Use scipy.special.expit for the inverse of the logit function #12914
Description
In a few places in the code the expit function is manually implemented. For instance,
sklearn/linear_model/base.py:_predict_proba_lrprob *= -1 np.exp(prob, prob) prob += 1 np.reciprocal(prob, prob)
- `sklearn/discriminant_analysis.py:predict_proba:L521
-
sklearn/utils/tests/test_extmath.py 447: return np.log(1 / (1 + np.exp(-x))) - (maybe other places I missed)
It might be better to use scipy.special.expit instead which is simpler and also appears to be faster,
In [3]: from scipy.special import expit
In [4]: import numpy as np
In [14]: y0 = np.random.rand((1000))
In [15]: %%timeit
...:
...: y = y0.copy()
...: y *= -1
...: np.exp(y, y)
...: y += 1
...: np.reciprocal(y, y)
...:
...:
24.8 µs ± 123 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [16]: %timeit y0.copy()
544 ns ± 2.05 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [17]: %timeit expit(y)
18 µs ± 16.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
@ZaydH would you be interesting in taking this, following your PR #12909?