Forget what I said. I didn't read your test properly.
On Thu, Jul 31, 2014 at 2:47 PM, Michael Eickenberg <
[email protected]> wrote:
> 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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>
>
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