赫斯特指数3

Question

我想确定时间序列是否是均值回复，但在计算赫斯特指数时遇到了一些问题。它应该打印0.5-ish，但是我得到了一个“nan”。我们将不胜感激

我收到以下错误/警告：

RuntimeWarning: divide by zero encountered in log
  poly = polyfit(log(lags), log(tau), 1)

下面是我正在编写的代码

import statsmodels.tsa.stattools as ts
from datetime import datetime

from pandas_datareader import DataReader
security = DataReader("GOOG", "yahoo", datetime(2000,1,1), datetime(2013,1,1))
ts.adfuller(security['Adj Close'], 1)



from numpy import cumsum, log, polyfit, sqrt, std, subtract
from numpy.random import randn

def hurst(ts):
    """Returns the Hurst Exponent of the time series vector ts"""

    lags = range(2, 100)

    tau = [sqrt(std(subtract(ts[lag:], ts[:-lag]))) for lag in lags]

    poly = polyfit(log(lags), log(tau), 1)


    return poly[0]*2.0


gbm = log(cumsum(randn(100000))+1000)
mr = log(randn(100000)+1000)
tr = log(cumsum(randn(100000)+1)+1000)

print ("Hurst(GBM):   %s" % hurst(gbm))
print ("Hurst(MR):    %s" % hurst(mr))
print ("Hurst(TR):    %s" % hurst(tr))
print ("Hurst(SECURITY):  %s" % hurst(security['Adj Close']))



print ("Hurst(GBM):   %s" % hurst(gbm))
print ("Hurst(MR):    %s" % hurst(mr))
print ("Hurst(TR):    %s" % hurst(tr))
print ("Hurst(SECURITY):  %s" % hurst(security['Adj Close']))
Hurst(GBM):   0.5039604262314196
Hurst(MR):    -2.3832407841923795e-05
Hurst(TR):    0.962521148986032
Hurst(SECURITY):  nan
__main__:11: RuntimeWarning: divide by zero encountered in log

Answer 1

我在发送序列时遇到了与ts参数相同的问题。 您只需发送一个列表，而不是一个系列或：

def hurst(ts):
    """Returns the Hurst Exponent of the time series vector ts"""
    ts = ts if not isinstance(ts, pd.Series) else ts.to_list()
    lags = range(2, 100)
    tau = [sqrt(std(subtract(ts[lag:], ts[:-lag]))) for lag in lags]
    poly = polyfit(log(lags), log(tau), 1)
    return poly[0]*2.0

NaN值也可能是一个问题，我会先检查dropna（）是否可以，然后再查看_list（）

Answer 2

根本原因是Series[<slice>]语法为每个片返回相应的索引，-运算符处理每个索引相等（而不是实际位置）

例如：

s = pd.Series(range(5))
s[2:] - s[:-2]
=>
0    NaN
1    NaN
2    0.0
3    NaN
4    NaN
dtype: float64

显然，这不是我们所期望的。看看为什么我们可以使用concat分别创建s[2:], s[:-2]的逐行数据帧

pd.concat([s[2:], s[:-2]], axis=1)
=>
    0   1
0   NaN 0.0
1   NaN 1.0
2   2.0 2.0
3   3.0 NaN
4   4.0 NaN

给定此输入，hurst函数中的tau = 方程的结果是（大部分）nan值的列表

本机使用Series的解决方案是使用Series.shift()而不是数组切片：

def hurst(ts):
  ... 

  # Calculate the array of the variances of the lagged differences
  tau = [sqrt((ts - ts.shift(-lag)).std()) for lag in lags]

  ...

或者，将Series.values传递给原始函数，该函数传递numpy数组