FIX Make eignvectors consistent#1
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Micky774 merged 7 commits intoMicky774:change_svdfrom Mar 13, 2022
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thomasjpfan
commented
Mar 9, 2022
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| @pytest.mark.parametrize("add_noise", [True, False]) | ||
| @pytest.mark.parametrize("seed", range(2)) | ||
| def test_fastica_simple_different_solvers(add_noise, seed): |
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This is a new test to check for consistency based on what you had.
thomasjpfan
commented
Mar 9, 2022
| outs[solver] = sources | ||
| assert ica.components_.shape == (2, 2) | ||
| assert sources.shape == (1000, 2) | ||
| _, _, sources_fun = fastica(m.T, fun=nl, algorithm=algo, random_state=seed) |
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This reverts the tests to what they were on upstream/main.
thomasjpfan
commented
Mar 9, 2022
| d, u = d[sort_indices], u[sort_indices] | ||
| # Resize and reorder to match svd | ||
| u = u[::-1, : min(X.shape) : -1] | ||
| d, u = d[sort_indices], u[:, sort_indices] |
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Based on eigh docs, the eigenvectors has shape (M, N) where N is the number of eigenvalues. Thus, I think the sorting should be done over axis=1.
Author
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Nevermind I think this works. |
Author
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Here is a code snippet to demonstrate: from scipy import linalg
import numpy as np
from numpy.random import default_rng
from numpy.testing import assert_allclose
def my_svd(X):
u, d = linalg.svd(X, full_matrices=False)[:2]
u *= np.sign(u[0])
return u, d
def my_eigh(X):
d, u = linalg.eigh(X.T.dot(X))
sort_indicies = np.argsort(d)[::-1]
d = d[sort_indicies]
u = u[:, sort_indicies]
u *= np.sign(u[0])
return u, np.sqrt(d)
rng = default_rng()
for i in range(10):
m, n = 9, 6
X = rng.standard_normal((m, n))
u_svd, d_svd = my_svd(X.T)
u_eigen, d_eigen = my_eigh(X)
assert_allclose(d_svd, d_eigen)
assert_allclose(u_svd, u_eigen) |
Author
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Although, we can use the align the eigenvectors based on the "maximum": from scipy import linalg
import numpy as np
from numpy.random import default_rng
from numpy.testing import assert_allclose
def my_svd(X):
u, d = linalg.svd(X, full_matrices=False)[:2]
max_abs_cols = np.argmax(np.abs(u), axis=0)
signs = np.sign(u[max_abs_cols, range(u.shape[1])])
u *= signs
return u, d
def my_eigh(X):
d, u = linalg.eigh(X.T.dot(X))
sort_indicies = np.argsort(d)[::-1]
d = d[sort_indicies]
u = u[:, sort_indicies]
max_abs_cols = np.argmax(np.abs(u), axis=0)
signs = np.sign(u[max_abs_cols, range(u.shape[1])])
u *= signs
return u, np.sqrt(d)
rng = default_rng()
for i in range(10):
m, n = 9, 6
X = rng.standard_normal((m, n))
u_svd, d_svd = my_svd(X.T)
u_eigen, d_eigen = my_eigh(X)
assert_allclose(d_svd, d_eigen)
assert_allclose(u_svd, u_eigen) |
thomasjpfan
commented
Mar 9, 2022
| u, d = linalg.svd(XT, full_matrices=False, check_finite=False)[:2] | ||
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| # Give consistent eigenvectors for both svd solvers | ||
| u *= np.sign(u[0]) |
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The alternative solution is:
max_abs_cols = np.argmax(np.abs(u), axis=0)
signs = np.sign(u[max_abs_cols, range(u.shape[1])])
u *= signsIf we want to be strictly the same as svd_flip.
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Here is a quick fix to make the eigenvectors consistent.