I am using SIMULINK and I needed to define a Rotation Matrix 3,3,N where N is the number of Robots which I am trying to simulate. To do that, because I am also using the Simulink Coder I had to define the signal related to this matrix as Variable Size and I had to define the upper-bound in the following way:
The problem is that when I want to use only one robot (I set n_robots to 1) I get the following error.
Cannot initialize dimensions of 'R' of 'test_pos_ctrl_target/rotation matrix to Euler angles' to [3x3x1]. When the number of dimensions of a matrix exceeds 2, the size of the trailing dimension must be greater than 1.
Someone could help me?
thanks a lot.
You can't have the last dimension as 1 because MATLAB treats any matrix of dimension [m,n,1] as [m,n]. See size() returns 1 where matrix dimension should not exist for more details.
Try defining R of size [n_robots,3,3] and then re-arrange the matrix inside your code (I assume you are using a MATLAB Function block).
Related
I am trying to solve some equations on Matlab using Binary Integer Programming.
I have 3 sets of equations:
Ma.X=1
Mp.X<=1
Mr.X<=m*
Where, Ma is a known matrix with size 5*12
X is unknown set with size 12*1
Also Mp is known matrix with size 5*12
and Mr is a known matrix with size 4*12.
1 in the equations is an unity matrix with size 5*1 in both sets(1&2)
m* is a given known matrix with size 4*1
I'm trying to use command bintprog but how to put 1 equality and 2 inequalities
to get values of X. Also I don't have Function f to insert, I just have set of equations. Given that X unknown values with values 1 or 0.
I tried this command bintprog([],Ma,One51,Mp,One51)
but it gives me The problem is infeasible. with zeros answer matrix.
Please help me to solve this on Matlab
The correct syntax for bintprog is X = bintprog(f,A,b,Aeq,beq).
If you don't have f (which means you just want any feasible point), you can set it to []. However, for the others, your syntax is slightly wrong.
I am assuming that the + in your constraints is actually a * since otherwise, the matrix algebra doesn't quite make sense.
Try this:
X = bintprog([],[Mp;Mr],[ones(5,1);mstar],Ma,ones(5,1))
If even then it tells you that the problem is infeasible; it might as well be true that there are no Xs that can satisfy all your constraints.
I solved a PDE using Matlab solver, pdepe. The initial condition is the solution of an ODE, which I solved in a different m.file. Now, I have the ODE solution in a matrix form of size NxM. How I can use that to be my IC in pdepe? Is that even possible? When I use for loop, pdepe takes only the last iteration to be the initial condition. Any help is appreciated.
Per the pdepe documentation, the initial condition function for the solver has the syntax:
u = icFun(x);
where the initial value of the PDE at a specified value of x is returned in the column vector u.
So the only time an initial condition will be a N x M matrix is when the PDE is a system of N unknowns with M spatial mesh points.
Therefore, an N x M matrix could be used to populate the initial condition, but there would need to be some mapping that associates a given column with a specific value of x. For instance, in the main function that calls pdepe, there could be
% icData is the NxM matrix of data
% xMesh is an 1xM row vector that has the spatial value for each column of icData
icFun = #(x) icData(:,x==xMesh);
The only shortcoming of this approach is that the mesh of the initial condition, and therefore the pdepe solution, is constrained by the initial data. This can be overcome by using an interpolation scheme like:
% icData is the NxM matrix of data
% xMesh is an 1xM row vector that has the spatial value for each column of icData
icFun = #(x) interp1(xMesh,icData',x,'pchip')';
where the transposes are present to conform to the interpretation of the data by interp1.
it is easier for u to use 'method of line' style to define different conditions on each mesh rather than using pdepe
MOL is also more flexible to use in different situation like 3D problem
just saying :))
My experience is that the function defining the initial conditions must return a column vector, i.e. Nx1 matrix if you have N equations. Even if your xmesh is an array of M numbers, the matrix corresponding to the initial condition is still Nx1. You can still return a spatially varying initial condition, and my solution was the following.
I defined an anonymous function, pdeic, which was passed as an argument to pdepe:
pdeic=#(x) pdeic2(x,p1,p2,p3);
And I also defined pdeic2, which always returns a 3x1 column vector, but depending on x, the value is different:
function u0=pdeic2(x,extrap1,extrap2,extrap3)
if x==extrap3
u0=[extrap1;0;extrap2];
else
u0=[extrap1;0;0];
end
So going back to your original question, my guess would be that you have to pass the solution of your ODE to what is named 'pdeic2' in my example, and depending on X, return a column vector.
In my current analysis, I am trying to multiply a matrix (flm), of dimension nxm, with the inverse of a matrix nxmxp, and then use this result to multiply it by the inverse of the matrix (flm).
