I am trying to register two volumes(vol1 and vol2). The size of the vol1 is 40x40x24. The size of the vol2 is 64 x64x11.
So far, I have extracted their features and then matched them. Now, I have a set of corresponding points in two volumes that is stored in pairs which is a matrix of size 100x6 (every row of pairs is [x y z X Y Z] where (x,y,z) are the coordinates of a voxel in vol1 and [X Y Z] is the corresponding voxel in vol2). Then, using RANSAC algorithm, I have calculated the 3D affine transform, T. For example, T is something like below:
T=
2.7791 0.8204 0.7561 -61.6055
-0.4418 2.2663 -1.9882 29.0375
-0.2120 0.6568 -0.7041 6.2702
0 0 0 1.0000
I have a couple of questions. First, does this affine transformation matrix look correct? It looks correct to me. If it is correct then why the function affine3d in MATLAB which calculates the 3D affine transform has the column of zeroes instead of the row of zeroes (like in T above)? It looks like the transpose of my transform T.
If my transform is correct then another problem occurs. I tried to resample vol1 using transform T, but the calculated voxel coordinates are negative!!! I am so confused. I do not know what causes this problem.
Here, is my code. Please let me know if you see any problem with it.
function [ vol1R ] = resampling(vol1, vol2, T)
[size1, size2, size3] = size(vol2);
vol1R = zeros(size1,size2,size3); % Initializing registered vol1
for i= 1:size1
for j= 1:size2
for k = 1:size3
V = [i;j;k;1]; % V is the coordinates of a voxel on the registered vol1
% correspoding to voxel v on the vol1
% V = T * v : Relationship between v and V
v = T\V; % Problem occurs here!!!!!!! v has some negative values
% v (coordinates in vol1)
% continue
end
end
end
The format of your affine matrix looks fine if you are using it to operate on column vectors, as in [X; Y; Z; 1] = T*[x; y; z; 1]. The reason MATLAB's example is transposed is because it operates on 1x4 row vectors, as in [X, Y, Z, 1] = [x, y, z, 1]*(T').
As for the the actual numbers in the matrix, I can't comment since I don't know the original images. I do notice that there is an x-translation of -61.6055, which seems pretty big for the sizes of the images you have. Do you know for sure this is the right answer?
I'd suggest maybe starting with a simple pair of images where you know the answer beforehand (maybe white squares on a black background drawn in Paint) to confirm your registration algorithm gives the right answer.
Related
I want to draw a contour plot for 3D data.
I have a force in x,y,z directions I want to plot the contour3 for that
the dimensions of the Fx = 21x21X21 same for Fy and Fz
I am finding force = f*vector(x,y,z)
Then
Fx(x,y,z) = force(1)
Fy(x,y,z) = force(2)
Fz(x,y,z) = force(3)
I did the following but it is not working with me ?? why and how can I plot that
FS = sqrt(Fx.^2 + Fy.^2 + Fz.^2);
x = -10:1:10;
[X,Y] = meshgrid(x);
for i=1:length(FS)
for j = 1:length(FS)
for k=1:length(FS)
contour3(X,Y,FS(i,j,k),10)
hold on
end
end
end
This is the error I am getting
Error using contour3 (line 129)
When Z is a vector, X and Y must also be vectors.
Your problem is that FS is not the same shape as X and Y.
Lets illustrate with a simple example:
X=[1 1 1
2 2 2
3 3 3];
Y=[1 2 3
1 2 3
1 2 3];
Z=[ 2 4 5 1 2 5 5 1 2];
Your data is probably something like this. How does Matlab knows which Z entry corresponds to which X,Y position? He doesnt, and thats why he tells you When Z is a vector, X and Y must also be vectors.
You could solve this by doing reshape(FS,size(X,1),size(X,2)) and will probably work in your case, but you need to be careful. In your example, X and Y don't seem programatically related to FS in any way. To have a meaningful contour plot, you need to make sure that FS(ii,jj,k)[ 1 ] corresponds to X(ii,jj), else your contour plot would not make sense.
Generally you'd want to plot the result of FS against the variables your are using to compute it, such as ii, jj or k, however, I dont know how these look like so I will stop my explanation here.
[ 1 ]: DO NOT CALL VARIABLES i and j IN MATLAB!
I'm not sure if this solution is what you want.
Your problem is that contour and contour3 are plots to represent scalar field in 2D objects. Note that ball is 2D object - every single point is defined by angles theta and phi - even it is an object in "space" not in "plane".
For representation of vector fields there is quiver, quiver3, streamslice and streamline functions.
If you want to use contour plot, you have to transform your data from vector field to scalar field. So your data in form F = f(x,y,z) must be transformed to form of H = f(x,y). In that case H is MxN matrix, x and y are Mx1 and Nx1 vectors, respectively. Then contour3(x,y,H) will work resulting in so-called 3D graph.
