First, I am aware that this can be done in sklearn - I'm intentionally trying to do it myself.

I am trying to extract the eigenvectors from np.linalg.eig to form principal components. I am able to do it but I think there's a more elegant way. The part that is making it tricky is that, according to the documentation, the eigenvalues resulting from np.linalg.eig are not necessarily ordered.

To find the first principal component (and second and so on) I am sorting the eigenvalues, then finding their original indexes, then using that to extract the right eigenvectors. I am intentionally reinventing the wheel a bit up to the point where I find the eigenvalues and eigenvectors, but not afterward. If there's any easier way to get from e_vals, e_vecs = np.linalg.eig(cov_mat) to the principal components I'm interested.

import numpy as np

x = 10 * np.random.rand(100)
y = 0.75 * x + 2 * np.random.randn(100)

centered_x = x - np.mean(x)
centered_y = y - np.mean(y)

X = np.array(list(zip(centered_x, centered_y))).T

def covariance_matrix(X):
    # I am aware of np.cov - intentionally reinventing
    n = X.shape[1]
    return (X @ X.T) / (n-1)

cov_mat = covariance_matrix(X)

e_vals, e_vecs = np.linalg.eig(cov_mat)

# The part below seems inelegant - looking for improvement
sorted_vals = sorted(e_vals, reverse=True)

index = [sorted_vals.index(v) for v in e_vals]

i = np.argsort(index)

sorted_vecs = e_vecs[:,i]

pc1 = sorted_vecs[:, 0]
pc2 = sorted_vecs[:, 1]

There is a nice trick to get the eigenvalues and eigenvectors of the covariance matrix without ever forming the matrix itself. This can be done using the singular value decomposition (SVD), as described in this post from Stats.SE. Not only is this more numerically stable, but the results are automatically sorted.

A python version might look like this:

def components(X):
    _, vals, vecs = np.linalg.svd(X - X.mean(axis=0), full_matrices=False)
    return vals**2/(len(X)-1), vecs

A few things to note:

  • As described in the linked post above, the data matrix is typically defined with dimensions as columns, i.e. the transpose of your X.
  • The principle values and components are typically sorted from largest to smallest, i.e. the reverse of yours.
  • The function above does not assume that X has been pre-centered.

So to get results comparable to yours, you would need to do:

vals, vecs = components(X.T)
e_vals, e_vecs = vals[::-1], vecs[::-1]

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