Vectorisation of a simple for loop
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David Schranz
el 28 de Sept. de 2016
Comentada: David Schranz
el 29 de Sept. de 2016
Will it be possible to write this code vectorised?
x=ones(10,1);
a=zeros(size(x));
b=zeros(size(x));
c=zeros(size(x));
for i = 1:size(x,1)-1
a(i+1) = a(i)+x(i)+b(i);
b(i+1) = b(i)+a(i)+c(i);
c(i+1) = c(i)+b(i)+c(i);
end
Any help would be greatly appreciated.
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Thorsten
el 28 de Sept. de 2016
A = [1 1 0 0 ; 1 1 1 0 ; 0 1 2 0; 1 0 0 1];
Q = cell2mat(arrayfun(@(i) A^i, 1:9, 'Uni', false));
Q = [zeros(1, 4); reshape(Q(end,:), 4, [])'];
a = Q(:,1); b = Q(:,2); c = Q(:,3);
Más respuestas (2)
Teja Muppirala
el 28 de Sept. de 2016
Just wondering, but what is your purpose in vectorizing that code? If it's to make it faster, I don't think you can do much better than what you've already got, especially if you have a recent version of MATLAB.
If it's to make it more readable, or more general (flexibly deal with more variables), so you don't have to keep writing every letter like this
a(i+1)=...
b(i+1)=...
c(i+1)=...
d(i+1)= ...
...
then you could use matrix equations.
x=ones(10,1); % External input
A = [1 1 0; 1 1 1; 0 1 2]; % State matrix
B = [1;0;0]; % Input matrix
v = zeros(length(B),length(x)); % v contains [a,b,c]
for i = 1:size(x,1)-1
v(:,i+1) = A*v(:,i) + B*x(i);
end
For many applications, matrix representation is usually how this type of calculation is expressed.
KSSV
el 28 de Sept. de 2016
It is very much possible:
https://in.mathworks.com/matlabcentral/answers/81775-recursive-vector-operation-without-for-loops
2 comentarios
Stephen23
el 28 de Sept. de 2016
Editada: Stephen23
el 28 de Sept. de 2016
@Dr. Siva Srinivas Kolukula: the solution presented in that answer, and indeed the filter command itself, is a fundamentally 1D solution, even if applied to an array. The question above has dependencies between the variables a, b, and c. How do you propose to represent these using a 1D filter?
KSSV
el 28 de Sept. de 2016
If I am not mistaken, a similar question appeared a week ago which was vectorised using filter...I searched for the question, but could not get it.
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