Counting average on huge matrix with conditional

The following code runs forever on 10^5 square matrix A. It runs a few seconds on 10^3 square matrix. Is there any way to speed it up?

   function [meanNV]=calcMeanNV_k(A,V,d)
%A: input matrix (n x n)
%V: eigenvectors of k-th largest eigenvalues (n x 1)
%d: degrees of each node (n x 1)
%meanNV: mean of neighbors vectors (n x 1)
m=size(A,2);
meanNV=zeros(113336);
 for i=1:m
     sumNode = 0;
     for j=1:m
        if A(i,j)==1
            sumNode=sumNode+V(j);
        end
     end
    [meanNV(i)]=sumNode/d(i); 
 end

[Brendan Hamm]

What exactly is this code supposed to be doing?

It seems as though you are finding the last 1 in each row of A. You are taking the corresponding value of V (at the location from A). You are then dividing this by the corresponding value in d.

• The size of a 10^5x10^5 double matrix would be 74.5GB. Do you have enough memory to support this?
• I am doubting that the code does what you are thinking it does, but I can't tell either because I don't know what A represents. It does exactly what I describe above.
• Does meanNV need to be a 113336x113336 matrix? You mention A has similar dimensions to this, but you are only storing m elements in meanNV (one for each row of A), so I think you mean to make this:

meanNV = zeros(m,1); % Save lots of space here.

• You can avoid for loops to improve speed. Use logical criteria. The variable below contains a one in every location where A(i,j) == 1 with no need for a loop.

idx = A == 1; % Or is A logical already?

• Now we can loop through the rows and find the last element with a 1:

for k = 1:m
   col = find(idx(k,:));
   col = col(end);
   meanNV(k) = V(col)/d(k);
end

[Amin Rois Sinung Nugroho]

Hi Brendan, Thanks for your reply.

I made a mistake in the previous code. The initialization of sumNode=0; should be outside of the inner for loop. A represents links between nodes. If node i and j are connected, then A(i,j) = 1. Otherwise, A(i,j) = 0.

I want to get the average of all the eigen vectors of neighbor nodes of i. Neighbor nodes are all nodes that have connection with node i, where A(i,j) = 1. It should accumulate all V(j) that has 1 in the corresponding location in A. Then divide the sum with d(i). d(i) represents the degree/total number of connections from node i.

Thank you.

Answer 1

[Brendan Hamm]

It seems this is likely what you are looking for then:

function [meanNV]=calcMeanNV_k(A,V,d)
%A: input matrix (n x n)
%V: eigenvectors of k-th largest eigenvalues (n x 1)
%d: degrees of each node (n x 1)
%meanNV: mean of neighbors vectors (n x 1)
m=size(A,2);
meanNV=zeros(113336);
idx = A == 1;
for k = 1:m
   meanNV = sum(V(idx(k,:)));
end
meanNV = meanNV./d

[Amin Rois Sinung Nugroho]

It works. But I change the 7th line to meanNV=zeros(m,1); Thanks.