Wilson分数区间的Python实现？

在阅读之后，我很好奇是否有人有一个Python实现的Wilson分数置信区间下限用于Bernoulli参数？

Reddit使用Wilson分数区间进行评论排名，可以找到一个解释和Python实现

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我认为这是一个错误的wilson调用，因为如果你有1向上0向下，你会得到NaN，因为你不能对负值进行

sqrt

在阅读本文中的ruby示例时，可以找到正确的示例：

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如果您希望直接从置信区间计算z，并且希望避免安装numpy/scipy，那么可以使用以下代码片段：

导入数学
def binconf（p，n，c=0.95）：
'''
根据正数和负数计算二项式置信区间
观察到负面事件。使用Wilson分数和近似值进行反演
正态累积密度函数。
参数
----------
p:int
观察到的积极事件数
n:int
观察到的负面事件数
c:可选[0,1]
信心百分比。e、 g.0.95表示95%的置信概率
成功介于两个返回值之间
退换商品
-------
θu低：浮动
置信区间的下界
θ_高：浮动
置信区间上界
'''
p、 n=浮动（p），浮动（n）
N=p+N
如果N==0.0：返回（0.0，1.0）
p=p/N
z=标准CDFI（1-0.5*（1-c））
a1=1.0/（1.0+z*z/N）
a2=p+z*z/（2*N）
a3=z*数学sqrt（p*（1-p）/N+z*z/（4*N*N））
报税表（a1*（a2-a3），a1*（a2+a3））
def erfi（x）：
“”“逆误差函数的近似值”“”
a=0.147#神奇！！！
a1=数学日志（1-x*x）
a2=(
2.0/（数学pi*a）
+a1/2.0
)
返回(
符号（x）*
数学sqrt（数学sqrt（a2*a2-a1/a）-a2）
)
def标志（x）：
如果x<0：返回-1
如果x==0：返回0
如果x>0：返回1
def normcdfi（p，mu=0.0，sigma2=1.0）：
“正态分布的逆CDF”
如果mu==0.0和sigma2==1.0：
返回数学sqrt（2）*erfi（2*p-1）
其他：
返回mu+math.sqrt（sigma2）*normcdfi（p）

要在不进行连续性校正的情况下获得Wilson CI，您可以在

statsmodels.stats.proportion

中使用

proportion\u confint

。要获得具有连续性校正的Wilson CI，可以使用以下代码

# 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))

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HTH

公认的解决方案似乎使用了硬编码的z值（最适合性能）

如果您希望从中获得ruby公式的直接python等价物和动态z值（基于置信区间）：

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如果你只是想发布一个链接，可以在评论中发布。如果您将其作为答案发布，请从内容中提供更多信息和/或拉出代码，以便不是每个人都需要关注该链接，并且即使链接失效，该答案也有价值。应更正此答案，以包括以下修改@Vladtn我刚刚用Gullevek的答案更新了它。如果还有其他问题，请告诉我。我只想补充一点，对于95%的置信区间，z分数应该是1.96，而不是1.6。@Wesley是的，我相信

1.0=85%

也是错误的，我已经更新了答案。。。如果n*p-cap*（1-p-cap）低于某个阈值，比如30-35，这里有一个更精确的值表，我会使用带df：（pos+neg）-2的t分布，而不是正态分布。不过只有我的两分钱

return ((phat + z*z/(2*n) - z * sqrt((phat*(1-phat)+z*z/(4*n))/n))/(1+z*z/n))

# 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))

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)