Truncated SVD returns unexpected 1*1 result
5 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Christopher McCausland
il 28 Gen 2022
Commentato: Christopher McCausland
il 30 Gen 2022
Hi,
I am currently prefoming SVD on six channels varying in time of the same type of biometric data. The recorded channels are similar as therefore they feature high correlation.
I want to take these six channels and reduce them to a single channel using SVD, I take my data, rescale it and then preform the svd, so far so good. Where I run into problems is the reconstruction. If I want the orignal data X = U*S*V '.Instead I am trying to preform a truncation by only taking the first PC as this accounts for over 95% of the information anyways.
Instead of PC1 returning a vector (1 channel * given number of samples) I am returned a single value (1*1). I am assuming I have a matrix the wrong way around but I can't see how I am getting this result (Though I know that I am the likely problem). I will attach some open source data too.
data = rescale(data(1:6,:),0,1); % Rescale the data for reobustness between patients
[U,S,V] = svd(data(:,:),'econ'); % Preform SVD
PC1 = V(:,1)'*data(1,:)'; % Atempt to eqivilent single channel
I also know I can compute something similar with the code below, but I would just really like to know where I am going wrong now!
[coeff,score,latent] = pca(data');
new_matrix_for_classification = score(:,1);
Thank you,
Christopher
0 Commenti
Risposta accettata
Matt J
il 28 Gen 2022
Modificato: Matt J
il 28 Gen 2022
Well, you're multiplying a row vector by a column vector, so a scalar is the only result that is possible.
Shouldn't a 1-component reconstruction be,
PC1=data*V(:,1);
5 Commenti
Matt J
il 29 Gen 2022
I have tried to move around the order of the data and V array and transpose them too but am still left with a 1*6 output
If you did, you should have gotten a 10000x1 output, as below.
data=rand(1e4,6);
[U,S,V] = svd(data,'econ'); % Preform SVD
PC1=data*V(:,1);
whos PC1
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Linear Algebra in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!