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A complicated matrix manipulation

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J AI
J AI on 19 Jun 2020
Commented: J AI on 19 Jun 2020
So I have a matrix, let's say
A = [1 2;
1 0;
3 2;
0 1;
2 1;
1 2];
[r,c] = size(A);
In general, what I would like to do is:
for i = 1:r-1
x = A(6,1)*A(i,1) + A(6,2)*A(i,2);
end
However, here is the issue: if A(i,1) and/or A(i,2) is zero, and let's assume just A(i,1) is 0 and A(i,2) is non-zero, then, instead I would like to do:
x = A(6,1)*A(i-1,1) + A(6,2)*A(i,2);
And in case, even A(i-1,1) is also zero, then, I would like to do:
x = A(6,1)*A(i-2,1) + A(6,2)*A(i,2);
and so on and so forth (same for A(i,2)). Note, A(1,1), A(1,2), A(6,1), A(6,2) are always non-zeros. I am trying to get my head around it, but so far no luck, although it seems it should have a simple and efficient way.
Your help is highly appreciated. Please let me know if anything seems confusing.
Thanks in advance!
  2 Comments
J AI
J AI on 19 Jun 2020
it was a typo..i added anotther row. thanks for pointing out

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Accepted Answer

Stephen Cobeldick
Stephen Cobeldick on 19 Jun 2020
Edited: Stephen Cobeldick on 19 Jun 2020
Try this reasonably "simple and efficient way":
r = size(A,1);
x = nan(r,1);
for k = 1:r
r1 = find(A(1:k,1),1,'last');
r2 = find(A(1:k,2),1,'last');
x(k) = A(6,1).*A(r1,1) + A(6,2).*A(r2,2);
end
  3 Comments
J AI
J AI on 19 Jun 2020
i agree for loops are clearer, but when you have a large-sized A with 10,000 columns, like i do, it is better getting the job done without a for loop. but as i said, i agree with you - in fact, i am not sure what's happening in the code you just provided so i will be figuring it out. thanks a lot man! really appreciate your help

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