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PCA on high dimensional data

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Ame ZL
Ame ZL on 27 Jul 2018
Commented: Anton Semechko on 28 Jul 2018
Hello,
I have a matrix X with 13952736 rows x 104 columns, of single data type values. I've been trying to run PCA, with a simple one line code that has worked before, but in return i'm having empty arrays results.
the code can't be simpler: [COEFF, SCORE, LATENT, TSQ , EXPLAINED] = pca (X)
I've also tried asking for less variables, but they also come out empty. [~, ~, LATENT, ~ , EXPLAINED] = pca (X)
So I imagine the problem is related with Matlab memory, but no error message is being displayed. I'm running it in Windows 10 Pro, 16 Gb RAM, i7-8550U processor.
any suggestions?
Thanks
SOLUTION SUMMARY:
Thanks to both Ben and Anton, for stopping by and helping!
It seems the problem was not related with memory, but one of my columns was filled with NaN from top to bottom.
By performing SVD step by step as Anton suggested it allowed me to spot this problems, and by eliminating the faulty column the pca() function worked fine again.
Hope this will help somebody in future.
Best wishes
Ame,
  2 Comments
Ame ZL
Ame ZL on 27 Jul 2018
Hi Ben, thanks very much for your reply.
Yeah I tried calculating the PCA on the first 60 columns (just to try), and it worked perfectly.
So I really believe it's a matter of memory as my computer goes to 95 to 100% on all CPU, RAM, and Disk use.
I wonder if there's any solution to this other than changing my laptop?

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Accepted Answer

Anton Semechko
Anton Semechko on 27 Jul 2018
Edited: Anton Semechko on 27 Jul 2018
A 13952736-by-104 data matrix (with observations along rows and variables along columns) will take up
13952736*104*8/2^30 = 10.8 GB
of memory when represented in 'double' format (i.e., 8 bits per element).
Are you able to load this entire matrix into your Matlab workspace?
If yes, then you can obtain PCA of X by performing singular value decomposition (SVD) of its 104-by-104 covariance matrix.
N=size(X,1);
Xo=mean(X,1);
X=bsxfun(@minus,X,Xo);
C=(X'*X)/(N-1);
[U,D]=svd(C,0);
D=diag(D);
Principal directions will be along columns of U, and D will contain singular values of C corresponding to U. Proportion of variance explained with the first k modes of U will be:
R2=cumsum(D)/sum(D);
figure('color','w')
plot(1:numel(D),R2)
xlabel('# principal modes')
ylabel('R^2')
To project X on the first k modes, do:
Y=X*U(:,1:k);
To whiten the loadings, do:
B=Y*diag(1./sqrt(D(1:k)));
  4 Comments
Anton Semechko
Anton Semechko on 28 Jul 2018
Glad to help, America.
And good job spotting those 'NaNs'. In general, it is a good practice to make sure the input data matrix does not contain any 'NaN's or 'Inf's prior to any type of data analysis. This can be do in one line of code:
sum(~isfinite(X(:)))>0

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