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Why do I get a wrong answer when I use FIND function in this case?

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Waseem AL Aqqad
Waseem AL Aqqad on 17 Jun 2021
Commented: Waseem AL Aqqad on 17 Jun 2021
I have a graph structure "G_dmg.Nodes.Load" which has typically 40 "-inf" values and 10 other numbers ~= -inf.
In my script:
inActive = find(G_dmg.Nodes.Load==-inf);
for ii = 1:length(inActive)
b = neighbors(G_dmg,inActive(ii));
idx = find(G_dmg.Nodes.Load(b)~= -inf);
Load = G_dmg.Nodes.Load(idx);
end
So, I'm trying to find the indices of inActive nodes' neighbors that do not have -inf Load values. FIND function is not giving me back the correct indices.
say b = [10 11 2 3 4] and the 10th, 11th entries in G_dmg.Nodes.Load ~=-inf but 2nd, 3rd, 4th have -inf values. idx variable in my script is giving me back 1 2 which are basically are the first and second entries in b.

Accepted Answer

Rik
Rik on 17 Jun 2021
When you subindex a variable before using find, it will return the indices relative to that subindexed variable. This is because find doesn't know about the indexing.
This might fix the problem:
idx=b(idx);

More Answers (1)

KSSV
KSSV on 17 Jun 2021
Edited: KSSV on 17 Jun 2021
Replace the line:
inActive = find(G_dmg.Nodes.Load==-inf);
with
idx = isinf(G_dmg.Nodes.Load) ;
inActive = find(sign(G_dmg.Nodes.Load).*idx == -1)
Follow the same for other lines where you are trying to find -inf.
You can achieve the above without loop.
  1 Comment
Waseem AL Aqqad
Waseem AL Aqqad on 17 Jun 2021
Thanks for telling me this technique.
Unfortunately, I have to use a loop here since the built in function "neighbors" accepts only scalar inputs, I can't pass the vector inActive to it.
I believe it has this ristriction as each node in a graph has different number of neighboring nodes. Of course, I can concatenate them like this:
b = [];
b = [b, neighbors(G_dmg,inActive(ii))];
But still, I cannot tell which nodes are neighbors to node X and which are neighbors to node Y.

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