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# How to operate on different rows of the same matrix?

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Marco Boesso on 28 Oct 2020
Closed: MATLAB Answer Bot on 20 Aug 2021
Good morning folks, I need your help to figure out a possible solution for my issue. I can't find a way out.
Let's consider this matrix A:
1 4 -2 0
2 0 -2 0
3 0 -3 0
4 2 -2 0
I want to operate on the rows, in order to do some operations. Let's stick, for easiness' sake, to this problem: I want to pair the first elements of two different rows. I.e.:
[1, 2]; [1, 3]; [1, 4]; [2, 3]; [2, 4]; [3, 4]
so, it's the first row with the second, the first with the third, the first with the fourth, the second with the third, the second with the fourth, so on...
I had the job done with nested for cycles, but... my supervisor wants me to use a different solution. I need help as I can't find a way out but I desperately need this!
Marco Boesso on 28 Oct 2020
Went for some trials, I figured out how to handle those strings by my own. Looks like I've got to compare each string manually.
Nevertheless, thanks for your help: it's been vital to allow me to go on with my work. Brilliant mate :D
Rik on 28 Oct 2020
You can fairly easily convert what you have to either style in my comment. I don't see why you would need to do manual work, that doesn't sound like the Matlab way.

Sudhakar Shinde on 28 Oct 2020
Try this:
[A(1,1) A(2,1)] %[1 2]
[A(1,1) A(3,1)] %[1 3]
[A(1,1) A(4,1)] %[1 4]
[A(2,1) A(3,1)] %[2 3]
[A(2,1) A(4,1)] %[2 4]
[A(3,1) A(4,1)] %[3 4]
Sudhakar Shinde on 28 Oct 2020
@rik, Thanks for your suggestion and it will surely helpful for OP. If A is extended, vectorized answer will surely helpful. Need to keep in mind that using a loops will increase computational complexity.
to store the result in cell :
B={};
for ind1=1:(size(A,1)-1)
for ind2=(ind1+1):size(A,1)
B{end+1}=A([ind1 ind2],1);
end
end
Rik on 28 Oct 2020
Or you don't use a nested loop with a dynamically exanding variable:
B=nchoosek(1:size(A,1),2);
B=mat2cell(B,ones(size(B,1),1),2);