PCA on high dimensional data
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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 个评论
Ben Frankel
2018-7-27
编辑:Ben Frankel
2018-7-27
The matrix X should take up about 6 GB of RAM. It depends on what else is in your computer's memory while MATLAB is running, but most likely that should fit in your RAM. Also, I've had MATLAB use too much memory before, but I didn't get unexpected empty results or crashes (my computer's memory just went to swap). The behavior might be different for different functions for all I know, though.
EDIT: Actually, the return values of pca will take up memory, and the temporary variables used in the pca function will take up memory as well. It's definitely possible that you are running out of memory.
采纳的回答
Anton Semechko
2018-7-27
编辑:Anton Semechko
2018-7-27
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 个评论
Anton Semechko
2018-7-28
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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