randomized_svd flip sign according to v and not u when transpose=True #5608
Description
From the randomized_svd doc:
flip_sign: boolean, (True by default)
The output of a singular value decomposition is only unique up to a
permutation of the signs of the singular vectors. If `flip_sign` is
set to `True`, the sign ambiguity is resolved by making the largest
loadings for each component in the left singular vectors positive.
Looking at svd_flip the sign flip is done by column so that the maximum of the column is ensured to be positive (u_based_decision is True by default).
Here is an example that shows that it is not the case when transpose=True (I guess in this case the svd_flip is according to v and not u):
import numpy as np
import sklearn.utils.extmath as extmath
mat = np.arange(10 * 8).reshape(10, -1)
def max_loading_is_positive(u, v):
return {
'u_based': (np.abs(u).max(axis=0) == u.max(axis=0)).all(),
'v_based': (np.abs(v).max(axis=1) == v.max(axis=1)).all()
}
u_flipped, _, v_flipped = extmath.randomized_svd(mat, 3, flip_sign=True)
u_flipped_with_transpose, _, v_flipped_with_transpose = extmath.randomized_svd(
mat, 3, flip_sign=True, transpose=True)
print('flipped:', max_loading_is_positive(u_flipped, v_flipped))
print('flipped_with_transpose:',
max_loading_is_positive(
u_flipped_with_transpose, v_flipped_with_transpose))Output:
flipped: {'u_based': True, 'v_based': False}
flipped_with_transpose: {'u_based': False, 'v_based': True}