Python implementation of the Wilson Score Interval?

Question

After reading How Not to Sort by Average Rating, I was curious if anyone has a Python implementation of a Lower bound of Wilson score confidence interval for a Bernoulli parameter?

Answer 1

Reddit uses the Wilson score interval for comment ranking, an explanation and python implementation can be found 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))

Answer 2

I think this one has a wrong wilson call, because if you have 1 up 0 down you get NaN because you can't do a sqrt on the negative value.

The correct one can be found when looking at the ruby example from the article How not to sort by average page:

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

Answer 3

To get the Wilson CI without continuity correction, you can use proportion_confint in statsmodels.stats.proportion. To get the Wilson CI with continuity correction, you can use the code below.

# 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

Answer 4

The accepted solution seems to use a hard-coded z-value (best for performance).

In the event that you wanted a direct python equivalent of the ruby formula from the blogpost with a dynamic z-value (based on the confidence interval):

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

If you'd like to actually calculate z directly from a confidence bound and want to avoid installing numpy/scipy, you can use the following snippet of code,

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)

Answer 6

There is a recent Scipy implementation: scipy.stats._result_classes.BinomTestResult.proportion_ci .

Answer 7

Here is a simplified (no need for numpy) and slightly improved (0 and n values for k do not cause a math domain error) version of the Wilson score confidence interval with continuity correction, from the original sourcecode written by batesbatesbates in another answer, and also a pure python no-numpy non-continuity correction version, with 2 equivalent ways to calculate (can be switched with eqmode argument, but both ways give the exact same non-continuity correction results):

import math

def propci_wilson_nocc(k, n, z=1.96, eqmode=0):
    # Calculates the Binomial Proportion Confidence Interval using the Wilson Score method without continuation correction
    # Equations eqmode == 1 from: https://en.wikipedia.org/w/index.php?title=Binomial_proportion_confidence_interval&oldid=1101942017#Wilson_score_interval
    # Equations eqmode == 0 from: https://www.evanmiller.org/how-not-to-sort-by-average-rating.html
    # The results should be close to:
    #    from statsmodels.stats.proportion import proportion_confint
    #    proportion_confint(k, n, alpha=0.05, method='wilson')
    #z=1.44 = 85%, 1.96 = 95%
    if n == 0:
        return 0
    p_hat = float(k) / n
    z2 = z**2
    if eqmode == 0:
        ci_l = (p_hat + z2/(2*n) - z*math.sqrt(max(0.0, (p_hat*(1 - p_hat) + z2/(4*n))/n))) / (1 + z2 / n)
    else:
        ci_l = (1.0 / (1.0 + z2/n)) * (p_hat + z2/(2*n)) - (z / (1 + z2/n)) * math.sqrt(max(0.0, (p_hat*(1 - p_hat)/n + z2/(4*(n**2)))))
    if eqmode == 0:
        ci_u = (p_hat + z2/(2*n) + z*math.sqrt(max(0.0, (p_hat*(1 - p_hat) + z2/(4*n))/n))) / (1 + z2 / n)
    else:
        ci_u = (1.0 / (1.0 + z2/n)) * (p_hat + z2/(2*n)) + (z / (1 + z2/n)) * math.sqrt(max(0.0, (p_hat*(1 - p_hat)/n + z2/(4*(n**2)))))
    return [ci_l, ci_u]

def propci_wilson_cc(n, k, z=1.96):
    # Calculates the Binomial Proportion Confidence Interval using the Wilson Score method with continuation correction
    # i.e. Method 4 in Newcombe [1]: R. G. Newcombe. Two-sided confidence intervals for the single proportion, 1998;
    # verified via Table 1
    # originally written by batesbatesbates https://stackoverflow.com/questions/10029588/python-implementation-of-the-wilson-score-interval/74021634#74021634
    p_hat = k/n
    q = 1.0-p
    z2 = z**2
    denom = 2*(n+z2)
    num = 2.0*n*p_hat + z2 - 1.0 - z*math.sqrt(max(0.0, z2 - 2 - 1.0/n + 4*p_hat*(n*q + 1)))
    ci_l = num/denom
    num2 = 2.0*n*p_hat + z2 + 1.0 + z*math.sqrt(max(0.0, z2 + 2 - 1.0/n + 4*p_hat*(n*q - 1)))
    ci_u = num2/denom
    if p_hat == 0:
        ci_l = 0.0
    elif p_hat == 1:
        ci_u = 1.0
    return [ci_l, ci_u]

Note that the returned value will always be bounded between [0.0, 1.0] (due to how p_hat is a ratio of k/n), this is why it's a score and not really a confidence interval, but it's easy to project back to a confidence interval by multiplying ci_l * n and ci_u * n, these values will be in the same domain as k and can be plotted alongside.

Answer 8

Here is a much more readable version for how to compute the Wilson Score interval without continuity correction, by Bartosz Mikulski:

from math import sqrt
def wilson(p, n, z = 1.96):
    denominator = 1 + z**2/n
    centre_adjusted_probability = p + z*z / (2*n)
    adjusted_standard_deviation = sqrt((p*(1 - p) + z*z / (4*n)) / n)
    
    lower_bound = (centre_adjusted_probability - z*adjusted_standard_deviation) / denominator
    upper_bound = (centre_adjusted_probability + z*adjusted_standard_deviation) / denominator
    return (lower_bound, upper_bound)