在阅读How Not to Sort by Average Rating之后，我很好奇是否有人对伯努利参数的Wilson分数置信区间的下限进行了Python实现？
答案 0 :(得分：29)
Reddit使用Wilson评分间隔进行评论排名，可以找到解释和python实现here
#Rewritten code from /r2/r2/lib/db/_sorts.pyx
from math import sqrt
def confidence(ups, downs):
n = ups + downs
if n == 0:
return 0
z = 1.0 #1.44 = 85%, 1.96 = 95%
phat = float(ups) / n
return ((phat + z*z/(2*n) - z * sqrt((phat*(1-phat)+z*z/(4*n))/n))/(1+z*z/n))
答案 1 :(得分：18)
我认为这个有一个错误的威尔逊电话，因为如果你有1个向下0，你会得到 NaN ，因为你不能对负值做sqrt。
在查看文章How not to sort by average page中的ruby示例时，可以找到正确的一个：
return ((phat + z*z/(2*n) - z * sqrt((phat*(1-phat)+z*z/(4*n))/n))/(1+z*z/n))
答案 2 :(得分：6)
要获得没有连续性更正的Wilson CI，您可以在proportion_confint中使用statsmodels.stats.proportion。要使Wilson CI具有连续性校正，您可以使用下面的代码。
# cf.
# [1] R. G. Newcombe. Two-sided confidence intervals for the single proportion, 1998
# [2] R. G. Newcombe. Interval Estimation for the difference between independent proportions: comparison of eleven methods, 1998
import numpy as np
from statsmodels.stats.proportion import proportion_confint
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #
def propci_wilson_cc(count, nobs, alpha=0.05):
# get confidence limits for proportion
# using wilson score method w/ cont correction
# i.e. Method 4 in Newcombe [1];
# verified via Table 1
from scipy import stats
n = nobs
p = count/n
q = 1.-p
z = stats.norm.isf(alpha / 2.)
z2 = z**2
denom = 2*(n+z2)
num = 2.*n*p+z2-1.-z*np.sqrt(z2-2-1./n+4*p*(n*q+1))
ci_l = num/denom
num = 2.*n*p+z2+1.+z*np.sqrt(z2+2-1./n+4*p*(n*q-1))
ci_u = num/denom
if p == 0:
ci_l = 0.
elif p == 1:
ci_u = 1.
return ci_l, ci_u
def dpropci_wilson_nocc(a,m,b,n,alpha=0.05):
# get confidence limits for difference in proportions
# a/m - b/n
# using wilson score method WITHOUT cont correction
# i.e. Method 10 in Newcombe [2]
# verified via Table II
theta = a/m - b/n
l1, u1 = proportion_confint(count=a, nobs=m, alpha=0.05, method='wilson')
l2, u2 = proportion_confint(count=b, nobs=n, alpha=0.05, method='wilson')
ci_u = theta + np.sqrt((a/m-u1)**2+(b/n-l2)**2)
ci_l = theta - np.sqrt((a/m-l1)**2+(b/n-u2)**2)
return ci_l, ci_u
def dpropci_wilson_cc(a,m,b,n,alpha=0.05):
# get confidence limits for difference in proportions
# a/m - b/n
# using wilson score method w/ cont correction
# i.e. Method 11 in Newcombe [2]
# verified via Table II
theta = a/m - b/n
l1, u1 = propci_wilson_cc(count=a, nobs=m, alpha=alpha)
l2, u2 = propci_wilson_cc(count=b, nobs=n, alpha=alpha)
ci_u = theta + np.sqrt((a/m-u1)**2+(b/n-l2)**2)
ci_l = theta - np.sqrt((a/m-l1)**2+(b/n-u2)**2)
return ci_l, ci_u
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #
# single proportion testing
# these come from Newcombe [1] (Table 1)
a_vec = np.array([81, 15, 0, 1])
m_vec = np.array([263, 148, 20, 29])
