不能重现python pandas/scipy中的偏度测量结果

Question

我正在尝试从scratch编写偏度度量计算。但不能匹配来自pandas attribute/ scipy.stats函数的值。

我已经看过了scipy.stats here的源代码。但我找不到我遗漏了什么。

from math import sqrt
from scipy import stats
import pandas as pd
  
def mean(values):
    return sum(values) / len(values)

def standard_dev(values): 
    vals_mean = mean(values)
    numerator = 0
    for val in values:
        numerator += (val - vals_mean) ** 2 
    return sqrt(numerator/len(values))

def skewness(values): 
    n = len(values)
    vals_mean = mean(values)
    thrid_moment = 0
    for val in values: 
        thrid_moment += (val - vals_mean)**3
    return (sqrt(n*(n-1))/ (n-2)) * (thrid_moment / standard_dev(values) ** 3)
  
values = [1,1,1,2,2,3,3,3,4,4,5] 
  
print(f'mean{mean(values)}')
# mean2.6363636363636362

print(f'standard_dev{standard_dev(values)}')
# standard_dev1.2984415324623364

print(f'skewness{skewness(values)}') 
# skewness2.5341000098031734

a = pd.Series(values)
a.std(ddof=0)
# 1.2984415324623364
a.skew()
# 0.23037272816392504

stats.skew(a, bias=False)
# 0.230372728163925
stats.skew(a, bias=True)
# 0.19768660009807223

Answer 1

我发现了错误。我没有对三阶矩的值进行归一化，也就是说没有除以len(values)。

下面是完整的版本：

from math import sqrt 
  
def mean(values):
    return sum(values) / len(values)

def moments(values, moment):
    vals_mean = mean(values)
    numerator = 0
    for val in values:
        numerator += (val - vals_mean) ** moment 
    return numerator/len(values)

def skewness(values): 
    n = len(values)
    m2 = moments(values, 2)
    m3 = moments(values, 3)
    return (sqrt(n*(n-1))/ (n-2)) * (m3 / m2 ** 1.5)
  
values = [1,1,1,2,2, 3,3,3, 4,4, 5] 
  
print(f'mean: {mean(values)}')
print(f'skewness: {skewness(values)}') 
# mean: 2.6363636363636362
# skewness: 0.23037272816392482