matrix formula for rotation:
For translation matrix the formula is given:
Everything is good. But there are two problems:
I could not imagine how to implement the formulas using MATLAB.
The formulas are shaped to find x',y' values but I already have got x,x',y,y' values. I need to find rotation angle (theta) and tx and ty.
I want to know the equivailence of x, x', y, y' in the the matrix.
I have got the following code:
rotationMatrix = [ cos(theta) sin(theta) 0 ; ...
-sin(theta) cos(theta) 0 ; ...
0 0 1];
translationMatrix = [ 1 0 tx; ...
0 1 ty; ...
0 0 1];
But as you can see, tx, ty, theta variables are not defined before used. How can I calculate theta, tx and ty?
PS: It is forbidden to use Image Processing Toolbox functions.
This is essentially a homography recovery problem. What you are doing is given co-ordinates in one image and the corresponding co-ordinates in the other image, you are trying to recover the combined translation and rotation matrix that was used to warp the points from the one image to the other.
You can essentially combine the rotation and translation into a single matrix by multiplying the two matrices together. Multiplying is simply compositing the two operations together. You would this get:
H = [cos(theta) -sin(theta) tx]
[sin(theta) cos(theta) ty]
[ 0 0 1]
The idea behind this is to find the parameters by minimizing the error through least squares between each pair of points.
Basically, what you want to find is the following relationship:
xi_after = H*xi_before
H is the combined rotation and translation matrix required to map the co-ordinates from the one image to the other. H is also a 3 x 3 matrix, and knowing that the lower right entry (row 3, column 3) is 1, it makes things easier. Also, assuming that your points are in the augmented co-ordinate system, we essentially want to find this relationship for each pair of co-ordinates from the first image (x_i, y_i) to the other (x_i', y_i'):
[p_i*x_i'] [h11 h12 h13] [x_i]
[p_i*y_i'] = [h21 h22 h23] * [y_i]
[ p_i ] [h31 h32 1 ] [ 1 ]
The scale of p_i is to account for homography scaling and vanishing points. Let's perform a matrix-vector multiplication of this equation. We can ignore the 3rd element as it isn't useful to us (for now):
p_i*x_i' = h11*x_i + h12*y_i + h13
p_i*y_i' = h21*x_i + h22*y_i + h23
Now let's take a look at the 3rd element. We know that p_i = h31*x_i + h32*y_i + 1. As such, substituting p_i into each of the equations, and rearranging to solve for x_i' and y_i', we thus get:
x_i' = h11*x_i + h12*y_i + h13 - h31*x_i*x_i' - h32*y_i*x_i'
y_i' = h21*x_i + h22*y_i + h23 - h31*x_i*y_i' - h32*y_i*y_i'
What you have here now are two equations for each unique pair of points. What we can do now is build an over-determined system of equations. Take each pair and build two equations out of them. You will then put it into matrix form, i.e.:
Ah = b
A would be a matrix of coefficients that were built from each set of equations using the co-ordinates from the first image, b would be each pair of points for the second image and h would be the parameters you are solving for. Ultimately, you are finally solving this linear system of equations reformulated in matrix form:
You would solve for the vector h which can be performed through least squares. In MATLAB, you can do this via:
h = A \ b;
A sidenote for you: If the movement between images is truly just a rotation and translation, then h31 and h32 will both be zero after we solve for the parameters. However, I always like to be thorough and so I will solve for h31 and h32 anyway.
NB: This method will only work if you have at least 4 unique pairs of points. Because there are 8 parameters to solve for, and there are 2 equations per point, A must have at least a rank of 8 in order for the system to be consistent (if you want to throw in some linear algebra terminology in the loop). You will not be able to solve this problem if you have less than 4 points.
