vt.T should be equal to the output of sklearn_pca(X.T).

There is a big difference in centered across samples or across features.
Sklearn PCA centers across samples and that is the standard afaik.

Hope that helps,
Michael


On Thu, Jul 31, 2014 at 2:41 PM, Deepak Pandian <[email protected]>
wrote:

> Hello Everyone,
>
> I assumed that doing a PCA on X is equivalent to performing a SVD  on
> a mean-centered X.
>
> For sklearn.pca the input matrix is of the shape, (n_samples,n_features).
>
> When I perform a SVD on  a matrix X shaped (n_features,n_samples)
> ,some of the eigen vectors arent matching with the pca.components_
>
>
> def sklearn_pca(X):
>     pca =PCA(n_components=4)
>     pca.fit(X)
>     return pca.components_
>
>
> def svd_pca(X):
>     #every sample is a row
>     X= X- np.mean(X,axis=0)
>     u,s,v = np.linalg.svd(X,full_matrices=False)
>     return u,s,v
>
> def svd_t(X):
>     #x.shape = n_features,n_samples
>     #every sample is a column
>     mean= np.mean(X,axis=1)
>     mean = mean[:,np.newaxis]
>     X = X -mean
>     u,s,v = np.linalg.svd(X,full_matrices=False)
>     return u,s,v
>
> def test_svd_pca():
>     rng = np.random.RandomState(1)
>     X  = rng.randn(4,10)
>     XT = np.copy(X)
>     PX= np.copy(X)
>     pca_comp = sklearn_pca(PX)
>     u,s,v = svd_pca(X)
>     ut,st,vt = svd_t(X.T)
>
>
> I expected pca.components_ to be equivalent to v and vt.T. While its
> same as v , I get  mismatches with vt.T
>
> Is it right to expect vt.T to be same as pca.components_? If not
> please help me clear my misunderstanding.
>
> --
> With Regards,
> Deepak
>
>
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