如何计算Python中的自协方差

Question

我想计算3个数组X1，X2和Y的自协方差，它们都是平稳的随机过程。 sciPy或其他库中是否有任何函数可以解决这个问题？

Answer 1

Statsmodels具有自相关和互协方差函数

http://statsmodels.sourceforge.net/devel/generated/statsmodels.tsa.stattools.acovf.html http://statsmodels.sourceforge.net/devel/generated/statsmodels.tsa.stattools.ccovf.html

加的相关函数和部分自相关 http://statsmodels.sourceforge.net/devel/tsa.html#descriptive-statistics-and-tests

Answer 2

根据自协方差系数为离散信号的标准估计，这可以由等式表示：

...其中x(i)是一个给定的信号（即特定1D向量），k代表x(i)信号由k样本移位，N是x(i)信号的长度，并且：

...这是简单的平均，我们可以这样写：

''' 
Calculate the autocovarriance coefficient. 
''' 

import numpy as np 

Xi = np.array([1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5]) 
N = np.size(Xi) 
k = 5 
Xs = np.average(Xi) 

def autocovariance(Xi, N, k, Xs): 
    autoCov = 0 
    for i in np.arange(0, N-k): 
     autoCov += ((Xi[i+k])-Xs)*(Xi[i]-Xs) 
    return (1/(N-1))*autoCov 

print("Autocovariance:", autocovariance(Xi, N, k, Xs)) 

如果你想标准化自协方差系数，这将成为自相关系数表示为：

...比你只需要添加到上面的代码只是两个额外的线路：

def autocorrelation(): 
    return autocovariance(Xi, N, k, Xs)/autocovariance(Xi, N, 0, Xs) 

这里完整的脚本：

''' 
Calculate the autocovarriance and autocorrelation coefficients. 
''' 

import numpy as np 

Xi = np.array([1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5]) 
N = np.size(Xi) 
k = 5 
Xs = np.average(Xi) 

def autocovariance(Xi, N, k, Xs): 
    autoCov = 0 
    for i in np.arange(0, N-k): 
     autoCov += ((Xi[i+k])-Xs)*(Xi[i]-Xs) 
    return (1/(N-1))*autoCov 

def autocorrelation(): 
    return autocovariance(Xi, N, k, Xs)/autocovariance(Xi, N, 0, Xs) 

print("Autocovariance:", autocovariance(Xi, N, k, Xs)) 
print("Autocorrelation:", autocorrelation()) 

Answer 3

获取样本自协方差：

# cov_auto_samp(X,delta)/cov_auto_samp(X,0) = auto correlation 
def cov_auto_samp(X,delta): 
    N = len(X) 
    Xs = np.average(X) 
    autoCov = 0.0 
    times = 0.0 
    for i in np.arange(0, N-delta): 
     autoCov += (X[i+delta]-Xs)*(X[i]-Xs) 
     times +=1 
    return autoCov/times 

Answer 4

对以前的答案进行了小小的调整，避免了python for循环，而是使用numpy数组操作。如果你有很多数据，这会更快。

def lagged_auto_cov(Xi,t): 
    """ 
    for series of values x_i, length N, compute empirical auto-cov with lag t 
    defined: 1/(N-1) * \sum_{i=0}^{N-t} (x_i - x_s) * (x_{i+t} - x_s) 
    """ 
    N = len(time_series) 

    # use sample mean estimate from whole series 
    Xs = np.mean(Xi) 

    # construct copies of series shifted relative to each other, 
    # with mean subtracted from values 
    end_padded_series = np.zeros(N+t) 
    end_padded_series[:N] = Xi - Xs 
    start_padded_series = np.zeros(N+t) 
    start_padded_series[t:] = Xi - Xs 

    auto_cov = 1./(N-1) * np.sum(start_padded_series*end_padded_series) 
    return auto_cov 

此针对@bluevoxel的代码，用一个时间序列的50000个数据点，计算用于滞后的单一固定值的自相关相比，蟒for循环代码平均约为30毫秒和使用numpy阵列的平均速度超过0.3毫秒（运行在我的笔记本电脑上）。