I'm performing scaling operation on grid created. But the dimensions seem improper for the scaling the grid. Any ideas how to do it?
Code:
% plot grid
[X,Y] = meshgrid(-1:0.1:5, 0:0.1:1);
X = X(:);
Y = Y(:);
plot(X,Y,'b.');
xlabel('X');
ylabel('Y');
sx = 0.75;
sy = 0.6;
Tscale = [sx 0 0;
0 sy 0;
0 0 1];
Scale_val=Tscale*[X Y].';
X_Scale=Scale_val(1,:);
Y_Scale=Scale_val(2,:);
figure, plot(X_Scale, Y_Scale);
error:
Error using *
Inner matrix dimensions must agree.
Error in: Scale_val=Tscale*[X Y].';
Scale_val=Tscale*[X Y].';
you were making 2 mistakes i think.
first of all your input coordinates are 2D with just X and Y, but you are trying to do a 3D transformation. second, the matrix multiplication format is not correct. this is what i think you were trying to do
Tscale = [sx 0 0;
0 sy 0;
0 0 1];
Scale_val=Tscale*[X'; Y'; zeros(1,length(X))];
X_Scale=Scale_val(1,:);
Y_Scale=Scale_val(2,:);
figure, plot(X_Scale, Y_Scale,'*');
Related
theta=acos((trace(R)-1)/2);
if trace(R)==3
vec = [0 0 0];
axang=[0 0 0 0];
vec(1)=R(3,2)-R(2,3);
vec(2)=R(1,3)-R(3,1);
vec(3)=R(2,1)-R(1,2);
vec=(1/(2*sin(theta)))*vec;
axang = [vec, theta];
elseif trace(R)==-1
vec=[0 0 0;0 0 0];
axang=[0 0 0 0;0 0 0 0];
X=[0 0];
Y=[0 0];
Z=[0 0];
Y(1)=sqrt((R(2,2)+1)/2);
Y(2)=-Y(1);
X(1)=R(2,1)/(2*Y(1));
X(2)=R(2,1)/(2*Y(2));
Z(1)=R(2,3)/(2*Y(1));
Z(2)=R(2,3)/(2*Y(2));
vec(1,:)=[X(1) Y(1) Z(1)];
vec(2,:)=[X(2) Y(2) Z(2)];
axang(1,:)=[vec(1,:), theta];
axang(2,:)=[vec(2,:), theta];
else
vec = [0 0 0];
axang=[0 0 0 0];
vec(1)=R(3,2)-R(2,3);
vec(2)=R(1,3)-R(3,1);
vec(3)=R(2,1)-R(1,2);
vec=(1/(2*sin(theta)))*vec;
axang = [vec, theta];
end
So this was my code but it didn't work when the rotation matrix is
R = [-1 0 0;
0 -1 0;
0 0 1]
What is wrong with the code ? axang is a vector that stores axis in the first three positions and the angle in the last position.
It seems to me that you are looking for a conversion of a rotation matrix to quaternions, which is a built-in feature of Matlab if you installed the Robotics System Toolbox, i.e. rotm2quat:
axang = rotm2quat(R)
Note that the output format is slightly different as documented by Matlab:
Unit quaternion, returned as an n-by-4 matrix containing n
quaternions. Each quaternion, one per row, is of the form q = [w x y
z], with w as the scalar number.
Therefore you may need to swap the columns as follows:
axang = axang(:, [2 3 4 1]);
In a similar vein to the above answer you may wish to look into the use of the MatLab tool Translation1 = se2(StructuringElement, TranslationOffset).
The variable TranslationOffset can be applied as an angle in the form of 60*pi/180 for example.
In case of trace(R)==-1, the sign of axis term may be flipped. To get rid of it, following steps compute the axis angle vector.
find X(1) = sqrt((R(1,1)+1)/2);
if it is not zero, compute Y(1) = R(1,2)/(2*X(1)) and Y(2) = R(1,2)/(2*X(2)) and Z(1) = R(1,3)/(2*X(1)) Z(2) = R(1,3)/(2*X(2));
If X(1) = 0, then find Y(1) = sqrt((R(2,2)+1)/2)
if Y(1) is not zero then find the other terms from Y(1)
else find Z(1) and find the other terms from Z(1)
I am trying to generate points within ellipsoid and then trying to fit with smooth ellipsoid surfaces. The goal is to fit with unknown data where i have to find value of a,b and c in 3 principal axes.
