Computing determinants of a 3D array
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Yanir Hainick
el 26 de Nov. de 2012
Let’s say I have an NxNxL array. L is typically 10^4-10^5, and N is typically 10^0-10^1.
My goal is to calculate a vector of length L, where the i-th cell contains the determinant of (:,:,i).
I currently use for-loop, as det(A) accepts 3D arrays with the last dimension being a singleton, so this code works:
for i = 1:L
Vec(i) = det(Mat(:,:,i));
end
However, it seems weird that i can't implement this in a vectorial fasion. Can anybody think of any way to get rid of the for loop? Note that the format of an 3D array is pretty stiff, i.e. i can't change the input to cell array and use cellfun.
Thanks!
Yanir.
2 comentarios
Matt J
el 26 de Nov. de 2012
Hopefully, the reason you're asking for this is not for the purpose of solving many linear systems.
Respuesta aceptada
Sean de Wolski
el 26 de Nov. de 2012
What's wrong with the for-loop?
n=100;
pages = 1e4;
X = rand(n,n,pages);
D = zeros(pages,1);
tic;
for ii = 1:pages
D(ii) = det(X(:,:,ii));
end
toc;
%Elapsed time is 1.805616 seconds.
1 comentario
James Tursa
el 27 de Nov. de 2012
My 2 cents to all:
The performance advantage of vectorized approaches to this is that one wishes to avoid the data copy involved with the X(:,:,ii) slices and the overhead of the loop. That being said, this data copy & loop overhead is in all likelihood swamped by the numerical calculations (and possible data copy) involved in the determinant calculation itself. So even if one were to get a vectorized one-liner to this that did not involve explicit slices, it probably wouldn't run significantly faster (if at all) than straight forward loops (the small size explicit code e.g. that Matt shows excepted).
Más respuestas (2)
Matt J
el 26 de Nov. de 2012
Editada: Matt J
el 8 de Abr. de 2020
For small N, it would be an advantage to vectorize the determinant formula explicitly (example for for N=2 below). For larger N, maybe you could do the same thing recursively.
%fake data
N=2;
L=1e5;
Mat=rand(N,N,L);
tic
Vec=zeros(1,L);
for ii=1:L
Vec(ii)=det(Mat(:,:,ii));
end
toc;
%Elapsed time is 0.194555 seconds.
Mat=reshape(Mat,[],L);
tic;
Vec =Mat(1,:).*Mat(4,:) - Mat(2,:).*Mat(3,:);
toc
%Elapsed time is 0.000378 seconds.
2 comentarios
Pi Ting
el 8 de Nov. de 2017
The line
Vec =Mat(1,:).*Mat(3,:) - Mat(2,:).*Mat(4,:);
should be
Vec =Mat(1,:).*Mat(4,:) - Mat(2,:).*Mat(3,:)?
Al in St. Louis
el 8 de Abr. de 2020
I had to use Pi Ting's expression to get the correct answers. This is exactly what I need to do!
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