If you rely on vector field You have to specify 6 vectors/matrices of the same size of corresponding x, y, z coordinates and Fx, Fy, Fz vector values.
In that case quiver3(x,y,z,Fx,Fy,Fz) will work resulting in 6D graph. Use it wisely!
As I comment the Ander's answer, you can use colourspace to get more dimensions, so You can create 5D or, theoretically, 6D, because you have x, y, z coordinates for position and R, G, B coordinates for the values. I'd recommend using static (x,y,R,G,B) for 5D graph and animated (x,y,t,R,G,B) for 6D. Use it wisely!
In the example I show all approaches mentioned above. i chose gravity field and calculate the plane 0.25 units below the centre of gravity.
Assume a force field defined in polar coordinates as F=-r/r^3; F=1/r^2.
Here both x and yare in range of -1;1 and same size N.
F is the MxMx3 matrix where F(ii,jj) is force vector corresponding to x(ii) and y(jj).
Matrix H(ii,jj) is the norm of F(ii,jj) and X, Y and Z are matrices of coordinates.
Last command ensures that F values are in (-1;1) range. The F./2+0.5 moves values of F so they fit into RGB range. The colour meaning will be:
black for (-1,-1,-1),
red for (1,-1,-1),
grey for (0,0,0)
Un-comment the type of plot You want to see. For quiver use resolution of 0.1, for other cases use 0.01.
clear all,close all
% Definition of coordinates
resolution=0.1;
x=-1:resolution:1;
y=x;
z=-.25;
%definition of matrices
F=zeros([max(size(x))*[1 1],3]); % matrix of the force
X=zeros(max(size(x))*[1 1]); % X coordinates for quiver3
Y=X; % Y coordinates for quiver3
Z=X+z; % Z coordinates for quiver3
% Force F in polar coordinates
% F=-1/r^2
% spherical -> cartesian transformation
for ii=1:max(size(x))
for jj=1:max(size(y))
% temporary variables for transformations
xyz=sqrt(x(ii)^2+y(jj)^2+z^2);
xy= sqrt(x(ii)^2+y(jj)^2);
sinarc=sin(acos(z/xyz));
%filling the quiver3 matrices
X(ii,jj)=x(ii);
Y(ii,jj)=y(jj);
F(ii,jj,3)=-z/xyz^2;
if xy~=0 % 0/0 error for x=y=0
F(ii,jj,2)=-y(jj)/xyz/xy*sinarc;
F(ii,jj,1)=-x(ii)/xyz/xy*sinarc;
end
H(ii,jj)=sqrt(F(ii,jj,1)^2+F(ii,jj,2)^2+F(ii,jj,3)^2);
end
end
F=F./max(max(max(F)));
% quiver3(X,Y,Z,F(:,:,1),F(:,:,2),F(:,:,3));
% image(x,y,F./2+0.5),set(gca,'ydir','normal');
% surf(x,y,Z,F./2+.5,'linestyle','none')
% surf(x,y,H,'linestyle','none')
surfc(x,y,H,'linestyle','none')
% contour3(x,y,H,15)
I'm trying to make a color plot in matlab using output data from another program. What I have are 3 vectors indicating the x-position, y-yposition (both in milliarcseconds, since this represents an image of the surroundings of a black hole), and value (which will be assigned a color) of every point in the desired image. I apparently can't use pcolor, because the values which indicate the color of each "pixel" are not in a matrix, and I don't know a way other than meshgrid to create a matrix out of the vectors, which didn't work due to the size of the vectors.
Thanks in advance for any help, I may not be able to reply immediately.
If we make no assumptions about the arrangement of the x,y coordinates (i.e. non-monotonic) and the sparsity of the data samples, the best way to get a nice image out of your vectors is to use TriScatteredInterp. Here is an example:
% samplesToGrid.m
function [vi,xi,yi] = samplesToGrid(x,y,v)
F = TriScatteredInterp(x,y,v);
[yi,xi] = ndgrid(min(y(:)):max(y(:)), min(x(:)):max(x(:)));
vi = F(xi,yi);
Here's an example of taking 500 "pixel" samples on a 100x100 grid and building a full image:
% exampleSparsePeakSamples.m
x = randi(100,[500 1]); y = randi(100,[500 1]);
v = exp(-(x-50).^2/50) .* exp(-(y-50).^2/50) + 1e-2*randn(size(x));
vi = samplesToGrid(x,y,v);
imagesc(vi); axis image
Gordon's answer will work if the coordinates are integer-valued, but the image will be spare.
You can assign your values to a matrix based on the x and y coordinates and then use imagesc (or a similar function).
% Assuming the X and Y coords start at 1
max_x = max(Xcoords);
max_y = max(Ycoords);
data = nan(max_y, max_x); % Note the order of y and x
indexes = sub2ind(size(data), max_y, max_x);
data(indexes) = Values;
imagesc(data); % note that NaN values will be colored with the minimum colormap value
I am charting the following data:
a=[...