Rinv should be equivalent to pc. But what i am getting pc in different order. So i have to find correct order to rotate my data to matlab coordinate.
a=3;
b=5;
c=1;
index=1;
for i=1:500000
x=10*rand-5;
y=10*rand-5;
z=10*rand-5;
if ((x^2/a^2) + (y^2/b^2) + (z^2/c^2) -1) <0
C(index,:)=[x,y,z];
index=index+1;
end
end
theta=pi/4; phi=pi/6; omega = pi/3;
Rx= [1 0 0; 0 cos(theta) -sin(theta); 0 sin(theta) cos(theta)];
Ry= [cos(phi) 0 sin(phi); 0 1 0; -sin(phi) 0 cos(phi)];
Rz= [cos(omega) -sin(omega) 0; sin(omega) cos(omega) 0; 0 0 1];
R= Rz*Ry*Rx;
Rinv = inv(R);
X = C*R;
[pc,val]=eig(X'*X); E=diag(val);
[sa,sb]= sort(pc*E, 'descend'); sb
order = [2,3,1];
nC= X*pc(:,order);
plot3(nC(:,1),nC(:,2),nC(:,3),'.')
hold on
[x, y, z] = ellipsoid(0,0,0,a,b,c,30);
hSurface=surf(x, y, z, 'FaceColor','blue','EdgeColor','none');
alpha(0.5)
Specially in this line nC= X*pc(:,order);
I am finding order manually. Can any one tells (1) how to find pc order correctly. (2) Value of a,b,c for unknown data set here "[x, y, z] = ellipsoid(0,0,0,a,b,c,30)"
Thanks
To compute the values of a,b,c from the eigen-space, you just need to compute the radius of the eigen-space projected ellipsoid. This is trivial because the ellipsoid is axis-oriented in eigen-space (which I assume is the only reason you are doing this in the first place). Just get the max and min in each direction!
nC= X*pc;
plot3(nC(:,1),nC(:,2),nC(:,3),'r.')
hold on
a2=(max(nC(:,1))-min(nC(:,1)))/2;
b2=(max(nC(:,2))-min(nC(:,2)))/2;
c2=(max(nC(:,3))-min(nC(:,3)))/2;
[x, y, z] = ellipsoid(0,0,0,a2,b2,c2,30);
hSurface=surf(x, y, z, 'FaceColor','blue','EdgeColor','none');
alpha(0.5)
axis equal
Note that this code actually makes order obsolete.
I'm writing a function in matlab which mimics the built-in 'imwarp' function (applying geometric transformation) without using any kind of loops. i'm in the final step when i have to call my function for bi-linear interpolation for every index in final 2D image.
I have 3 arrays here , 'pts' have homogenized vectors (x,y,1) for which i interpolate and 'row' and 'cols' have x and y coordinates respectively for resultant image where interpolated intensity value would be placed.
finalImage (rows(1,:),cols(1,:))=bilinear(pts(:,:),im);
Kindly correct my syntax here to do it properly. thanks in advance.
The following is a simple implementation of applying an affine transformation to an image. Some of the matrices may be reversed because I did this from memory. I don't know exactly how you are formatting your pts array so I figure a working example is the best I can do. The interp2 function applies bilinear interpolation, the bilinear function performs the bilinear transform which describes analog filters as digital filters. This is not what you want.
P.S. You have to make sure to use the inverse transform when applying image warping (that is, define the point you want to sample in the input image for each point in the output image). If you perform the forward transform (i.e. define the point in the output image that each point in the input image maps to) then you will end up with some serious aliasing effects and potentially holes in the output image.
Hope this helps. Let me know if you have questions.
img = double(imread('rice.png'))/255;
theta = 30; % rotate 30 degrees
R = [cosd(theta) -sind(theta) 0; ...
sind(theta) cosd(theta) 0; ...
0 0 1];
sx = 15; % skew by 15 degrees in x
Skx = [1 tand(sx) 0; ...
0 1 0; ...
0 0 1];
% Translate by 1/2 size of image
tx = -size(img, 2)/2;
ty = -size(img, 1)/2;
T = [1 0 tx; ...
0 1 ty; ...
0 0 1];
% Scale image down by 1/2
sx = 0.5;
sy = 0.5;
S = [sx 0 0; ...
0 sy 0; ...