for (a,m) in zip(a_vec,m_vec):
l1, u1 = proportion_confint(count=a, nobs=m, alpha=0.05, method='wilson')
l2, u2 = propci_wilson_cc(count=a, nobs=m, alpha=0.05)
print(a,m,l1,u1,' ',l2,u2)
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #
# difference in proportions testing
# these come from Newcombe [2] (Table II)
a_vec = np.array([56,9,6,5,0,0,10,10],dtype=float)
m_vec = np.array([70,10,7,56,10,10,10,10],dtype=float)
b_vec = np.array([48,3,2,0,0,0,0,0],dtype=float)
n_vec = np.array([80,10,7,29,20,10,20,10],dtype=float)
print('\nWilson without CC')
for (a,m,b,n) in zip(a_vec,m_vec,b_vec,n_vec):
l, u = dpropci_wilson_nocc(a,m,b,n,alpha=0.05)
print('{:2.0f}/{:2.0f}-{:2.0f}/{:2.0f} ; {:6.4f} ; {:8.4f}, {:8.4f}'.format(a,m,b,n,a/m-b/n,l,u))
print('\nWilson with CC')
for (a,m,b,n) in zip(a_vec,m_vec,b_vec,n_vec):
l, u = dpropci_wilson_cc(a,m,b,n,alpha=0.05)
print('{:2.0f}/{:2.0f}-{:2.0f}/{:2.0f} ; {:6.4f} ; {:8.4f}, {:8.4f}'.format(a,m,b,n,a/m-b/n,l,u))
HTH
答案 3 :(得分：5)
如果您想直接从置信区域计算z并希望避免安装numpy / scipy，可以使用以下代码片段，
import math
def binconf(p, n, c=0.95):
'''
Calculate binomial confidence interval based on the number of positive and
negative events observed. Uses Wilson score and approximations to inverse
of normal cumulative density function.
Parameters
----------
p: int
number of positive events observed
n: int
number of negative events observed
c : optional, [0,1]
confidence percentage. e.g. 0.95 means 95% confident the probability of
success lies between the 2 returned values
Returns
-------
theta_low : float
lower bound on confidence interval
theta_high : float
upper bound on confidence interval
'''
p, n = float(p), float(n)
N = p + n
if N == 0.0: return (0.0, 1.0)
p = p / N
z = normcdfi(1 - 0.5 * (1-c))
a1 = 1.0 / (1.0 + z * z / N)
a2 = p + z * z / (2 * N)
a3 = z * math.sqrt(p * (1-p) / N + z * z / (4 * N * N))
return (a1 * (a2 - a3), a1 * (a2 + a3))
def erfi(x):
"""Approximation to inverse error function"""
a = 0.147 # MAGIC!!!
a1 = math.log(1 - x * x)
a2 = (
2.0 / (math.pi * a)
+ a1 / 2.0
)
return (
sign(x) *
math.sqrt( math.sqrt(a2 * a2 - a1 / a) - a2 )
)
def sign(x):
if x < 0: return -1
if x == 0: return 0
if x > 0: return 1
def normcdfi(p, mu=0.0, sigma2=1.0):
"""Inverse CDF of normal distribution"""
if mu == 0.0 and sigma2 == 1.0:
return math.sqrt(2) * erfi(2 * p - 1)
else:
return mu + math.sqrt(sigma2) * normcdfi(p)
答案 4 :(得分：4)
接受的解决方案似乎使用硬编码的z值（最适合性能）。
如果您希望使用动态z值（基于置信区间）直接python等同于the blogpost的ruby公式：
import math
import scipy.stats as st
def ci_lower_bound(pos, n, confidence):
if n == 0:
return 0
z = st.norm.ppf(1 - (1 - confidence) / 2)
phat = 1.0 * pos / n
return (phat + z * z / (2 * n) - z * math.sqrt((phat * (1 - phat) + z * z / (4 * n)) / n)) / (1 + z * z / n)