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.
Related
I am trying to rotate an "orbit" using rotation matrix given below:
[cos(angle) -sin(angle) 0;
sin(angle) cos (angle) 0;
0 0 1 ]
First thing I thought I should do was to use sphere():
[x y z] = sphere;
Then concatenate x, y, and z together in a vector:
xyz = [x; y; z];
rotatMat = [cos(angle) -sin(angle) 0; sin(angle) cos (angle) 0; 0 0 1 ];
multiply rotation matrix and xyz to rotate an orbit:
rotated = rotatMat .* xyz;
However, xyz turns out to be 62x22 dimension while my rotatMat is only 3x3 so I cannot multiply them together.
How can i fix this issue?
Thank you in advance.
You have to use the * operator for matrix multiplication, and not .* which is for element-wise multiplication.
Moreover, your xyz matrix should be of size n-by-3 (and not 62-by-22) and you have to use xyz*rotatMat' to match the dimensions correctly. Alternatively, you can have xyz of size 3-by-n and use the syntax rotatMat*xyz.
Best,
xyz = [x(:) y(:) z(:)].'; %'// put x, y, z as rows of a matrix
xyz_rotated = rotatMat*xyz %// use matrix multiplication
x_rotated = reshape(xyz_rotated(1,:), size(x)); %// reshape transformed rows
y_rotated = reshape(xyz_rotated(2,:), size(x));
z_rotated = reshape(xyz_rotated(3,:), size(x));
I have been struggling with this problem for a while and I would appreciate if anyone can help me out. I am able to generate a 10 by 10 matrix and have it randomly assign "1"s in the matrix. My goal is to plot a "star" at the location of each element in the vector that has a value of "1", but I can't seem to figure out how to map the vector to a x-y coordinate system. The code I wrote below generates a plot of 100 stars at each cell and also generates a vector "v", but I don't know how I can link the plot to the vector that instead of having 100 "star"s in my plot, I have however many that there is a value of "1" at the corresponding location of the element.
Thanks!!
David
davidtongg#gmail.com
close all
clear all
clc
a=10;b=10;
v = zeros(a,b);
xy = int32(randi(a, 100, 2));
z = randi(1, 100, 1); % 100 values.
indexes = sub2ind([a, b], xy(:,1), xy(:,2))
v(indexes) = z
m=length(v);
ctr=0;
for i=1:m^2
x_cor(i)=(i-(floor(i/m)*m))*200-100;
y_cor(i)=(floor(i/m)+1)*200-100;
for j=1:m
if i==j*m
x_cor(i)=((i-(floor(i/m)*m))*200-100)+(2*m*100);
y_cor(i)=(floor(i/m))*200-100;
end
end
end
figure(1)
plot(x_cor,y_cor,'*');
grid on
I may of course have misinterpreted this because that code is confusingly complicated, but this is what I think you're after.
For an axb matrix with a random number of ones:
v = randi([0 1], a, b);
Or for a specific number n of ones, in random locations:
v = zeros(a, b);
idx = randi([1 numel(v)], n, 1);
v(idx) = 1; % linear indexing into a matrix
Then to plot them in arbitrarily scaled coordinates:
[y x] = find(v);
x = x * xscale + xoffset;
y = y * yscale + yoffset;
plot(x, y, '*');
Or the really cheaty way:
spy(v);
You can do it easily taking into account that plot(A) , where A is a matrix, plots the columns of the matrix vs their index, and that NaNs are not plotted:
v =[ 1 0 0 0
1 1 0 0
0 0 0 1
1 1 1 1
0 1 1 0 ]; %// example data
v2 = double(v); %// create copy; will be overwritten
v2(~v2) = NaN; %// change zeros to NaNs
plot(bsxfun(#plus, fliplr(v2.'), 0:size(v,1)-1) ,'b*')
%'// transpose and flip from left to right.
%// Add 1 incrementally to each column to have all of them "stacked" in the plot
axis([0 size(v,2)+1 0 size(v,1)+1]) %// set axis limits
set(gca,'xtick',1:size(v,2),'ytick',1:size(v,1)) %// set ticks
grid
I am trying to transform an image using a 3D transformation matrix and assuming my camera is orthonormal.
I am defining my homography using the plane-induced homography formula H=R-t*n'/d (with d=Inf so H=R) as given in Hartley and Zisserman Chapter 13.
What I am confused about is when I use a rather modest rotation, the image seems to be distorting much more than I expect (I'm sure I'm not confounding radians and degrees).
What could be going wrong here?