0 0 1];
% translate, scale, rotate, skew, then translate back
A = inv(T)*Skx*R*S*T;
% create meshgrid points
[x, y] = meshgrid(1:size(img,2), 1:size(img,1));
% reshape so we can apply matrix op
V = [reshape(x, 1, []); reshape(y, 1, []); ones(1, numel(x))];
Vq = inv(A)*V;
% probably not necessary for these transformations but project back to the z=1 plane
Vq(1,:) = Vq(1,:) ./ V(3,:);
Vq(2,:) = Vq(2,:) ./ V(3,:);
% reshape back into a meshgrid
xq = reshape(Vq(1,:), size(img));
yq = reshape(Vq(2,:), size(img));
% use interp2 to perform bilinear interpolation
imgnew = interp2(x, y, img, xq, yq);
% show the resulting image
imshow(imgnew);
I was asked to perform an image rotation about an arbitrary point. The framework they provided was in matlab so I had to fill a function called MakeTransformMat that receives the angle of rotation and the point where we want to rotate.
As I've seen in class to do this rotation first we translate the point to the origin, then we rotate and finally we translate back.
The framework asks me to return a Transformation Matrix. Am I right to build that matrix as the multiplication of the translate-rotate-translate matrices? otherwise, what am I forgetting?
function TransformMat = MakeTransformMat(theta,center_y,center_x)
%Translate image to origin
trans2orig = [1 0 -center_x;
0 1 -center_y;
0 0 1];
%Rotate image theta degrees
rotation = [cos(theta) -sin(theta) 0;
sin(theta) cos(theta) 0;
0 0 1];
%Translate back to point
trans2pos = [1 0 center_x;
0 1 center_y;
0 0 1];
TransformMat = trans2orig * rotation * trans2pos;
end
This worked for me. Here I is the input image and J is the rotated image
[height, width] = size(I);
rot_deg = 45; % Or whatever you like (in degrees)
rot_xc = width/2; % Or whatever you like (in pixels)
rot_yc = height/2; % Or whatever you like (in pixels)
T1 = maketform('affine',[1 0 0; 0 1 0; -rot_xc -rot_yc 1]);
R1 = maketform('affine',[cosd(rot_deg) sind(rot_deg) 0; -sind(rot_deg) cosd(rot_deg) 0; 0 0 1]);
T2 = maketform('affine',[1 0 0; 0 1 0; width/2 height/2 1]);
tform = maketform('composite', T2, R1, T1);
J = imtransform(I, tform, 'XData', [1 width], 'YData', [1 height]);
Cheers.
I've answered a very similar question elsewhere: Here is the link.
In the code linked to, the point about which you rotate is determined by how the meshgrid is defined.
Does that help? Have you read the Wikipedia page on rotation matrices?
I have a 3-D grayscale volume corresponding to ultrasound data. In Matlab this 3-D volume is simply a 3-D matrix of MxNxP. The structure I'm interested in is not oriented along the z axis, but along a local coordinate system already known (x'y'z'). What I have up to this point is something like the figure shown below, depicting the original (xyz) and the local coordinate systems (x'y'z'):
I want to obtain the 2-D projection of this volume (i.e. an image) through a specific plane on the local coordinate system, say at z' = z0. How can I do this?
If the volume was oriented along the z axis this projection could be readily achieved. i.e. if the volume, in Matlab, is V, then:
projection = sum(V,3);
thus, the projection can be computed just as the sum along the 3rd dimension of the array. However with a change of orientation the problem becomes more complicated.
I've been looking at radon transform (2D, that applies only to 2-D images and not volumes) and also been considering ortographic projections, but at this point I'm clueless as to what to do!
Thanks for any advice!
New attempt at solution:
Following the tutorial http://blogs.mathworks.com/steve/2006/08/17/spatial-transformations-three-dimensional-rotation/ and making some small changes, I might have something which could help you. Bear in mind, I have little or no experience with volumetric data in MATLAB, so the implementation is quite hacky.
In the below code I use tformarray() to rotate the structure in space. First, the data is centered, then rotated using rotationmat3D to produce the spacial transformation, before the data is moved back to its original position.
