Simultaneously inverting many matrices
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Dear all, I have many 2-by-2 matrices (which are covariance matrices). I want to invert them all. I'm curious if there's an efficient way of doing this. I thought, maybe, you create a cell, in which each element is one of these matrices, and then use cellfun() in some way to do it. Quintessentially, my question is, is there a way of simultaneously inverting many matrices? I'd appreciate any and all comments. Thank you very much in advance!
Best, John
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Más respuestas (4)
Tohru Kikawada
el 28 de Abr. de 2019
Editada: Tohru Kikawada
el 28 de Abr. de 2019
3 votos
You can leverage Symbolic Math Toolbox to vectorize the calculation.
% Define size of matrices
M=2;
N=10000;
A=rand(M,M,N);
% Calculate the inverse matrices in a loop
invA_loop = zeros(size(A));
tic
for k = 1:N
invA_loop(:,:,k) = inv(A(:,:,k));
end
disp('Elapsed time in calculation in a loop:');
toc
% Calculate the inverse matrices in a vectorization
As = sym('a', [M,M]); % Define an MxM matrix as a symbolic variable
invAs = reshape(inv(As),[],1); % Solve inverse matrix in symbol
invAfh = matlabFunction(invAs,'Vars',As); % Convert the symbolic function to an anonymous function.
tic
invA_sym = reshape(invAfh(A(1,1,:),A(2,1,:),A(1,2,:),A(2,2,:)),M,M,N);
disp('Elapsed time in the vectorized calculation:');
toc
% Max difference between the results in the loop and the vectorization
disp('Max difference between the elements of the results:');
disp(max(abs(invA_loop(:)-invA_sym(:))))
Results:
Elapsed time in calculation in a loop:
Elapsed time is 0.093938 seconds
Elapsed time in the vectorized calculation:
Elapsed time is 0.001396 seconds
Max difference between the elements of the results:
3.0323e-09
5 comentarios
Lucas Azevedo
el 8 de Mayo de 2020
Could you please tell me why there is a different between the elements of the results?
Tohru Kikawada
el 9 de Mayo de 2020
Because function inv performs an LU decomposition which is different from the vectorized one. Please see the documentation for more details.
Lucas Azevedo
el 9 de Mayo de 2020
Humm, I see. By the way, I am using your vectorization and now my code is 2x faster! Thank you!
Alec Jacobson
el 14 de Nov. de 2020
%invA_sym = reshape(invAfh(A(1,1,:),A(2,1,:),A(1,2,:),A(2,2,:)),M,M,N);
Acell = reshape(num2cell(A,3),1,[]);
invA_sym = reshape(invAfh(Acell{:}),M,M,N);
so the code above works for M≠2
Tohru Kikawada
el 31 de En. de 2021
Alec, this is great! Thanks for your extension!
James Tursa
el 18 de Jun. de 2015
2 votos
You might look at this FEX submission by Bruno Luong for solving 2x2 or 3x3 systems:
Hugo
el 18 de Jun. de 2015
In my experience, using cells is rather slow. Since your matrices are 2x2, then you could simple arrange them in a 3D array, with the first dimension representing the index of each matrix. Let's call this matrix M, which will be of size Nx2x2, N denoting the number of matrices you want to invert.
Now recall that the inverse of a matrix A=[A11,A12;A21,A22] can be computed as
[A22, -A12; -A21, A11] /(A11*A22-A12*A21)
You can implement this for all matrices as follows:
Minv = reshape([M(:,4),-M(:,2),-M(:,3),M(:,1)]./repmat(M(:,1).*M(:,4)-M(:,2).*M(:,3),1,4),N,2,2);
Hope this helps
Hugo
Azzi Abdelmalek
el 18 de Jun. de 2015
0 votos
Using cellfun will not do it simultaneously. The for loop can be faster. But if you have a Parallel Computing Toolbox, you can do it with parfor
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