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I am trying to do PCA using the SVD method manually and compare the results with the results computed by PCA on sklearn.

Let's say that our matrix is A = [[3 0],[0 -2]]

Following the SVD method, I first compute:

1) A^T A = [[9 0], [0 4]] with A^T the transpose matrix of A

2) Then I compute the eigen value given the fact that σ = sqrt(λ) det(A^T A - λ I ) = 0 with I the identity matrix. I found σ1 = 3 and σ2 = 2

3) I can compute now Σ whose value are the σ values.

4) Now I compute the V matrix. To do that I solve : A^T A v = λ v for each λ value. I got : V = [[1 0], [0 1]]

5) To compute U, we can use the formula ui = Avi/σi

U = [ [1 0], [0 -1]]

Then I tried to compute the projection of A (the new k features with k = 2 in this case) using the formula Y = U^T A but the results I got are different from what I got on the pca from sklearn.

My results : Y = [[3 0 ],[0 2]]

sklearn results : [[ 1.80277564e+00 -1.11022302e-16] [-1.80277564e+00 1.11022302e-16]]

Does anyone know where is my mistake ? Does the pca give us different results depending of the method used ?

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You should perform the steps above on centered data. You'll need to subtract a vector of each feature's sample mean from your observations.

The steps are as follows.

  1. Compute the covariance matrix $\Sigma$. I assume that your data matrix is observations x features, so this would be $\Sigma = \frac{1}{N}(A-\bar{A})^{T} (A-\bar{A})$.
  2. Compute the eigenvalue decomposition of $\Sigma$. Collect the eigenvectors as the columns of a matrix $V$. Equivalently, you can find the singular value decomposition $\Sigma = V^{T} D V$.

  3. Compute the projection of the (centered) data onto the eigenvectors. This gives the final reduction $\hat{A} = (A-\bar{A}) V$.

Here is my python code. I've used the eigenvector decomposition.

import numpy as np
from sklearn.decomposition import PCA

#use sklearn pca
A=np.array([[3, 0], [0, -2]])
pca = PCA()
pca.fit(A)
print("Sklearn PCA output:")
print(pca.transform(A))

#And here is by hand.
Cov=np.matmul((A-A.mean(axis=0)).T, (A-A.mean(axis=0)))/A.shape[0]
#Get the eigenvectors from covariance matrix
V = np.linalg.eig(Cov)[1]
#perform the projection
print("Manual PCA output:")
print(np.matmul(A-A.mean(axis=0), V))

And the output.

Sklearn PCA output:
[[-1.80277564e+00 -1.11022302e-16]
 [ 1.80277564e+00  1.11022302e-16]]
Manual PCA output:
[[ 1.80277564  0.        ]
 [-1.80277564  0.        ]]

Note that sklearn is doing the centering behind the scene. You can verify that fitting the PCA on the centered and original data give the same result.

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