I have two 3D meshes and on these meshes exists points that are mapped to each other and points with no mapping. For these points I want to calculate a mapping.
I want to do this by calculating a transformation Matrix for each point of one mesh that is already mapped to a point in the other mesh. Afterwards I want to calculate with an quadratic optimization the best possible mapping for all other points.
f(p) = T_i * p
(I'm using homogeneous coordinates - so every transfomation matrix should look like this:
1 0 0 a_i
0 1 0 b_i
0 0 1 c_i
0 0 0 1
Because of some not existing mathematics knowledge I stuck at this point, could somebody tell what exactly I need to feed matlab with that it gives me back an optimized transformation matrix, preferably with a quadratic optimization.
(and maybe an elegant way to do the transformation matrix calculation - actually I do it with a loop that runs over all points and calcs line by line the a_i, b_i, c_i and transforms them to a new matrix..., is there a function in matlab?)
Thanks in advance.
Related
I have a 3D array of dimension (NX,NY,NZ) which represents a variable in physical space, let's say velocities, taken from a simulation in a 3D domain.
1) I want to Fourier-transform only the dimensions X and Z, how should I use the built-in function fft in this case? At some point I want to also get back to the physical space, but only on X, so the same question applies.
2) I read that FFTW uses only 2*N/3 points, should I specify NX and NZ as the number of retained modes or fewer?
3) When using the FFTW package, is there any issue with the coefficient in front of the integral defining the Fourier transformation? Does this package assume that my domain is 2pix2pix2pi?
1°) The function for 2D FFT is fft2, and it will by default apply to the two first dimensions of the array. That is, fft2(velocities) will give you a 3D array with NZ Fourier transforms along dimensions X and Y
In order to do the FFT along other dimensions, you have to manually decompose the 2D FFT as two 1D FFTs. fft will work by default along dimension 1 and produce as many samples as there were in the input. fft(X[],n) does the same, but along dimension n.
Thus, you may compute a 2D FFT of your 3D array, along dimensions X and Z with the command:
my_FFT = fft(fft(velocities),[],3);
2°) There will be as many samples out as samples in.
3°) I believe the normalization by the size of the array is fully applied on the reverse transform, and not at all on the direct transform.
fft([1 0 0 0 0 0])
ans =
1 1 1 1 1 1
To maintain normalization, a coefficient sqrt(NX*NZ) should be applied (multiply when doing FFT, divide when doing an IFFT).
I have two black and white images and I need to calculate the mutual information.
Image 1 = X
Image 2 = Y
I know that the mutual information can be defined as:
MI = entropy(X) + entropy(Y) - JointEntropy(X,Y)
MATLAB already has built-in functions to calculate the entropy but not to calculate the joint entropy. I guess the true question is: How do I calculate the joint entropy of two images?
Here is an example of the images I'd like to find the joint entropy of:
X =
0 0 0 0 0 0
0 0 1 1 0 0
0 0 1 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
Y =
0 0 0 0 0 0
0 0 0.38 0.82 0.38 0.04
0 0 0.32 0.82 0.68 0.17
0 0 0.04 0.14 0.11 0
0 0 0 0 0 0
To calculate the joint entropy, you need to calculate the joint histogram between two images. The joint histogram is essentially the same as a normal 1D histogram but the first dimension logs intensities for the first image and the second dimension logs intensities for the second image. This is very similar to what is commonly referred to as a co-occurrence matrix. At location (i,j) in the joint histogram, it tells you how many intensity values we have encountered that have intensity i in the first image and intensity j in the second image.
What is important is that this logs how many times we have seen this pair of intensities at the same corresponding locations. For example, if we have a joint histogram count of (7,3) = 2, this means that when we were scanning both images, when we encountered the intensity of 7, at the same corresponding location in the second image, we encountered the intensity of 3 for a total of 2 times.
Constructing a joint histogram is very simple to do.
First, create a 256 x 256 matrix (assuming your image is unsigned 8-bit integer) and initialize them to all zeroes. Also, you need to make sure that both of your images are the same size (width and height).
Once you do that, take a look at the first pixel of each image, which we will denote as the top left corner. Specifically, take a look at the intensities for the first and second image at this location. The intensity of the first image will serve as the row while the intensity of the second image will serve as the column.
Find this location in the matrix and increment this spot in the matrix by 1.
Repeat this for the rest of the locations in your image.
After you're done, divide all entries by the total number of elements in either image (remember they should be the same size). This will give us the joint probability distribution between both images.
