How do i make different matrices run through the same code?

I have a program that runs a matrix through it. Now i want another matrix to run through it in a loop so that the first one runs, gives me the output and the other one runs and gives me the output. I use operations of this first matrix that i called 'D' (like sqrt(D(1,:)) so i don't know if the other matrixes that i run through the program have to be called 'D' also or how i would do it.

2 comentarios

Rik
Rik el 20 de Abr. de 2018

That depends. Terminology is important here: is your program a script or a function? Can you provide a MWE?

For example, I have this piece of code

m0=[0;0;0;0];
dn = sqrt((D(1,:)-m0(2)).^2+(D(2,:)-m0(3)).^2+(D(3,:)-m0(4)).^2);

in which 'D' is a 4x4 matrix. I would normally run the program through this line of code and get 'dn', now i want to run another matrix through the same code. they all are simple operations with the matrix

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If your 2D matrices are of the same dimension they can be combined into a single 3D matrix:

D(:,:,1)=matrixA
D(:,:,2)=matrixB
D(:,:,3)=matrixC

Each matrix can be returned in the loop by indexing the third dimension

    D(:,:,1)
    D(:,:,2)
    D(:,:,3)

If the matrices are of differing dimensions, they can be combined into a cell array

D={matrixA,matrixB,matrixC}

again each matrix can be obtained by indexing the cell array:

D{1}
D{2}
D{3}
...

7 comentarios

And how would the array look like for n-matrices?
D(:,:,1) .... D(:,:,n)
or do you mean n dimensional, rather than the 2D matrices in your example?
My matrices are of different dimensions so i would have to combine them into the cell array, how would the array look like for n-matrices? And how would i run the program for each matrix?
D{1} .. D{n}
Typically, you'd use the loop variable to index each matrix out of the cell array D{loopCounter}. Once you had removed each matrix from the cell array, you'd process them in the same way as you do at the moment.

so i would use a for like

for D=D{1}:D{n}

and make it run through it?

for iter=1:n
originalMatrix=D{iter}
...
Oh ok, thanks!

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