I've attached my code and example output.
n = [0;0;-1];
d = Inf;
im = imread('cameraman.tif');
rotations = [0 0.01 0.1 1 10];
for ind = 1:length(rotations)
theta = rotations(ind)*pi/180;
R = [ 1 0 0 ;
0 cos(theta) -sin(theta);
0 sin(theta) cos(theta)];
t = [0;0;0];
H = R-t*n'/d;
tform = maketform('projective',H');
imT = imtransform(im,tform);
subplot(1,5,ind) ;
imshow(imT)
title(['Rot=' num2str(rotations(ind)) 'deg']);
axis square
end
The formula H = R-t*n'/d has one assumption which is not met in your case:
This formula implies that you are using pinhole camera model with focal length=1
But in your case, for your camera to be more real and for your code to work, you should set the focal length to some positive number much greater than 1. (focal length is the distance from your camera center to the image plane)
To do this you can define a calibration matrix K which handles the focal length. You just need to change your formula to
H=K R inv(K) - 1/d K t n' inv(K)
in which K is a 3-by-3 identity matrix whose two first elements along the diagonal are set to the focal length (e.g. f=300). The formula can be easily derived if you assume a projective camera.
Below is the corrected version of your code, in which the angles make sense.
n = [0;0;-1];
d = Inf;
im = imread('cameraman.tif');
rotations = [0 0.01 0.1 30 60];
for ind = 1:length(rotations)
theta = rotations(ind)*pi/180;
R = [ 1 0 0 ;
0 cos(theta) -sin(theta);
0 sin(theta) cos(theta)];
t = [0;0;0];
K=[300 0 0;
0 300 0;
0 0 1];
H=K*R/K-1/d*K*t*n'/K;
tform = maketform('projective',H');
imT = imtransform(im,tform);
subplot(1,5,ind) ;
imshow(imT)
title(['Rot=' num2str(rotations(ind)) 'deg']);
axis square
end
You can see the result in the image below:
You can also rotate the image around its center. For it to happen you should set the image plane origin to the center of the image which I think is not possible with that method of matlab (maketform).
You can use the method below instead.
imT=imagehomog(im,H','c');
Note that if you use this method, you'll have to change some settings in n, d, t and R to get the appropriate result.
That method can be found at: https://github.com/covarep/covarep/blob/master/external/voicebox/imagehomog.m
The result of the program with imagehomog and some changes in n, d, t , and R is shown below which seems more real.
New settings are:
n = [0 0 1]';
d = 2;
t = [1 0 0]';
R = [cos(theta), 0, sin(theta);
0, 1, 0;
-sin(theta), 0, cos(theta)];
Hmm... I'm not 100% percent on this stuff, but it was an interesting question and relevant to my work, so I thought I'd play around and give it a shot.
EDIT: I tried this once using no built-ins. That was my original answer. Then I realized that you could do it your way pretty easily:
The easy answer to your question is to use the correct rotation matrix about the z-axis:
R = [cos(theta) -sin(theta) 0;
sin(theta) cos(theta) 0;
0 0 1];
Here's another way to do it (my original answer):
I'm going to share what I did; hopefully this is useful to you. I only did it in 2D (though that should be easy to expand to 3D). Note that if you want to rotate the image in plane, you will need to use a different rotation matrix that you have currently coded. You need to rotate about the Z-axis.
I did not use those matlab built-ins.
I referred to http://en.wikipedia.org/wiki/Rotation_matrix for some info.
im = double(imread('cameraman.tif')); % must be double for interpn
[x y] = ndgrid(1:size(im,1), 1:size(im,2));
rotation = 10;
theta = rotation*pi/180;
% calculate rotation matrix
R = [ cos(theta) -sin(theta);
sin(theta) cos(theta)]; % just 2D case
% calculate new positions of image indicies
tmp = R*[x(:)' ; y(:)']; % 2 by numel(im)
xi = reshape(tmp(1,:),size(x)); % new x-indicies
yi = reshape(tmp(2,:),size(y)); % new y-indicies
imrot = interpn(x,y,im,xi,yi); % interpolate from old->new indicies
imagesc(imrot);
My own question now is: "How do you change the origin about which you are rotating the image? Clearly, I'm rotating about (0,0), the top left corner.
EDIT 2 In response to the asker's comment, I've tried again.
This time I fixed a couple of things. Now I'm using the same transformation matrix (about x) as in the original question.
I rotated about the center of the image by redoing the way i do the ndgrids (put 0,0,0) in the center of the image. I also decided to show 3 planes of the image. This was not in the original question. The middle plane is the plane of interest. To get just the middle plane, you can leave out the zero-padding and redefine the 3rd ndgrid option to be just 1 instead of -1:1.
im = double(imread('cameraman.tif')); % must be double for interpn
im = padarray(im, [0 0 1],'both');
[x y z] = ndgrid(-floor(size(im,1)/2):floor(size(im,1)/2)-1, ...
-floor(size(im,2)/2):floor(size(im,2)/2)-1,...
