def test_log1p(self):
"""log1p should give same results as cephes"""
p_s = [
1e-10,
1e-5,
0.1,
0.8,
0.9,
0.95,
0.999,
0.9999999,
1,
1.000000001,
1.01,
2,
]
exp = [
9.9999999995e-11,
9.99995000033e-06,
0.0953101798043,
0.587786664902,
0.641853886172,
0.667829372576,
0.692647055518,
0.69314713056,
0.69314718056,
0.69314718106,
0.698134722071,
1.09861228867,
]
for p, e in zip(p_s, exp):
self.assertFloatEqual(log1p(p), e)
def bdtri(k, n, y):
"""Inverse of binomial distribution.
Finds binomial p such that sum of terms 0-k reaches cum probability y.
"""
y = fix_rounding_error(y)
if y < 0.0 or y > 1.0:
raise ZeroDivisionError("y must be between 1 and 0.")
if k < 0 or n <= k:
raise ZeroDivisionError("k must be between 0 and n")
dn = n - k
if k == 0:
if y > 0.8:
p = -expm1(log1p(y - 1.0) / dn)
else:
p = 1.0 - y**(1.0 / dn)
else:
dk = k + 1
p = incbet(dn, dk, 0.5)
if p > 0.5:
p = incbi(dk, dn, 1.0 - y)
else:
p = 1.0 - incbi(dn, dk, y)
return p
def bdtrc(k, n, p):
"""Complement of binomial distribution, k+1 through n.
Uses formula bdtrc(k, n, p) = betai(k+1, n-k, p)
See Cephes docs for details.
"""
p = fix_rounding_error(p)
if (p < 0) or (p > 1):
raise ValueError("Binomial p must be between 0 and 1.")
if (k < 0) or (n < k):
raise ValueError("Binomial k must be between 0 and n.")
if k == n:
return 0
dn = n - k
if k == 0:
if p < 0.01:
dk = -expm1(dn * log1p(-p))
else:
dk = 1 - pow(1.0 - p, dn)
else:
dk = k + 1
dk = betai(dk, dn, p)
return dk