pandas rolling_std和np.std在数组窗口上的区别

Question

我有一些关于pandas.stats.moments的rolling_std函数的问题.奇怪的是,与应用于数组滚动窗口的numpy.std函数相比,使用此功能得到的结果不同.

这是重现此错误的代码:

# import the modules
import numpy as np
import pandas as pd

# define timeseries and sliding window size
timeseries = np.arange(10)
periods = 4

# output of different results
pd.stats.moments.rolling_std(timeseries, periods)
[np.std(timeseries[max(i-periods+1,0):i+1]) for i in np.arange(10)]

产量:

#pandas
array([        nan,         nan,         nan,  1.29099445,  1.29099445,
    1.29099445,  1.29099445,  1.29099445,  1.29099445,  1.29099445])
#numpy
[0.0, 0.5, 0.81649658092772603, 1.1180339887498949, 1.1180339887498949, 1.1180339887498949, 1.1180339887498949, 1.1180339887498949, 1.1180339887498949, 1.1180339887498949]

如果我手工计算这个结果似乎是正确的.有没有人遇到这个或有解释？

Answer 1

Pandas' rolling_std使用默认的delta自由度计算ddof,等于1,在该方面更像R.虽然numpy的std的默认ddof为0.在指定时ddof=1,您将获得相同的结果np.std

>>> [np.std(timeseries[max(i-periods+1,0):i+1], ddof=1) for i in np.arange(10)]
[nan, 0.70710678118654757, 1.0, 1.2909944487358056, 1.2909944487358056, 1.2909944487358056, 1.2909944487358056, 1.29099444873580
56, 1.2909944487358056, 1.2909944487358056]

或ddof=0为rolling_std:

>>> pd.stats.moments.rolling_std(timeseries, periods, ddof=0)
array([        nan,         nan,         nan,  1.11803399,  1.11803399,
        1.11803399,  1.11803399,  1.11803399,  1.11803399,  1.11803399])