As I have never used tformarray before, I handeled datapoints falling outside the defined region after rotation by simply padding the data matrix (NxMxP) with zeros all around. If anyone know a better way, please let us know :)
The code:
%Synthetic dataset, 25x50x25
blob = flow();
%Pad to allow for rotations in space. Bad solution,
%something better might be possible to better understanding
%of tformarray()
blob = padarray(blob,size(blob));
f1 = figure(1);clf;
s1=subplot(1,2,1);
p = patch(isosurface(blob,1));
set(p, 'FaceColor', 'red', 'EdgeColor', 'none');
daspect([1 1 1]);
view([1 1 1])
camlight
lighting gouraud
%Calculate center
blob_center = (size(blob) + 1) / 2;
%Translate to origin transformation
T1 = [1 0 0 0
0 1 0 0
0 0 1 0
-blob_center 1];
%Rotation around [0 0 1]
rot = -pi/3;
Rot = rotationmat3D(rot,[0 1 1]);
T2 = [ 1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1];
T2(1:3,1:3) = Rot;
%Translation back
T3 = [1 0 0 0
0 1 0 0
0 0 1 0
blob_center 1];
%Total transform
T = T1 * T2 * T3;
%See http://blogs.mathworks.com/steve/2006/08/17/spatial-transformations-three-dimensional-rotation/
tform = maketform('affine', T);
R = makeresampler('linear', 'fill');
TDIMS_A = [1 2 3];
TDIMS_B = [1 2 3];
TSIZE_B = size(blob);
TMAP_B = [];
F = 0;
blob2 = ...
tformarray(blob, tform, R, TDIMS_A, TDIMS_B, TSIZE_B, TMAP_B, F);
s2=subplot(1,2,2);
p2 = patch(isosurface(blob2,1));
set(p2, 'FaceColor', 'red', 'EdgeColor', 'none');
daspect([1 1 1]);
view([1 1 1])
camlight
lighting gouraud
The arbitrary visualization below is just to confirm that the data is rotated as expected, plotting a closed surface when the data passed the value '1'. With blob2, you should know be able to project by using simple sums.
figure(2)
subplot(1,2,1);imagesc(sum(blob,3));
subplot(1,2,2);imagesc(sum(blob2,3));
Assuming you have access to the coordinate basis R=[x' y' z'], and that those vectors are orthonormal, you can simply extract the representation in this basis by multiplying your data with the the 3x3 matrix R, where x',y',z' are column vectors.
With the data stored in D (Nx3), you can get the representation with R, by multiplying by it:
Dmarked = D*R;
and now D = Dmarked*inv(R), so going back and forth is stragihtforward.
The following code might provide help to see the transformation. Here I create a synthetic dataset, rotate it, and then rotate it back. Doing sum(DR(:,3)) would then be your sum along z'
%#Create synthetic dataset
N1 = 250;
r1 = 1;
dr1 = 0.1;
dz1 = 0;
mu1 = [0;0];
Sigma1 = eye(2);
theta1 = 0 + (2*pi).*rand(N1,1);
rRand1 = normrnd(r1,dr1,1,N1);
rZ1 = rand(N1,1)*dz1+1;
D = [([rZ1*0 rZ1*0] + repmat(rRand1',1,2)).*[sin(theta1) cos(theta1)] rZ1];
%Create roation matrix
rot = pi/8;
R = rotationmat3D(rot,[0 1 0]);
% R = 0.9239 0 0.3827
% 0 1.0000 0
% -0.3827 0 0.9239
Rinv = inv(R);
%Rotate data
DR = D*R;
%#Visaulize data
f1 = figure(1);clf
subplot(1,3,1);
plot3(DR(:,1),DR(:,2),DR(:,3),'.');title('Your data')
subplot(1,3,2);
plot3(DR*Rinv(:,1),DR*Rinv(:,2),DR*Rinv(:,3),'.r');
view([0.5 0.5 0.2]);title('Representation using your [xmarked ymarked zmarked]');
subplot(1,3,3);
plot3(D(:,1),D(:,2),D(:,3),'.');
view([0.5 0.5 0.2]);title('Original data before rotation');
If you have two normalized 3x1 vectors x2 and y2 corresponding to your local coordinate system (x' and y').
Then, for a position P, its local coordinate will be xP=P'x2 and yP=P'*y2.
So you can try to project your volume using accumarray:
[x y z]=ndgrid(1:M,1:N,1:P);
posP=[x(:) y(:) z(:)];
xP=round(posP*x2);
yP=round(posP*y2);
xP=xP+min(xP(:))+1;
yP=yP+min(yP(:))+1;
V2=accumarray([xP(:),yP(:)],V(:));
If you provide your data, I will test it.