One would be inclined to do this with for loops, but as it is commonly known, for loops are notoriously slow and should be avoided if at all possible. However, you can easily do this in MATLAB in the following way without loops. Let's assume that im1 and im2 are the first and second images you want to compare to. What we can do is convert im1 and im2 into vectors. We can then use accumarray to help us compute the joint histogram. accumarray is one of the most powerful functions in MATLAB. You can think of it as a miniature MapReduce paradigm. Simply put, each data input has a key and an associated value. The goal of accumarray is to bin all of the values that belong to the same key and do some operation on all of these values. In our case, the "key" would be the intensity values, and the values themselves are the value of 1 for every intensity value. We would then want to add up all of the values of 1 that map to the same bin, which is exactly how we'd compute a histogram. The default behaviour for accumarray is to add all of these values. Specifically, the output of accumarray would be an array where each position computes the sum of all values that mapped to that key. For example, the first position would be the summation of all values that mapped to the key of 1, the second position would be the summation of all values that mapped to the key of 2 and so on.
However, for the joint histogram, you want to figure out which values map to the same intensity pair of (i,j), and so the keys here would be a pair of 2D coordinates. As such, any intensities that have an intensity of i in the first image and j in the second image in the same spatial location shared between the two images go to the same key. Therefore in the 2D case, the output of accumarray would be a 2D matrix where each element (i,j) contains the summation of all values that mapped to key (i,j), similar to the 1D case that was mentioned previously which is exactly what we are after.
In other words:
indrow = double(im1(:)) + 1;
indcol = double(im2(:)) + 1; %// Should be the same size as indrow
jointHistogram = accumarray([indrow indcol], 1);
jointProb = jointHistogram / numel(indrow);
With accumarray, the first input are the keys and the second input are the values. A note with accumarray is that if each key has the same value, you can simply assign a constant to the second input, which is what I've done and it's 1. In general, this is an array with the same number of rows as the first input. Also, take special note of the first two lines. There will inevitably be an intensity of 0 in your image, but because MATLAB starts indexing at 1, we need to offset both arrays by 1.
Now that we have the joint histogram, it's really simple to calculate the joint entropy. It is similar to the entropy in 1D, except now we are just summing over the entire joint probability matrix. Bear in mind that it will be very likely that your joint histogram will have many 0 entries. We need to make sure that we skip those or the log2 operation will be undefined. Let's get rid of any zero entries now:
indNoZero = jointHistogram ~= 0;
jointProb1DNoZero = jointProb(indNoZero);
Take notice that I searched the joint histogram instead of the joint probability matrix. This is because the joint histogram consists of whole numbers while the joint probability matrix will lie between 0 and 1. Because of the division, I want to avoid comparing any entries in this matrix with 0 due to numerical roundoff and instability. The above will also convert our joint probability matrix into a stacked 1D vector, which is fine.
As such, the joint entropy can be calculated as:
jointEntropy = -sum(jointProb1DNoZero.*log2(jointProb1DNoZero));
If my understanding of calculating entropy for an image in MATLAB is correct, it should calculate the histogram / probability distribution over 256 bins, so you can certainly use that function here with the joint entropy that was just calculated.
What if we have floating-point data instead?
So far, we have assumed that the images that you have dealt with have intensities that are integer-valued. What if we have floating point data? accumarray assumes that you are trying to index into the output array using integers, but we can still certainly accomplish what we want with this small bump in the road. What you would do is simply assign each floating point value in both images to have a unique ID. You would thus use accumarray with these IDs instead. To facilitate this ID assigning, use unique - specifically the third output from the function. You would take each of the images, put them into unique and make these the indices to be input into accumarray. In other words, do this instead:
[~,~,indrow] = unique(im1(:)); %// Change here
[~,~,indcol] = unique(im2(:)); %// Change here
%// Same code
jointHistogram = accumarray([indrow indcol], 1);
jointProb = jointHistogram / numel(indrow);
indNoZero = jointHistogram ~= 0;
jointProb1DNoZero = jointProb(indNoZero);
jointEntropy = -sum(jointProb1DNoZero.*log2(jointProb1DNoZero));
Note that with indrow and indcol, we are directly assigning the third output of unique to these variables and then using the same joint entropy code that we computed earlier. We also don't have to offset the variables by 1 as we did previously because unique will assign IDs starting at 1.
Aside
You can actually calculate the histograms or probability distributions for each image individually using the joint probability matrix. If you wanted to calculate the histograms / probability distributions for the first image, you would simply accumulate all of the columns for each row. To do it for the second image, you would simply accumulate all of the rows for each column. As such, you can do:
histogramImage1 = sum(jointHistogram, 1);
histogramImage2 = sum(jointHistogram, 2);
After, you can calculate the entropy of both of these by yourself. To double check, make sure you turn both of these into PDFs, then compute the entropy using the standard equation (like above).
