Using Numpy (np.linalg.svd) for Singular Value Decomposition

问题

Im reading Abdi & Williams (2010) "Principal Component Analysis", and I'm trying to redo the SVD to attain values for further PCA.

The article states that following SVD:

X = P D Q^t

I load my data in a np.array X.

X = np.array(data)
P, D, Q = np.linalg.svd(X, full_matrices=False)
D = np.diag(D)

But i do not get the above equality when checking with

X_a = np.dot(np.dot(P, D), Q.T)

X_a and X are the same dimensions, but the values are not the same. Am I missing something, or is the functionality of the np.linalg.svd function not compatible somehow with the equation in the paper?

回答1:

TL;DR: numpy's SVD computes X = PDQ, so the Q is already transposed.

SVD decomposes the matrix X effectively into rotations P and Q and the diagonal matrix D. The version of linalg.svd() I have returns forward rotations for P and Q. You don't want to transform Q when you calculate X_a.

import numpy as np
X = np.random.normal(size=[20,18])
P, D, Q = np.linalg.svd(X, full_matrices=False)
X_a = np.matmul(np.matmul(P, np.diag(D)), Q)
print(np.std(X), np.std(X_a), np.std(X - X_a))

I get: 1.02, 1.02, 1.8e-15, showing that X_a very accurately reconstructs X.

If you are using Python 3, the @ operator implements matrix multiplication and makes the code easier to follow:

import numpy as np
X = np.random.normal(size=[20,18])
P, D, Q = np.linalg.svd(X, full_matrices=False)
X_a = P @ diag(D) @ Q
print(np.std(X), np.std(X_a), np.std(X - X_a))
print('Is X close to X_a?', np.isclose(X, X_a).all())

回答2:

From the scipy.linalg.svd docstring, where (M,N) is the shape of the input matrix, and K is the lesser of the two:

Returns
-------
U : ndarray
    Unitary matrix having left singular vectors as columns.
    Of shape ``(M,M)`` or ``(M,K)``, depending on `full_matrices`.
s : ndarray
    The singular values, sorted in non-increasing order.
    Of shape (K,), with ``K = min(M, N)``.
Vh : ndarray
    Unitary matrix having right singular vectors as rows.
    Of shape ``(N,N)`` or ``(K,N)`` depending on `full_matrices`.

Vh, as described, is the transpose of the Q used in the Abdi and Williams paper. So just

X_a = P.dot(D).dot(Q)

should give you your answer.

回答3:

I think there are still some important points for those who use SVD in Python/linalg library. Firstly, https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.svd.html is a good reference for SVD computation function.

Taking SVD computation as A= U D (V^T), For U, D, V = np.linalg.svd(A), this function returns V in V^T form already. Also D contains eigenvalues only, hence it has to be shaped into matrix form. Hence the reconstruction can be formed with

import numpy as np
U, D, V = np.linalg.svd(A)
A_reconstructed = U @ np.diag(D) @ V

The point is that, If A matrix is not a square but rectangular matrix, this won't work, you can use this instead

import numpy as np
U, D, V = np.linalg.svd(A)
m, n = A.shape
A_reconstructed = U[:,:n] @ np.diag(D) @ V[:m,:]

or you may use 'full_matrices=False' option in the SVD function;

import numpy as np
U, D, V = np.linalg.svd(A,full_matrices=False)
A_reconstructed = U @ np.diag(D) @ V

来源：https://stackoverflow.com/questions/24913232/using-numpy-np-linalg-svd-for-singular-value-decomposition

标签

python

numpy

pca