def test_log1mexp():
vals = np.array([-1, 0, 1e-20, 1e-4, 10, 100, 1e20])
vals_ = vals.copy()
# import mpmath
# mpmath.mp.dps = 1000
# [float(mpmath.log(1 - mpmath.exp(-x))) for x in vals]
expected = np.array([
np.nan,
-np.inf,
-46.051701859880914,
-9.210390371559516,
-4.540096037048921e-05,
-3.720075976020836e-44,
0.0,
])
actual = at.log1mexp(-vals).eval()
npt.assert_allclose(actual, expected)
actual_ = log1mexp_numpy(-vals, negative_input=True)
npt.assert_allclose(actual_, expected)
# Check that input was not changed in place
npt.assert_allclose(vals, vals_)
def log1mexp(x, *, negative_input=False):
r"""Return log(1 - exp(-x)).
This function is numerically more stable than the naive approach.
For details, see
https://cran.r-project.org/web/packages/Rmpfr/vignettes/log1mexp-note.pdf
References
----------
.. [Machler2012] Martin Mächler (2012).
"Accurately computing `\log(1-\exp(- \mid a \mid))` Assessed by the Rmpfr package"
"""
if not negative_input:
warnings.warn(
"pymc.math.log1mexp will expect a negative input in a future "
"version of PyMC.\n To suppress this warning set `negative_input=True`",
FutureWarning,
stacklevel=2,
)
x = -x
return at.log1mexp(x)
def logdiffexp(a, b):
"""log(exp(a) - exp(b))"""
return a + at.log1mexp(b - a)
def test_log1mexp_numpy_integer_input():
assert np.isclose(log1mexp_numpy(-2, negative_input=True),
at.log1mexp(-2).eval())