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FIX Make eignvectors consistent #1

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Merged
Micky774 merged 7 commits into from
Mar 13, 2022
Merged

FIX Make eignvectors consistent #1

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66dd707
TST Check solvers
thomasjpfan Mar 9, 2022
e6e9602
TST Create new tests
thomasjpfan Mar 9, 2022
9cdbef9
CLN Slightly better
thomasjpfan Mar 9, 2022
c5ef5a9
TST Adjust seed
thomasjpfan Mar 9, 2022
77422c8
DOC Adds comment
thomasjpfan Mar 9, 2022
deaab6e
FIX Give a random state
thomasjpfan Mar 9, 2022
c077bfe
FIX Give a random state
thomasjpfan Mar 9, 2022
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7 changes: 4 additions & 3 deletions sklearn/decomposition/_fastica.py
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Original file line number Diff line number Diff line change
Expand Up @@ -574,12 +574,13 @@ def g(x, fun_args):
)
d[degenerate_idx] = eps # For numerical issues
d = np.sqrt(d, d)
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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@thomasjpfan thomasjpfan Mar 9, 2022

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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.

else:
u, d = linalg.svd(XT, full_matrices=False, check_finite=False)[:2]

# Give consistent eigenvectors for both svd solvers
u *= np.sign(u[0])
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@thomasjpfan thomasjpfan Mar 9, 2022

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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 *= signs

If we want to be strictly the same as svd_flip.


K = (u / d).T[:n_components] # see (6.33) p.140
del u, d
X1 = np.dot(K, XT)
Expand Down
74 changes: 48 additions & 26 deletions sklearn/decomposition/tests/test_fastica.py
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Expand Up @@ -119,36 +119,24 @@ def g_test(x):
assert_almost_equal(np.dot(s2_, s2) / n_samples, 1, decimal=1)

# Test FastICA class
outs = {}
for solver in ("eigh", "svd"):
_, _, sources_fun = fastica(
m.T, fun=nl, algorithm=algo, random_state=0, svd_solver=solver
)
ica = FastICA(fun=nl, algorithm=algo, random_state=0, svd_solver=solver)
sources = ica.fit_transform(m.T)
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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@thomasjpfan thomasjpfan Mar 9, 2022

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This reverts the tests to what they were on upstream/main.

ica = FastICA(fun=nl, algorithm=algo, random_state=seed)
sources = ica.fit_transform(m.T)
assert ica.components_.shape == (2, 2)
assert sources.shape == (1000, 2)

assert_array_almost_equal(sources_fun, sources)
assert_array_almost_equal(sources, ica.transform(m.T))
assert_array_almost_equal(sources_fun, sources)
assert_array_almost_equal(sources, ica.transform(m.T))

assert ica.mixing_.shape == (2, 2)
assert ica.mixing_.shape == (2, 2)

for fn in [np.tanh, "exp(-.5(x^2))"]:
ica = FastICA(fun=fn, algorithm=algo, svd_solver=solver)
with pytest.raises(ValueError):
ica.fit(m.T)
for fn in [np.tanh, "exp(-.5(x^2))"]:
ica = FastICA(fun=fn, algorithm=algo)
with pytest.raises(ValueError):
ica.fit(m.T)

with pytest.raises(TypeError):
FastICA(fun=range(10), svd_solver=solver).fit(m.T)

# Check equality up to column parity
for A in (outs["eigh"], outs["svd"]):
for c in range(A.shape[1]):
if A[0, c] < 0:
A[:, c] *= -1
assert_array_almost_equal(outs["eigh"], outs["svd"])
with pytest.raises(TypeError):
FastICA(fun=range(10)).fit(m.T)


def test_fastica_nowhiten():
Expand Down Expand Up @@ -399,3 +387,37 @@ def test_fastica_output_shape(whiten, return_X_mean, return_n_iter):
assert len(out) == expected_len
if not whiten:
assert out[0] is None


@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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@thomasjpfan thomasjpfan Mar 9, 2022 •
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This is a new test to check for consistency based on what you had.

"""Test FastICA is consistent between svd_solvers."""
rng = np.random.RandomState(seed)
n_samples = 1000
# Generate two sources:
s1 = (2 * np.sin(np.linspace(0, 100, n_samples)) > 0) - 1
s2 = stats.t.rvs(1, size=n_samples, random_state=rng)
s = np.c_[s1, s2].T
center_and_norm(s)
s1, s2 = s

# Mixing angle
phi = 0.6
mixing = np.array([[np.cos(phi), np.sin(phi)], [np.sin(phi), -np.cos(phi)]])
m = np.dot(mixing, s)

if add_noise:
m += 0.1 * rng.randn(2, 1000)

center_and_norm(m)

outs = {}
for solver in ("svd", "eigh"):
ica = FastICA(random_state=0, svd_solver=solver)
sources = ica.fit_transform(m.T)
outs[solver] = sources
assert ica.components_.shape == (2, 2)
assert sources.shape == (1000, 2)

assert_array_almost_equal(outs["eigh"], outs["svd"])