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# compute the adjacency matrix from the correlation

Posted on 2011-02-12
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is there a formula to compute the adjacency matrix from the correlation matrix?
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Expert Comment

ID: 34879328
Is your adjacency matrix for a non-directed, unweighted graph? (all 0s and 1s and symmetrical about the main diagonal)
A correlation matrix usually shows how related things are. This will not give you what are directly connected.

My guess is the answer is no. I see no way that you could get an adjacency matrix from a correlation matrix; however, if you describe in more detail, perhaps we can help get the same result some other way.
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Author Comment

ID: 34881748
Actually we can compute the adjacency matrix.
But as usual the problem is the threshold.

so what i did is the following:
if value[i,j] > 0.7 --> adj[i,j] = 0

but we go back to the problem of threshold selection, i just tried 0.7 it worked ok.
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LVL 37

Expert Comment

ID: 34882804
By "it worked ok" do you mean it gave the exact answer or that it got close? Is there complete separation or will you have some error no matter what threshold you choose?

Is this related to a previous question that explains what you are computing?

If there is no guarantee that you can correctly classify each cell in the matrix, then which is worse, false positives or false negatives?
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Author Comment

ID: 34900667
Sorry for being late i was very busy:
By "it worked ok" do you mean it gave the exact answer or that it got close?,  Is there complete separation or will you have some error no matter what threshold you choose?
Well it wont give me an exact question of course, since it's a threshold that i decided to choose, so i wont be able to have exact results. I will only have A VERY close result or close if the threshold is extracted from the Dataset with a specific method that i don't know how. If u know R-cran i'll send u a link for a function that do that:
http://www.gipsa-lab.inpg.fr/~sophie.achard/brainwaver/choose.thresh.nbedges.html
If there is no guarantee that you can correctly classify each cell in the matrix, then which is worse, false positives or false negatives?

Concerning this part of the question, i didn't really think about it. Can u give me an example on how to do it? but till now i am just picking a threshold manually.

Well it's a very hard problem to solve.
that's another link if u are interested: http://www.slideshare.net/athanasios/itab2010thresholding-correlation-matrices

The problem is that i don't have much time, but getting more ideas is really good to improve my classification.

Thank you,

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Author Comment

ID: 34900678
I hope to keep the subject open, and keep u updated, or even if u have ideas it will be great.

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LVL 37

Accepted Solution

TommySzalapski earned 2000 total points
ID: 34902420
>>>>If there is no guarantee that you can correctly classify each cell in the matrix, then which is worse, false positives or false negatives?
>>Concerning this part of the question, i didn't really think about it. Can u give me an example on how to do it? but till now i am just picking a threshold manually.

Any time you are making a classifier (which this basically is) you need to think about whether you want mostly false negatives, mostly false positives, or just minimize the two.
In other words, is your main goal to get as few 0s marked as 1s as possible even at the cost of missing some extra 1s or the other way around? Or do you just want to land as close to the middle as you can.

As another example. I'm working on a classifier for automatic skin cancer diagnosis.
If I set my threshold to .5 then maybe 2% of the patients are told they might have cancer when they do not so we sent them in to get a biopsy. Also 2% are told they are fine when they actually have cancer.
That's 4% error. Not too bad. However, telling someone "You're fine, go on home" when he has cancer is very very bad.
If I drop the theshold to .35 let's say. Then I end up sending 15% of the people in for biopses when they don't need them but I nail 100% of the real cancers.
That's only 85% accuracy; however, in this case it's much better.
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