0.1, 0.7, 0.00284643369242828;...
0.1, 0.71, 0.00284643369242828;...]
such that column 1 never surpasses approximately 10
also such that column 2 goes from .7 to 1.
Column 3 seems ok
When i chart my surface using surf(a) it looks like this:
it appears not to be properly considering what should be x and y.
anything seem weird there?
I think you need to try one of two things: either break out your height column into its own rectangular matrix Z and use surf(Z) to plot each point relative to its location in the matrix (so your x- and y-axes will not be scaled the way you want), or you can put your desired x- and y-coordinates in their own vectors, and plot the matrix Z (defined at every point (xi, yj) for all i in N and j in M where x is N elements long and y is M elements long) with surf(x,y,Z).
x = 0.1:0.1:10; % or whatever increment you need
y = 0.7:0.01:1; % or whatever increment you need
Z = zeros(length(x),length(y); % initialized to the correct size, fill with data
I think you are going to have to regenerate your Z-data so that it is in a rectangular matrix that is (elements in x) by (elements in y) in dimension.
EDIT: You do not need to recreate your data. If you know that you have n unique elements in x and m unique elements in y, then you can use:
X = reshape(data(:,1),m,n);
Y = reshape(data(:,2),m,n);
Z = reshape(data(:,3),m,n);
surf(X,Y,Z);
And that should give you what you are looking for.
I use 2D dataset like below,
37.0235000000000 18.4548000000000
28.4454000000000 15.7814000000000
34.6958000000000 20.9239000000000
26.0374000000000 17.1070000000000
27.1619000000000 17.6757000000000
28.4101000000000 15.9183000000000
33.7340000000000 17.1615000000000
34.7948000000000 18.2695000000000
34.5622000000000 19.3793000000000
36.2884000000000 18.4551000000000
26.1695000000000 16.8195000000000
26.2090000000000 14.2081000000000
26.0264000000000 21.8923000000000
35.8194000000000 18.4811000000000
to create a 3D histogram.
How can I find the histogram value of a point on a grid? For example, if [34.7948000000000 18.2695000000000] point is given, I would like to find the corresponding value of a histogram for a given point on the grid.
I used this code
point = feat_vec(i,:); // take the point given by the data set
X = centers{1}(1,:); // take center of the bins at one dimension
Y = centers{2}(1,:); // take center of the bins at other dim.
distanceX = abs(X-point(1)); // find distance to all bin centers at one dimension
distanceY = abs(Y-point(2)); // find distance to center points of other dimension
[~,indexX] = min(distanceX); // find the index of minimum distant center point
[~,indexY] = min(distanceY); // find the index of minimum distant center point for other dimension
You could use interp2 to accomplish that!
If X (1-D Vector, length N) and Y (1-D vector, length M) determine discrete coordinate on the axes where your histogram has defined values Z (matrix, size M x N). Getting value for one particular point with coordinates (XI, YI) could be done with:
% generate grid
[XM, YM] = meshgrid(X, Y);
% interpolate desired value
ZI = interp2(XM, YM, Z, XI, YI, 'spline')
In general, this kind of problem is interpolation problem. If you would want to get values for multiple points, you would have to generate grid for them in similar fashion done in code above. You could also use another interpolating method, for example linear (refer to linked documentation!)
I think you mean this:
[N,C] = hist3(X,...) returns the positions of the bin centers in a
1-by-2 cell array of numeric vectors, and does not plot the histogram.
That being said, if you have a 2D point x=[x1, x2], you are only to look up the closest points in C, and take the corresponding value in N.
In Matlab code:
[N, C] = hist3(data); % with your data format...
[~,indX] = min(abs(C{1}-x(1)));
[~,indY] = min(abs(C{2}-x(2)));
result = N(indX,indY);
done. (You can make it into your own function say result = hist_val(data, x).)
EDIT:
I just saw, that my answer in essence is just a more detailed version of #Erogol's answer.
I am trying to create a 2-D grid from a vector.
So, for example I have:
x = 1:1:10;
z = 2:2:20;
Now, I want to create a grid which has x on both side of the grid cell and z as grid cell value.
I tried doing it as :
[X,Y] = meshgrid(x, x);
newZ = griddata(x, x ,z, X, Y);
But this gives me error:
The underlying triangulation is empty - the points may be
collinear.
Need help solving this.
In a high level, griddata() takes a 2d surface with variable z-value at each point as the first part of the input, and the query points as the second part of the input. To be more specific, when we look into the definition of the function:
vq = griddata(x,y,v,xq,yq)
x and y specifies the range of x and y values, v is like z-value in a plane, and xq and yq together are query points. Here, v (in your case, z) is expected to be a 2d matrix, to be more specific, the size of v is [length(x), length(y)], whereas in your case, you put z as a vector. Matlab generates the warning since the size doesn't match.
For your reference: http://www.mathworks.com/help/matlab/ref/griddata.html?refresh=true