How do I finally compute Mutual Information?
To finally compute Mutual Information, you're going to need the entropy of the two images. You can use MATLAB's built-in entropy function, but this assumes that there are 256 unique levels. You probably want to apply this for the case of there being N distinct levels instead of 256, and so you can use what we did above with the joint histogram, then computing the histograms for each image in the aside code above, and then computing the entropy for each image. You would simply repeat the entropy calculation that was used jointly, but apply it to each image individually:
%// Find non-zero elements for first image's histogram
indNoZero = histogramImage1 ~= 0;
%// Extract them out and get the probabilities
prob1NoZero = histogramImage1(indNoZero);
prob1NoZero = prob1NoZero / sum(prob1NoZero);
%// Compute the entropy
entropy1 = -sum(prob1NoZero.*log2(prob1NoZero));
%// Repeat for the second image
indNoZero = histogramImage2 ~= 0;
prob2NoZero = histogramImage2(indNoZero);
prob2NoZero = prob2NoZero / sum(prob2NoZero);
entropy2 = -sum(prob2NoZero.*log2(prob2NoZero));
%// Now compute mutual information
mutualInformation = entropy1 + entropy2 - jointEntropy;
Hope this helps!
I know there are many plotting documents for Matlab online and I am pretty sure that it has been asked many times. I aplogize in advance for any inconvenience.
I am dealing with a new distribution and I need to draw 3D plot for different values of parameters (I can do it with Excel or any other programs, however, since my other graphs is drawn with MATLAB, and I need to put this 3D in Matlab, too, to publish it as an article). I calculated the result using MATLAB loops, however, plotting gives me the hardest time. I had no other choice but to ask for your assistance. I have these equations for different alphas and betas with a constant sigma and calculate Galton's Skewness and Moor's Kurtosis given with the last two equations.
median=sqrt(2*(sigma^2)*beta*gammaincinv(0.5,alpha));
q1=sqrt(2*(sigma^2)*beta*gammaincinv((6/8),alpha));
q3=sqrt(2*(sigma^2)*beta*gammaincinv((2/8),alpha));
q4=sqrt(2*(sigma^2)*beta*gammaincinv((7/8),alpha));
q5=sqrt(2*(sigma^2)*beta*gammaincinv((5/8),alpha));
q6=sqrt(2*(sigma^2)*beta*gammaincinv((3/8),alpha));
q7=sqrt(2*(sigma^2)*beta*gammaincinv((1/8),alpha));
galtonskewness=(q1-2*median+q3)/(q1-q3);
moorskurtosis=(q4-q5+q6-q7)/(q1-q3);
Let's assume that,
sigma=1
beta=[0.1 0.2 0.5 1 2 5];
alpha=[0.1 0.2 0.5 1 2 5];
I have used mesh(X,Y,Z) for the same range of alphas and betas with the same increment but I take the error "these values cannot be complex". I just want to draw something like the one below.
It must be something easy that I am missing out, but I do not understand where the mistake is. I appreciate any help. Thank you!
I ran the above code for a 2D mesh of points for alpha and beta between 0.1 and 5 for both dimensions and I got results for both.
I suspect it's due to your alpha and beta declaration. You are only providing a few points, and if you try to use mesh, it won't get good results. Therefore, define a meshgrid of points for both alpha and beta, then vectorize your MATLAB code to produce the kurotsis and skewness curves. Only under certain situations should you use for loops. In general, you should avoid using them whenever possible.
How meshgrid works is that given a range of X and Y values, it will produce two (or three if you want 3D co-ordinates) arrays where each location in each array gives you the spatial co-ordinate at that particular location. Therefore, if we did something like:
[X,Y] = meshgrid(1:3, 1:3);
This is what we get:
X =
1 2 3
1 2 3
1 2 3
Y =
1 1 1
2 2 2
3 3 3
Notice that in a 2D grid, for the top-left corner, (x,y) = (1,1), and so for the corresponding location in X, we get 1 and Y we get 1. If you do the same logic for any other position in the 2D grid, you simply look at the X and Y values in each array and it will tell you what the component is for each dimension.
As such, instead of looping through all possible points in your grid, generate them all using meshgrid, then vectorize the computation by calculating your values all at once rather than individually. Once you do this, you have the right structure to be able to put this into mesh.
