suppose that we have some vector y in matlab and let us suppose that there is such linear relationship between values of y
of course first vector can be easily implemented by
y(L:N-1), but related to matrix, we dont have index y[0] in matlab, so how can i solve this issue? there is another picture
should i use index y1 and instead of index y(l-1),start from y(l)?
Your equation is zero-based, while Matlab is one-based in terms of array indexing. Therefore, the best solution is to add 1 every index in the equation: the vector on the left hand side runs from L+1:N, instead of L:N-1, and the elements in the matrix from 1 to L or N-L (whichever is larger), instead of 0 to L-1 or N-L-1.
Related
So I have these three vectors:
And I have to find out for what value of k these three vectors are linearly dependent. I have tried using rref and linsolve with syms for this but that did not work out. I'm relatively new to MatLab and matrices so please keep that in mind.
I know in order to check if vectors are linearly dependent that c1...cn have to be non-zero.
I also want to know how you can use variables in general when solving these types of equations in MatLab.
A set of vectors (at least if you have n vectors in n dimensions) is linearly dependent if the matrix constructed from them is singular, i.e. if its determinant is 0. If you have the Symbolic Math Toolbox, you can construct a symbolic matrix:
syms k;
M = [1 k 0; -1 1 2; 0 0 3];
det(M)
This will tell you that det(M)==3*k+3, which you can solve by hand. But generally, you can ask matlab to solve it:
solve(det(M)==0,k);
which will tell you the answer is -1. So unless k==-1, these vectors are linearly independent (i.e. they comprise a basis of the Euclidean space R^3).
Update: If you don't have the Symbolic Math Toolbox, you could still try to find a numerical solution. First define a function
detfun=#(k) det([1 k 0; -1 1 2; 0 0 3]);
that for any value of k will give you the determinant of your matrix, for instance detfun(3) gives 12. Then you can use fsolve to find a numerical solution to the equation detfun(k)==0, by calling
fsolve(detfun,0)
in which the second argument, 0, refers to the starting point of the search performed by fsolve. This will tell you that the answer is k==-1, but a single call to fsolve will only give you a single solution. If your function has multiple roots, you have to play around with the starting points to find more of them. In this case, you can know that your function (i.e. det(M(k)) is linear in k, so it has a unique root.
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.
I used matlab to solve linear programming with the common [x, fval] = linprog(f,a,b) and I got the solution. My problem is I want to find the binary vector for the variables(x), for example, the values of (x) after I solve the linear problem were 13,0, 8,0,5,8,0,4,0,0 and I would like to obtain the vector(h) 1,0,1,0,1,1,0,1,0,0 which is represents the binary vector for x. I mean when the value of x greater than 0 we put in h 1 and when the value of x less than or equal 0 put 0 in the vector h?
Thanks.
What about
binvect=x>0;
In Matlab it is as easy as doing that, he will do give you a vector of all the x that fill the condition (>0)
I have two matrices X and Y, both of order mxn. I want to create a new matrix O of order mxm such that each i,j th entry in this new matrix is computed by applying a function to ith and jth row of X and Y respectively. In my case m = 10000 and n = 500. I tried using a loop but it takes forever. Is there an efficient way to do it?
I am targeting two functions dot product -- dot(row_i, row_j) and exp(-1*norm(row_i-row_j)). But I was wondering if there is a general way so that I can plugin any function.
Solution #1
For the first case, it looks like you can simply use matrix multiplication after transposing Y -
X*Y'
If you are dealing with complex numbers -
conj(X*ctranspose(Y))
Solution #2
For the second case, you need to do a little more work. You need to use bsxfun with permute to re-arrange dimensions and employ the raw form of norm calculations and finally squeeze to get a 2D array output -
squeeze(exp(-1*sqrt(sum(bsxfun(#minus,X,permute(Y,[3 2 1])).^2,2)))
If you would like to avoid squeeze, you can use two permute's -
exp(-1*sqrt(sum(bsxfun(#minus,permute(X,[1 3 2]),permute(Y,[3 1 2])).^2,3)))
I would also advise you to look into this problem - Efficiently compute pairwise squared Euclidean distance in Matlab.
In conclusion, there isn't a common most efficient way that could be employed for every function to ith and jth row of X. If you are still hell bent on that, you can use anonymous function handles with bsxfun, but I am afraid it won't be the most efficient technique.
For the second part, you could also use pdist2:
result = exp(-pdist2(X,Y));
Please see the following issue:
P=rand(4,4);
for i=1:size(P,2)
for j=1:size(P,2)
[r,p]=corr(P(:,i),P(:,j))
end
end
Clearly, the loop will cause the number of correlations to be doubled (i.e., corr(P(:,1),P(:,4)) and corr(P(:,4),P(:,1)). Does anyone have a suggestion on how to avoid this? Perhaps not using a loop?
Thanks!
I have four suggestions for you, depending on what exactly you are doing to compute your matrices. I'm assuming the example you gave is a simplified version of what needs to be done.
First Method - Adjusting the inner loop index
One thing you can do is change your j loop index so that it only goes from 1 up to i. This way, you get a lower triangular matrix and just concentrate on the values within the lower triangular half of your matrix. The upper half would essentially be all set to zero. In other words:
for i = 1 : size(P,2)
for j = 1 : i
%// Your code here
end
end
Second Method - Leave it unchanged, but then use unique
You can go ahead and use the same matrix like you did before with the full two for loops, but you can then filter the duplicates by using unique. In other words, you can do this:
[Y,indices] = unique(P);
Y will give you a list of unique values within the matrix P and indices will give you the locations of where these occurred within P. Note that these are column major indices, and so if you wanted to find the row and column locations of where these locations occur, you can do:
[rows,cols] = ind2sub(size(P), indices);
Third Method - Use pdist and squareform
Since you're looking for a solution that requires no loops, take a look at the pdist function. Given a M x N matrix, pdist will find distances between each pair of rows in a matrix. squareform will then transform these distances into a matrix like what you have seen above. In other words, do this:
dists = pdist(P.', 'correlation');
distMatrix = squareform(dists);
Fourth Method - Use the corr method straight out of the box
You can just use corr in the following way:
[rho, pvals] = corr(P);
corr in this case will produce a m x m matrix that contains the correlation coefficient between each pair of columns an n x m matrix stored in P.
Hopefully one of these will work!
this works ?
for i=1:size(P,2)
for j=1:i
Since you are just correlating each column with the other, then why not just use (straight from the documentation)
[Rho,Pval] = corr(P);
I don't have the Statistics Toolbox, but according to http://www.mathworks.com/help/stats/corr.html,
corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient between each pair of columns in the n-by-p matrix X.