如何正确发现熊猫的偏度和峰度？

Question

我想知道如何正确计算熊猫的偏度和峰度。Pandas为skew()和kurtosis()值提供了一些值，但它们似乎与scipy.stats值有很大不同。相信熊猫和scipy.stats哪个好呢？

下面是我的代码：

import numpy as np
import scipy.stats as stats
import pandas as pd

np.random.seed(100)
x = np.random.normal(size=(20))

kurtosis_scipy = stats.kurtosis(x)
kurtosis_pandas = pd.DataFrame(x).kurtosis()[0]

print(kurtosis_scipy, kurtosis_pandas)
# -0.5270409758168872
# -0.31467107631025604

skew_scipy = stats.skew(x)
skew_pandas = pd.DataFrame(x).skew()[0]

print(skew_scipy, skew_pandas)
# -0.41070929017558555
# -0.44478877631598901

版本：

print(np.__version__, pd.__version__, scipy.__version__)
1.11.0 0.20.0 0.19.0

Answer 1

bias=False

print(
    stats.kurtosis(x, bias=False), pd.DataFrame(x).kurtosis()[0],
    stats.skew(x, bias=False), pd.DataFrame(x).skew()[0],
    sep='\n'
)

-0.31467107631025515
-0.31467107631025604
-0.4447887763159889
-0.444788776315989

Answer 2

熊猫计算种群峰度的无偏估计。在维基百科上查找公式：https://www.wikiwand.com/en/Kurtosis

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从头开始计算峰度

import numpy as np
import pandas as pd
import scipy

x = np.array([0, 3, 4, 1, 2, 3, 0, 2, 1, 3, 2, 0,
              2, 2, 3, 2, 5, 2, 3, 999])
xbar = np.mean(x)
n = x.size
k2 = x.var(ddof=1) # default numpy is biased, ddof = 0
sum_term = ((x-xbar)**4).sum()
factor = (n+1) * n / (n-1) / (n-2) / (n-3)
second = - 3 * (n-1) * (n-1) / (n-2) / (n-3)

first = factor * sum_term / k2 / k2

G2 = first + second
G2 # 19.998428728659768

使用numpy/scipy计算峰度

scipy.stats.kurtosis(x,bias=False) # 19.998428728659757

使用pandas计算峰度

pd.DataFrame(x).kurtosis() # 19.998429

同样，您也可以计算偏斜度。