Calculate the sum of all the relations between a matrix components
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Hi, does anyone know how I can calculate the sum of all the relations between a matrix components? For example by having a 3*3 matrix like:
a=[a11,a12,a13;a21,a22,a23;a31,a32,a33];
I want to calculate a relation between all the components such that:
r11=((a11-a12)/(a11+a12) + (a11-a13)/(a11+a13) + (a11-a21)/(a11+a12) + (a11-a22)/(a11+a22) + (a11-a23)/(a11+a23)+...
(a11-a31)/(a11+a31) + (a11-a32)/(a11+a32) + (a11-a33)/(a11+a33))/n;
n=8; %Number of matrix components-1 in this case
I want to do this for every components of the matrix (each component has interaction with every other components) so that I have:
r12,r13,r21,r22,r23,r31,r32,r33
I used for loop but it takes a long time and long code to calculate the results (my real matrix is 101*101). Is there any simple way to do that? Thank you.
1 Comment
John D'Errico
on 26 Jun 2021
I don't see why a well written loop would take that long of a time here. Perhaps your problem is poorly written code? For example, are you naming individual variables r11, r12, etc?
Accepted Answer
DGM
on 27 Jun 2021
This could be done various ways. You could do this with loops if it's easier to understand. I'll just do this:
% example inputs
A = [1 2 3; 4 5 6; 7 8 9];
n = numel(A)-1;
F = @(x) sum((x-A)./(x+A),'all')/n; % define a function to calculate each sum
R = arrayfun(F,A) % calculate all of them
Using numbered variable names is a great way to cause problems for yourself. If you can embed indexing information within the variable name, then you can just use an array and index into it like normal.
3 Comments
DGM
on 8 Jul 2021
I'm just going to use a loop.
A = [1 2 3; 4 5 6; 7 8 9];
B = [1 2 3; 4 5 6; 7 8 9]*10;
C = [1 2 3; 4 5 6; 7 8 9]*100;
% if you use a cell array, the relative matrix sizes don't matter
D = {A,B,C};
R = cell(size(D));
for d = 1:numel(D)
thismat = D{d};
n = numel(thismat)-1;
F = @(x) sum((x-thismat)./(x+thismat),'all')/n;
R{d} = arrayfun(F,thismat);
end
celldisp(R)
Of course, these example results are identical because the inputs are all proportional.
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