Execution of for loop and indexing of strucutres

Hi,

I am trying to run the following command inside a for loop:

if A(p,1)== B(:,1)

My question: Is every row of B is being compared to A in one running of the loop?

Regards, Waqas

[Azzi Abdelmalek]

Not clear

[James Tursa]

Can you show the loop indexing line also?

[Waqas Syed]

Sorry for the vague question...Here is the code snippet.

n=0;
for p=1:length(MainWPR_mpc.branch)
  for q=1:length(Reus.bus)    
    if MainWPR_mpc.branch(p,1)== Reus.bus(:,1)
      if MainWPR_mpc.branch(p,2)== Reus.bus(:,1)
        Reus.branch(n,:)= MainWPR_mpc.branch(p,:);
        n=n+1;
      end
    end    
  end
end

[James Tursa]

Can you describe, in words, what this test is supposed to do?

if MainWPR_mpc.branch(p:1)== Reus.bus(:,1)

The (p:1) indexing definitely does not look right. Nor does the (p:2) indexing in the following line.

[Waqas Syed]

Sorry, it was supposed to be a comma instead of a colon. Apologies for that.

Answer 1

[Image Analyst]

MainWPR_mpc.branch(p,1) == Reus.bus(:,1)

produces a boolean vector, a list of true false values, not just a single one. So if you do that, you have to decide exactly what you're testing. What if you get 3 true and 5 false, like [1,0,0,1,1,0,0,0]? What do you want to do then? If you want ALL to be true, use all(). If you want to check if ANY of them is true, use any:

if all(MainWPR_mpc.branch(p,1)== Reus.bus(:,1))
if any(MainWPR_mpc.branch(p,1)== Reus.bus(:,1))

I don't know which way you want it.

Answer 2

[James Tursa]

Is p the loop index?

A(p:1) could be empty or could be a scalar depending on what p is. B(:,1) is the first column of B. In an "if" test, the == would mean that all of the elemental results would need to be non-zero for the if-test to evaluate as true. I.e., all of the elements of the first B column would have to equal A(p:1) for the if-test to evaluate as true.

That being said, A(p:1) doesn't look right for what you probably intended and I would re-examine your code.

[Waqas Syed]

Sorry, it was supposed to be a comma instead of a colon. Apologies for that.