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Extract principal components

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4

$begingroup$

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

np.random.seed(0)
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]

python reinventing-the-wheel numpy

share|improve this question

edited 10 mins ago

Jamal♦

30.4k11121227

asked 23 hours ago

jss367jss367

222310

$endgroup$

add a comment | 

4

$begingroup$

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

np.random.seed(0)
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]

python reinventing-the-wheel numpy

share|improve this question

edited 10 mins ago

Jamal♦

30.4k11121227

asked 23 hours ago

jss367jss367

222310

$endgroup$

add a comment | 

$begingroup$

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

np.random.seed(0)
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]

python reinventing-the-wheel numpy

share|improve this question

edited 10 mins ago

Jamal♦

30.4k11121227

asked 23 hours ago

jss367jss367

222310

$endgroup$

import numpy as np

np.random.seed(0)
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]

share|improve this question

edited 10 mins ago

Jamal♦

30.4k11121227

asked 23 hours ago

jss367jss367

222310

share|improve this question

edited 10 mins ago

Jamal♦

30.4k11121227

asked 23 hours ago

jss367jss367

222310

edited 10 mins ago

Jamal♦

30.4k11121227

edited 10 mins ago

Jamal♦

30.4k11121227

asked 23 hours ago

jss367jss367

222310

asked 23 hours ago

jss367jss367

222310