Therefore, try doing this instead:
%// Define meshgrid of points
[alpha,beta] = meshgrid(0.1:0.1:5, 0.1:0.1:5);
%// From your code
sigma = 1;
%// Calculate quantities - Notice that this is all vectorized
med=sqrt(2*(sigma^2)*beta.*gammaincinv(0.5,alpha));
q1=sqrt(2*(sigma^2)*beta.*gammaincinv((6/8),alpha));
q3=sqrt(2*(sigma^2)*beta.*gammaincinv((2/8),alpha));
q4=sqrt(2*(sigma^2)*beta.*gammaincinv((7/8),alpha));
q5=sqrt(2*(sigma^2)*beta.*gammaincinv((5/8),alpha));
q6=sqrt(2*(sigma^2)*beta.*gammaincinv((3/8),alpha));
q7=sqrt(2*(sigma^2)*beta.*gammaincinv((1/8),alpha));
galtonskewness=(q1-2*med+q3)./(q1-q3);
moorskurtosis=(q4-q5+q6-q7)./(q1-q3);
%// Show our meshes
figure;
mesh(alpha, beta, galtonskewness);
figure;
mesh(alpha, beta, moorskurtosis);
Also take note that I renamed your median variable to med. MATLAB has a function called median and so you don't want to unintentionally shadow over this function with a variable of the same name.
This is what I get:
Take note that I'm not getting the plots that you have placed in your post. It may be because I'm choosing the wrong variables to define the mesh, or perhaps your equations may be incorrect. Double check what you know in theory to what you have here in code and try again.
This should hopefully give you enough to start with though!
I am using a for loop to calculate the electric potential on a subset of the xy-plane (a square grid). Here is the code:
L=2;
for i=1:L
for j=1:L
for k=1:L
V(i,j,k)= -10;
end
end
end
where L is the length of the subset of the xy-plane. The difficulty I am having, however, is that I want the z component of the electric potential to be zero, I just want to the region in the xy-plane to be nonzero. The reason why I am using three dimensions is because I am going to eventually introduce an object, which is at a different electric potential relative to the plane, that is above the plane.
What I tried was taking a simple two dimensional matrix:
a =
1 1 1
1 1 1
and tried replacing the ones in the second column with zeros, which I did by typing a(:,2)=0, and matlab gave me
a =
1 0 1
1 0 1
I then tried to generalize this to a 3 dimensional matrix, but ran into some difficulty. Could someone help me?
I assume you want to set the 2nd component of a 3 dimensional matrix to zero.
You can do this in the same way as you do for 2 dimensional case.
A = ones(3,3,3) % Never use For Loops the way you did for operating on matrices.
A(:,2,:) = 0
%allocate the matrix:
V=nan(L,L,L)
%fill z=0 with intended potential. Assign a scalar to have identical
%values or a matrix to set individually
V(:,:,1)=-10
%set all other numbers to zero:
V(:,:,2:end)=0
You could merge the first and third step by allocating with zeros(L,L,L), but I think this way it's more obvious.
I am looking for a way to align 3D shapes. I have matrices AB, A, and B, each contains a 3D item. Matrix AB is composed of A and B.
I would like to find the transformation (rotation, translation) needed for A and B to rebuild AB.
Do you have any methodology, or useful functions that could help me ?
Regards.
I think of the following approach:
First obtain transform matrices that will take care of the relevant degrees of freedom. Then use a non-linear optimization to minimize the binary 3D difference.
About the degrees of freedom:
Each object at a (x,y,z) position and (theta,phi,psi) angluar orientation (Euler). So one would think you'll need 6+6=12 degrees of freedom (or dimensions) to perform a search. However, the dimensionality is lower, because you have the constraint that the 2 bodies must be present, so it is enough to look for the relative spatial separation (dx,dy,dz) and the relative angular difference (theta,phi,psi) . In addition, we can ask whether there is symmetry in the shapes, for example, if on of the shapes is a sphere, angles are irrelevant, etc..
So for each degree of freedom I'd expect to have a relevant 3x3 transformation matrices, Rtheta,Rphi,Rpsi,Tx,Ty,Tz. where Rtheta looks something like:
Rtheta = [cos(theta) 0 -sin(theta) 0
0 1 0 0
sin(theta) 0 cos(theta) 0
0 0 0 1];
These matrices can be combined a general rotation matrix
Rot=Rtheta*Rphi*Rpsi;
and the similarly obtain the translation matrix Trans=Tx*Ty*Tz, Then
tform_Trans = maketform('affine', Trans);
tform_Rot = maketform('affine', Rot);
tform_RT = maketform('composite',tform_Rot,tform_Trans);
and the transform is applied using the function
tformarray(A, tform_RT, R, TDIMS_A, TDIMS_B, TSIZE_B, TMAP_B, F)
Check out a more detailed account of this in the following link.