-1:1);
rotation = 1;
theta = rotation*pi/180;
% calculate rotation matrix
R = [ 1 0 0 ;
0 cos(theta) -sin(theta);
0 sin(theta) cos(theta)];
% calculate new positions of image indicies
tmp = R*[x(:)'; y(:)'; z(:)']; % 2 by numel(im)
xi = reshape(tmp(1,:),size(x)); % new x-indicies
yi = reshape(tmp(2,:),size(y)); % new y-indicies
zi = reshape(tmp(3,:),size(z));
imrot = interpn(x,y,z,im,xi,yi,zi); % interpolate from old->new indicies
figure;
subplot(3,1,1);imagesc(imrot(:,:,1)); axis image; axis off;
subplot(3,1,2);imagesc(imrot(:,:,2)); axis image; axis off;
subplot(3,1,3);imagesc(imrot(:,:,3)); axis image; axis off;
You are performing rotations around the x-axis: in your matrix, the 1st component (x) is left unchanged by the rotation matrix. This is confirmed by the perspective deformations from your examples.
The actual amount of deformation will then depend on the distance between the camera and the image plane (or more accurately on its value relative to the focal length of the camera). It can be important when the cameraman image plane is located near the camera.
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.
I would like to reproduce the following figure in MATLAB:
There are two classes of points with X and Y coordinates. I'd like to surround each class with an ellipse with one parameter of standard deviation, which determine how far the ellipse will go along the axis.
The figure was created with another software and I don't exactly understand how it calculates the ellipse.
Here is the data I'm using for this figure. The 1st column is class, 2nd - X, 3rd - Y. I can use gscatter to draw the points itself.
A = [
0 0.89287 1.54987
0 0.69933 1.81970
0 0.84022 1.28598
0 0.79523 1.16012
0 0.61266 1.12835
0 0.39950 0.37942
0 0.54807 1.66173
0 0.50882 1.43175
0 0.68840 1.58589
0 0.59572 1.29311
1 1.00787 1.09905
1 1.23724 0.98834
1 1.02175 0.67245
1 0.88458 0.36003
1 0.66582 1.22097
1 1.24408 0.59735
1 1.03421 0.88595
1 1.66279 0.84183
];
gscatter(A(:,2),A(:,3),A(:,1))
FYI, here is the SO question on how to draw ellipse. So, we just need to know all the parameters to draw it.
Update:
I agree that the center can be calculated as the means of X and Y coordinates. Probably I have to use principal component analysis (PRINCOMP) for each class to determine the angle and shape. Still thinking...
Consider the code:
%# generate data
num = 50;
X = [ mvnrnd([0.5 1.5], [0.025 0.03 ; 0.03 0.16], num) ; ...
mvnrnd([1 1], [0.09 -0.01 ; -0.01 0.08], num) ];
G = [1*ones(num,1) ; 2*ones(num,1)];
gscatter(X(:,1), X(:,2), G)
axis equal, hold on
for k=1:2
%# indices of points in this group
idx = ( G == k );
%# substract mean
Mu = mean( X(idx,:) );
X0 = bsxfun(#minus, X(idx,:), Mu);
%# eigen decomposition [sorted by eigen values]
[V D] = eig( X0'*X0 ./ (sum(idx)-1) ); %#' cov(X0)
[D order] = sort(diag(D), 'descend');
D = diag(D);
V = V(:, order);
t = linspace(0,2*pi,100);
e = [cos(t) ; sin(t)]; %# unit circle
VV = V*sqrt(D); %# scale eigenvectors
e = bsxfun(#plus, VV*e, Mu'); %#' project circle back to orig space
%# plot cov and major/minor axes
plot(e(1,:), e(2,:), 'Color','k');
%#quiver(Mu(1),Mu(2), VV(1,1),VV(2,1), 'Color','k')
%#quiver(Mu(1),Mu(2), VV(1,2),VV(2,2), 'Color','k')
end
EDIT
If you want the ellipse to represent a specific level of standard deviation, the correct way of doing is by scaling the covariance matrix:
STD = 2; %# 2 standard deviations
conf = 2*normcdf(STD)-1; %# covers around 95% of population
scale = chi2inv(conf,2); %# inverse chi-squared with dof=#dimensions
Cov = cov(X0) * scale;
[V D] = eig(Cov);
I'd try the following approach:
Calculate the x-y centroid for the center of the ellipse (x,y in the linked question)
Calculate the linear regression fit line to get the orientation of the ellipse's major axis (angle)
Calculate the standard deviation in the x and y axes
Translate the x-y standard deviations so they're orthogonal to the fit line (a,b)
I'll assume there is only one set of points given in a single matrix, e.g.
B = A(1:10,2:3);
you can reproduce this procedure for each data set.
Compute the center of the ellipsoid, which is the mean of the points. Matlab function: mean
Center your data. Matlab function bsxfun
Compute the principal axis of the ellipsoid and their respective magnitude. Matlab function: eig
The successive steps are illustrated below:
Center = mean(B,1);
Centered_data = bsxfun(#minus,B,Center);
[AX,MAG] = eig(Centered_data' * Centered_data);
The columns of AX contain the vectors describing the principal axis of the ellipsoid while the diagonal of MAG contains information on their magnitude.
To plot the ellipsoid, scale each principal axis with the square root of its magnitude.