python - 威尔逊分数区间的 Python 实现？

Question

在阅读了 How Not to Sort by Average Rating之后，我很好奇是否有人对 Bernoulli 参数的 Wilson 得分置信区间的下限进行了 Python 实现？

Answer 1

Reddit 使用 Wilson 得分区间进行评论排名，解释和 python 实现可以在这里找到

#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))

Answer 2

我认为这个有一个错误的威尔逊电话，因为如果你有 1 up 0 down 你会得到NaN因为你不能sqrt对负值做 a 。

查看文章如何不按平均页面排序的 ruby​​ 示例时可以找到正确的示例：

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

Answer 3

要获得没有连续性校正的 Wilson CI，您可以使用proportion_confint. statsmodels.stats.proportion要获得具有连续性校正的 Wilson CI，您可以使用以下代码。

# 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))

高温高压

Answer 4

公认的解决方案似乎使用硬编码的 z 值（最适合性能）。

如果您想要博客文章中具有动态 z 值（基于置信区间）的 ruby​​ 公式的直接 python 等效项：

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)

Answer 5

如果您想直接从置信区间实际计算 z 并且希望避免安装 numpy/scipy，您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)