If you want some MATLAB code, let's assume that your points are stored in sourcePoints and targetPoints. sourcePoints are from the first image and targetPoints are for the second image. Obviously, there should be the same number of points between both images. It is assumed that both sourcePoints and targetPoints are stored as M x 2 matrices. The first columns contain your x co-ordinates while the second columns contain your y co-ordinates.
numPoints = size(sourcePoints, 1);
%// Cast data to double to be sure
sourcePoints = double(sourcePoints);
targetPoints = double(targetPoints);
%//Extract relevant data
xSource = sourcePoints(:,1);
ySource = sourcePoints(:,2);
xTarget = targetPoints(:,1);
yTarget = targetPoints(:,2);
%//Create helper vectors
vec0 = zeros(numPoints, 1);
vec1 = ones(numPoints, 1);
xSourcexTarget = -xSource.*xTarget;
ySourcexTarget = -ySource.*xTarget;
xSourceyTarget = -xSource.*yTarget;
ySourceyTarget = -ySource.*yTarget;
%//Build matrix
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
%//Build RHS vector
b = [xTarget; yTarget];
%//Solve homography by least squares
h = A \ b;
%// Reshape to a 3 x 3 matrix (optional)
%// Must transpose as reshape is performed
%// in column major format
h(9) = 1; %// Add in that h33 is 1 before we reshape
hmatrix = reshape(h, 3, 3)';
Once you are finished, you have a combined rotation and translation matrix. If you want the x and y translations, simply pick off column 3, rows 1 and 2 in hmatrix. However, we can also work with the vector of h itself, and so h13 would be element 3, and h23 would be element number 6. If you want the angle of rotation, simply take the appropriate inverse trigonometric function to rows 1, 2 and columns 1, 2. For the h vector, this would be elements 1, 2, 4 and 5. There will be a bit of inconsistency depending on which elements you choose as this was solved by least squares. One way to get a good overall angle would perhaps be to find the angles of all 4 elements then do some sort of average. Either way, this is a good starting point.
References
I learned about homography a while ago through Leow Wee Kheng's Computer Vision course. What I have told you is based on his slides: http://www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf. Take a look at slides 30-32 if you want to know where I pulled this material from. However, the MATLAB code I wrote myself :)
I have two images which one of them is the Original image and the second one is Transformed image.
I have to find out how many degrees Transformed image was rotated using 3x3 transformation matrix. Plus, I need to find how far translated from origin.
Both images are grayscaled and held in matrix variables. Their sizes are same [350 500].
I have found a few lecture notes like this.
Lecture notes say that I should use the following matrix formula for rotation:
For translation matrix the formula is given:
Everything is good. But there are two problems:
I could not imagine how to implement the formulas using MATLAB.
The formulas are shaped to find x',y' values but I already have got x,x',y,y' values. I need to find rotation angle (theta) and tx and ty.
I want to know the equivailence of x, x', y, y' in the the matrix.
I have got the following code:
rotationMatrix = [ cos(theta) sin(theta) 0 ; ...
-sin(theta) cos(theta) 0 ; ...
0 0 1];
translationMatrix = [ 1 0 tx; ...
0 1 ty; ...
0 0 1];
But as you can see, tx, ty, theta variables are not defined before used. How can I calculate theta, tx and ty?
PS: It is forbidden to use Image Processing Toolbox functions.
This is essentially a homography recovery problem. What you are doing is given co-ordinates in one image and the corresponding co-ordinates in the other image, you are trying to recover the combined translation and rotation matrix that was used to warp the points from the one image to the other.
You can essentially combine the rotation and translation into a single matrix by multiplying the two matrices together. Multiplying is simply compositing the two operations together. You would this get:
H = [cos(theta) -sin(theta) tx]
[sin(theta) cos(theta) ty]
[ 0 0 1]
The idea behind this is to find the parameters by minimizing the error through least squares between each pair of points.
