Hi - here's some additional code - I *think* this is the relevant code for what
you're asking about.
Cheers.
% temporary version of vars, useful for various null tests below
grotvars=inormal(vars); grotvars(:,std(grotvars)<1e-10)=[];
grotvars(:,sum(isnan(grotvars)==0)<20)=[];
% permutation testing
grotRp=zeros(Nperm,Nkeep+1); clear grotRpval; nullNETr=[]; nullSMr=[];
nullNETv=[]; nullSMv=[];
for j=1:Nperm
j
[grotAr,grotBr,grotRp(j,1:end-1),grotUr,grotVr,grotstatsr]=canoncorr(uu1,uu2(PAPset(:,j),:));
grotRp(j,end)=mean(grotRp(j,1:end-1));
nullNETr=[nullNETr corr(grotUr(:,1),NET)']; nullSMr=[nullSMr
corr(grotVr(:,1),grotvars(PAPset(:,j),:),'rows','pairwise')'];
nullNETv=[nullNETv sum(corr(grotUr,NET).^2,2)]; nullSMv=[nullSMv
sum(corr(grotVr,grotvars(PAPset(:,j),:),'rows','pairwise').^2,2)];
end
Ncca=sum(grotRpval<0.05) % number of significant CCA components
% how does the % variance explained compare with the null scenario?
grot1=[ sum(corr(grotU,NET).^2,2) prctile(nullNETv,5,2) mean(nullNETv,2)
prctile(nullNETv,95,2) sum(corr(uu1,NET).^2,2) ] * 100 / size(NET,2);
grot2=[ sum(corr(grotV,grotvars,'rows','pairwise').^2,2) prctile(nullSMv,5,2)
mean(nullSMv,2) prctile(nullSMv,95,2) ...
sum(corr(uu2,grotvars,'rows','pairwise').^2,2) ] * 100 / size(grotvars,2);
I=1:20; figure;
subplot(2,1,1); hold on;
for i=1:length(I)
rectangle('Position',[i-0.5 grot1(i,2) 1
grot1(i,4)-grot1(i,2)],'FaceColor',[0.8 0.8 0.8],'EdgeColor',[0.8 0.8 0.8]);
end
plot(grot1(I,3),'k'); plot(grot1(I,1),'b'); plot(grot1(I,1),'b.'); %
plot(grot1(I,5),'g'); % turned off showing the PCA equivalent plots
subplot(2,1,2); hold on;
for i=1:length(I)
rectangle('Position',[i-0.5 grot2(i,2) 1
grot2(i,4)-grot2(i,2)],'FaceColor',[0.8 0.8 0.8],'EdgeColor',[0.8 0.8 0.8]);
end
plot(grot2(I,3),'k'); plot(grot2(I,1),'b'); plot(grot2(I,1),'b.'); %
plot(grot2(I,5),'g');
> On 27 Apr 2017, at 09:00, Eleanor Wong <eleanor.w...@utah.edu> wrote:
>
> Hi HCP group,
>
> I have a question regarding how to compute the variance explained after
> running CCA on the HCP data using the code from
> http://www.fmrib.ox.ac.uk/datasets/HCP-CCA/
> <http://www.fmrib.ox.ac.uk/datasets/HCP-CCA/>. I ran it on the 461 subject
> subset with dimensionality reduced to 100 dimensions. The result from the
> code is very similar to the weights provided on the webpage for the HCP900
> dataset, but I was unable to compute the % variance explained in the graphs
> of figure 1c) of paper
> http://www.nature.com/neuro/journal/v18/n11/full/nn.4125.html
> <http://www.nature.com/neuro/journal/v18/n11/full/nn.4125.html>. I tried
> looking at the variance of the scores on the top CCA component divided by the
> sum of the variance of the scores across all components for the subject
> measures and the network data, with the Matlab command:
> var(grotU(:,1))/sum(var(grot(U)) but that did not give similar results.
>
> Thank you for your help.
>
> Best,
> Eleanor
> _______________________________________________
> HCP-Users mailing list
> HCP-Users@humanconnectome.org
> http://lists.humanconnectome.org/mailman/listinfo/hcp-users
>
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Stephen M. Smith, Professor of Biomedical Engineering
Head of Analysis, Oxford University FMRIB Centre
FMRIB, JR Hospital, Headington, Oxford OX3 9DU, UK
+44 (0) 1865 222726 (fax 222717)
st...@fmrib.ox.ac.uk http://www.fmrib.ox.ac.uk/~steve
<http://www.fmrib.ox.ac.uk/~steve>
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