Hurst 指数变为 nan - Python 3

Question

我想确定时间序列是否均值回归，但我运行在计算 Hurst 指数时遇到了一些问题。它应该打印 0.5-ish，但我得到的是 "nan"。我们将不胜感激。

我得到以下 error/warning:

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

下面是我正在处理的代码。

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 Series 作为 ts 参数时遇到了同样的问题。您所要做的就是发送列表而不是系列或：

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 值也可能是个问题，我会在 to_list()

之前检查 dropna() 是否可以

Answer 2

根本原因是 Series[<slice>] 语法 returns 每个切片的相应索引，而 - 运算符在每个索引相等（而不是实际位置）上工作。

示例：

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] 的逐行数据框。

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 tau = 方程的结果是（大部分）nan 值的列表。

原生使用 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，该函数传递一个 numpy 数组。

Hurst 指数变为 nan - Python 3

Hurst Exponent turns nan - Python 3

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