I was trying using the following code:
flm = repmat(Data.fm.flm(chan,:),[1 1 morder]); %chan -> is a vector 1by3
A = (flm(:,:,:)/A_inv(:,:,:))/flm(:,:,:);
However. due to the problem of dimensions, I am getting the following error message:
Error using ==> mrdivide
Inputs must be 2-D, or at least one
input must be scalar.
To compute elementwise RDIVIDE, use
RDIVIDE (./) instead.
I have no idea on how to proceed without using a for loop, so anyone as any suggestion?
I think you are looking for a way to conveniently multiply matrices when one is of higher dimensionality than the other. In that case you can use bxsfun to automatically 'expand' the smaller matrix.
x = rand(3,4);
y = rand(3,4,5);
bsxfun(#times,x,y)
It is quite simple, and very efficient.
Make sure to check out doc bsxfun for more examples.
Inside a MATLAB function I have built a matrix A, whose dimensions M and N are set as parameters of the function. I would like to plot all the columns of this matrix, given a vector of indices B with length M. Hence, I use these lines:
figure
plot(B,A)
I specified figure as the MATLAB function returns more different plots.
My problem is that the program plots just two columns of the matrix with different colours (blue and violet). Where is my mistake?
Thank you for your attention.
go for
plot(repmat(B,1,N),A);
or
plot(repmat(B,N,1),A);
(depending on your rows/columns). You need to have same size matrices in plot.
Moreover, if B are just consecutive indexes, you may want to consider Plot(A) (or Plot(A')).
I noticed that there was an error which caused the overlap of the different curves, so the way which I used to plot the colums of a matrix is valid. However, the method proposed by Acorbe is a possibility, too.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
MATLAB is running out of memory but it should not be
I want to perform PCA analysis on a huge data set of points. To be more specific, I have size(dataPoints) = [329150 132] where 328150 is the number of data points and 132 are the number of features.
I want to extract the eigenvectors and their corresponding eigenvalues so that I can perform PCA reconstruction.
However, when I am using the princomp function (i.e. [eigenVectors projectedData eigenValues] = princomp(dataPoints); I obtain the following error :
>> [eigenVectors projectedData eigenValues] = princomp(pointsData);
Error using svd
Out of memory. Type HELP MEMORY for your options.
Error in princomp (line 86)
[U,sigma,coeff] = svd(x0,econFlag); % put in 1/sqrt(n-1) later
However, if I am using a smaller data set, I have no problem.
How can I perform PCA on my whole dataset in Matlab? Have someone encountered this problem?
Edit:
I have modified the princomp function and tried to use svds instead of svd, but however, I am obtaining pretty much the same error. I have dropped the error bellow :
Error using horzcat
Out of memory. Type HELP MEMORY for your options.
Error in svds (line 65)
B = [sparse(m,m) A; A' sparse(n,n)];
Error in princomp (line 86)
[U,sigma,coeff] = svds(x0,econFlag); % put in 1/sqrt(n-1) later
Solution based on Eigen Decomposition
You can first compute PCA on X'X as #david said. Specifically, see the script below:
sz = [329150 132];
X = rand(sz);
[V D] = eig(X.' * X);
Actually, V holds the right singular vectors, and it holds the principal vectors if you put your data vectors in rows. The eigenvalues, D, are the variances among each direction. The singular vectors, which are the standard deviations, are computed as the square root of the variances:
S = sqrt(D);
Then, the left singular vectors, U, are computed using the formula X = USV'. Note that U refers to the principal components if your data vectors are in columns.
U = X*V*S^(-1);
Let us reconstruct the original data matrix and see the L2 reconstruction error:
X2 = U*S*V';
L2ReconstructionError = norm(X(:)-X2(:))
It is almost zero:
L2ReconstructionError =
6.5143e-012
If your data vectors are in columns and you want to convert your data into eigenspace coefficients, you should do U.'*X.
This code snippet takes around 3 seconds in my moderate 64-bit desktop.
Solution based on Randomized PCA
Alternatively, you can use a faster approximate method which is based on randomized PCA. Please see my answer in Cross Validated. You can directly compute fsvd and get U and V instead of using eig.
You may employ randomized PCA if the data size is too big. But, I think the previous way is sufficient for the size you gave.
My guess is that you have a huge data set. You don't need all of the svd coefficients. In this case, use svds instead of svd :
Taken directly from Matlab help:
s = svds(A,k) computes the k largest singular values and associated singular vectors of matrix A.
From your question, I understand that you don't call svd directly. But you might as well take a look at princomp (It is editable!) and alter the line that calls it.
You probably needed to calculate an n by n matrix in your computation somehow that is to say:
329150 * 329150 * 8btyes ~ 866GB`
of space which explains why you're getting a memory error. There seems to be an efficient way to calculate pca using princomp(X, 'econ') which I suggest you give it a try.
More on this in stackoverflow and mathworks..
Manually compute X'X (132x132) and svd on it. Or find NIPALS script.