Basically, what you want to find is the following relationship:
xi_after = H*xi_before
H is the combined rotation and translation matrix required to map the co-ordinates from the one image to the other. H is also a 3 x 3 matrix, and knowing that the lower right entry (row 3, column 3) is 1, it makes things easier. Also, assuming that your points are in the augmented co-ordinate system, we essentially want to find this relationship for each pair of co-ordinates from the first image (x_i, y_i) to the other (x_i', y_i'):
[p_i*x_i'] [h11 h12 h13] [x_i]
[p_i*y_i'] = [h21 h22 h23] * [y_i]
[ p_i ] [h31 h32 1 ] [ 1 ]
The scale of p_i is to account for homography scaling and vanishing points. Let's perform a matrix-vector multiplication of this equation. We can ignore the 3rd element as it isn't useful to us (for now):
p_i*x_i' = h11*x_i + h12*y_i + h13
p_i*y_i' = h21*x_i + h22*y_i + h23
Now let's take a look at the 3rd element. We know that p_i = h31*x_i + h32*y_i + 1. As such, substituting p_i into each of the equations, and rearranging to solve for x_i' and y_i', we thus get:
x_i' = h11*x_i + h12*y_i + h13 - h31*x_i*x_i' - h32*y_i*x_i'
y_i' = h21*x_i + h22*y_i + h23 - h31*x_i*y_i' - h32*y_i*y_i'
What you have here now are two equations for each unique pair of points. What we can do now is build an over-determined system of equations. Take each pair and build two equations out of them. You will then put it into matrix form, i.e.:
Ah = b
A would be a matrix of coefficients that were built from each set of equations using the co-ordinates from the first image, b would be each pair of points for the second image and h would be the parameters you are solving for. Ultimately, you are finally solving this linear system of equations reformulated in matrix form:
You would solve for the vector h which can be performed through least squares. In MATLAB, you can do this via:
h = A \ b;
A sidenote for you: If the movement between images is truly just a rotation and translation, then h31 and h32 will both be zero after we solve for the parameters. However, I always like to be thorough and so I will solve for h31 and h32 anyway.
NB: This method will only work if you have at least 4 unique pairs of points. Because there are 8 parameters to solve for, and there are 2 equations per point, A must have at least a rank of 8 in order for the system to be consistent (if you want to throw in some linear algebra terminology in the loop). You will not be able to solve this problem if you have less than 4 points.
If you want some MATLAB code, let's assume that your points are stored in sourcePoints and targetPoints. sourcePoints are from the first image and targetPoints are for the second image. Obviously, there should be the same number of points between both images. It is assumed that both sourcePoints and targetPoints are stored as M x 2 matrices. The first columns contain your x co-ordinates while the second columns contain your y co-ordinates.
numPoints = size(sourcePoints, 1);
%// Cast data to double to be sure
sourcePoints = double(sourcePoints);
targetPoints = double(targetPoints);
%//Extract relevant data
xSource = sourcePoints(:,1);
ySource = sourcePoints(:,2);
xTarget = targetPoints(:,1);
yTarget = targetPoints(:,2);
%//Create helper vectors
vec0 = zeros(numPoints, 1);
vec1 = ones(numPoints, 1);
xSourcexTarget = -xSource.*xTarget;
ySourcexTarget = -ySource.*xTarget;
xSourceyTarget = -xSource.*yTarget;
ySourceyTarget = -ySource.*yTarget;
%//Build matrix
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
%//Build RHS vector
b = [xTarget; yTarget];
%//Solve homography by least squares
h = A \ b;
%// Reshape to a 3 x 3 matrix (optional)
%// Must transpose as reshape is performed
%// in column major format
h(9) = 1; %// Add in that h33 is 1 before we reshape
hmatrix = reshape(h, 3, 3)';
Once you are finished, you have a combined rotation and translation matrix. If you want the x and y translations, simply pick off column 3, rows 1 and 2 in hmatrix. However, we can also work with the vector of h itself, and so h13 would be element 3, and h23 would be element number 6. If you want the angle of rotation, simply take the appropriate inverse trigonometric function to rows 1, 2 and columns 1, 2. For the h vector, this would be elements 1, 2, 4 and 5. There will be a bit of inconsistency depending on which elements you choose as this was solved by least squares. One way to get a good overall angle would perhaps be to find the angles of all 4 elements then do some sort of average. Either way, this is a good starting point.
References
I learned about homography a while ago through Leow Wee Kheng's Computer Vision course. What I have told you is based on his slides: http://www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf. Take a look at slides 30-32 if you want to know where I pulled this material from. However, the MATLAB code I wrote myself :)