find a correlation

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Niki
Niki 2011-9-5
I have a Matrix
X=[0.231914928 3.126057882 -1.752476846
-0.779092587 2.143243132 -1.944363312
-1.744892449 1.206497824 -2.267829067
-0.276817947 1.774687601 -1.768924258
-0.367233254 1.697905199 -1.508506912
-0.367233254 1.697905199 -1.508506912
-1.378240769 0.814907572 -1.700393377
-2.389248284 -0.060411815 -1.892279842
-1.333033116 0.860977013 -1.831972668
0.135041386 1.40613207 -1.333067858]
Y=[0.253549664
-0.231692981
0.768395971
2.988670669
-0.038625616
-0.038625616
-0.525155376
-1.011685136
0.961463336
3.181738034]
At first I want to calculate the correlation coefficient between all X columns which can be done like this
[R]=corrcoef(X)
then I want to see which pair of columns has the highest correlation together for example column 1 with 2 ? 1 with 3? 2 with 3?
then the one that has correlation more than 0.5 lets say for example columns 1 and 2 , then check their correlation with y and say which one is more correlated
  2 个评论
Daniel Shub
Daniel Shub 2011-9-5
Mohammad, in general, I think your questions are great questions for Answers, but often I feel that I do not understand your questions. I think it would be easier for those of us who chose to answer questions, if you could spend a little more time composing your questions.
Niki
Niki 2011-9-5
Thanks Daniel for your comment, For sure, I will do my best , Thanks

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采纳的回答

Grzegorz Knor
Grzegorz Knor 2011-9-5
For highest correlation:
maxR = max(max(triu(R,1))) % highest correlation
[row,col] = find(R==maxR,1,'first')
The greatest correlation occurs between the columns row an col.
Second question:
[r w] = max([corr(X(:,row),Y) corr(X(:,col),Y)])
w equal to 1 means that X(:,row) has a higher correlation than X(:,col). w equal to 2 mean that X(:,col) has a higher correlation than X(:,row).
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Niki
Niki 2011-9-5
if you put the command, I can accept your answer
Niki
Niki 2011-9-5
corr(X(:,unique([row;col])),Y)

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更多回答（2 个）

the cyclist
the cyclist 2011-9-5
It wasn't perfectly clear to me if you wanted to find correlation with Y for all the columns that had r>0.5, or only the highest. This does all of them. Maybe you could tailor this to what you need.
r = corrcoef(X); % Correlation coefficiant
[i j] = find(r>0.5); % Indices of r > 0.5
indexToOffDiagonalElementsWithHighCorrelation = (i~=j); % Only use off-diagonal elements
XColumnsWithHighCorrelation = unique(i(indexToOffDiagonalElementsWithHighCorrelation))
for nx = 1:numel(XColumnsWithHighCorrelation)
rxy{nx} = corrcoef(X(:,XColumnsWithHighCorrelation(nx)),Y);
disp(rxy{nx})
end
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Oleg Komarov
Oleg Komarov 2011-9-5
use triu to zero out the upper diagonal.
Niki
Niki 2011-9-5
Thanks Oleg, I did not know " Triu " :D
Andrei put a command for that, Thanks

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Andrei Bobrov
Andrei Bobrov 2011-9-5
[v id]= max(triu(corrcoef([X,Y]),1))
Variant last
R = triu(corrcoef([X,Y]),1)
Rx = R(1:end-1,1:end-1)
Rx05 = Rx.*(Rx>.5)
[ix jx] = find(Rx05==max(Rx05(:)))
cYX = R([ix,jx],4)
[vXY xi]= max(cYX)
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Niki
Niki 2011-9-5
I think you can not reach to the answer with this command please check this out
for example we have
>> X=rand(10);
>> Y=rand(1,10);
then we perform
>>[R]=corrcoef(X);
then if you perform
>>[v id]= max(triu(corrcoef([X,Y]),1))
v is different with R , which I only can see one value similar, could you please tell me what is happening with this command ?
Niki
Niki 2011-9-5
Andrei, I like your comment very much, Thanks

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