使用Python估计自相关

Question

我想对下面显示的信号执行自相关.两个连续点之间的时间是2.5ms(或400Hz的重复率).

这是我想要使用的估计自相关的等式(取自http://en.wikipedia.org/wiki/Autocorrelation,部分估计):

在python中查找我的数据估计自相关的最简单方法是什么？有什么类似于numpy.correlate我可以使用的东西吗？

或者我应该只计算均值和方差？

---

编辑:

在unutbu的帮助下,我写道:

from numpy import *
import numpy as N
import pylab as P

fn = 'data.txt'
x = loadtxt(fn,unpack=True,usecols=[1])
time = loadtxt(fn,unpack=True,usecols=[0]) 

def estimated_autocorrelation(x):
    n = len(x)
    variance = x.var()
    x = x-x.mean()
    r = N.correlate(x, x, mode = 'full')[-n:]
    #assert N.allclose(r, N.array([(x[:n-k]*x[-(n-k):]).sum() for k in range(n)]))
    result = r/(variance*(N.arange(n, 0, -1)))
    return result

P.plot(time,estimated_autocorrelation(x))
P.xlabel('time (s)')
P.ylabel('autocorrelation')
P.show()

Answer 1

我不认为这个特定计算有NumPy函数.我是这样写的:

def estimated_autocorrelation(x):
    """
    http://stackoverflow.com/q/14297012/190597
    http://en.wikipedia.org/wiki/Autocorrelation#Estimation
    """
    n = len(x)
    variance = x.var()
    x = x-x.mean()
    r = np.correlate(x, x, mode = 'full')[-n:]
    assert np.allclose(r, np.array([(x[:n-k]*x[-(n-k):]).sum() for k in range(n)]))
    result = r/(variance*(np.arange(n, 0, -1)))
    return result

assert语句用于检查计算并记录其意图.

当您确信此函数的行为符合预期时,您可以注释掉该assert语句,或者运行您的脚本python -O.(该-O标志告诉Python忽略断言语句.)

Answer 2

我从pandas autocorrelation_plot()函数中获取了一部分代码.我用R检查了答案,并且值完全匹配.

import numpy
def acf(series):
    n = len(series)
    data = numpy.asarray(series)
    mean = numpy.mean(data)
    c0 = numpy.sum((data - mean) ** 2) / float(n)

    def r(h):
        acf_lag = ((data[:n - h] - mean) * (data[h:] - mean)).sum() / float(n) / c0
        return round(acf_lag, 3)
    x = numpy.arange(n) # Avoiding lag 0 calculation
    acf_coeffs = map(r, x)
    return acf_coeffs

Answer 3

statsmodels包添加了一个内部使用的自相关函数np.correlate(根据statsmodels文档).

请参阅:http: //statsmodels.sourceforge.net/stable/generated/statsmodels.tsa.stattools.acf.html#statsmodels.tsa.stattools.acf

Answer 4

我写我的最新编辑的方法是比现在还要快scipy.statstools.acf用fft=True,直到试样尺寸变得非常大.

错误分析如果您想调整偏差并获得高度准确的错误估计:请查看我的代码,该代码由Ulli Wolff (或UW原创)实施本文Matlab

功能测试

• a = correlatedData(n=10000)来自这里的常规
• gamma() 来自同一个地方 correlated_data()
• acorr() 是我的功能如下
• estimated_autocorrelation 在另一个答案中找到
• acf() 来自 from statsmodels.tsa.stattools import acf

计时

%timeit a0, junk, junk = gamma(a, f=0)                            # puwr.py
%timeit a1 = [acorr(a, m, i) for i in range(l)]                   # my own
%timeit a2 = acf(a)                                               # statstools
%timeit a3 = estimated_autocorrelation(a)                         # numpy
%timeit a4 = acf(a, fft=True)                                     # stats FFT

## -- End pasted text --
100 loops, best of 3: 7.18 ms per loop
100 loops, best of 3: 2.15 ms per loop
10 loops, best of 3: 88.3 ms per loop
10 loops, best of 3: 87.6 ms per loop
100 loops, best of 3: 3.33 ms per loop

编辑...我再次检查保持l=40并更改n=10000为n=200000样本FFT方法开始获得一点牵引力和statsmodelsfft实现只是边缘...(顺序是相同的)

## -- End pasted text --
10 loops, best of 3: 86.2 ms per loop
10 loops, best of 3: 69.5 ms per loop
1 loops, best of 3: 16.2 s per loop
1 loops, best of 3: 16.3 s per loop
10 loops, best of 3: 52.3 ms per loop

编辑2:我改变了我的例程并重新测试了FFT n=10000和forn=20000

a = correlatedData(n=200000); b=correlatedData(n=10000)
m = a.mean(); rng = np.arange(40); mb = b.mean()
%timeit a1 = map(lambda t:acorr(a, m, t), rng)
%timeit a1 = map(lambda t:acorr.acorr(b, mb, t), rng)
%timeit a4 = acf(a, fft=True)
%timeit a4 = acf(b, fft=True)

10 loops, best of 3: 73.3 ms per loop   # acorr below
100 loops, best of 3: 2.37 ms per loop  # acorr below
10 loops, best of 3: 79.2 ms per loop   # statstools with FFT
100 loops, best of 3: 2.69 ms per loop # statstools with FFT

履行

def acorr(op_samples, mean, separation, norm = 1):
    """autocorrelation of a measured operator with optional normalisation
    the autocorrelation is measured over the 0th axis

    Required Inputs
        op_samples  :: np.ndarray :: the operator samples
        mean        :: float :: the mean of the operator
        separation  :: int :: the separation between HMC steps
        norm        :: float :: the autocorrelation with separation=0
    """
    return ((op_samples[:op_samples.size-separation] - mean)*(op_samples[separation:]- mean)).ravel().mean() / norm

4x加速可以在下面实现.你必须小心,只传递op_samples=a.copy(),因为它会修改数组a通过a-=mean其他方式:

op_samples -= mean
return (op_samples[:op_samples.size-separation]*op_samples[separation:]).ravel().mean() / norm

完整性检查

示例错误分析

这有点超出了范围,但是如果没有集成的自相关时间或积分窗口计算,我就不会为重做这个数字而烦恼.底部图中清楚地显示了具有错误的自相关