Question

很可能我现在让自己看起来非常愚蠢，但以下是从fft自相关得到间隔[1，-1]缩放输出的正确方法吗？缩放应该是numpy.correlate（x，x，mode =“same”）将结果缩放到[1，-1]间隔。

def autocorrelate(x):
  fftx = fft(x)
  fftx_mean = np.mean(fftx)
  fftx_std = np.std(fftx)

  ffty = np.conjugate(fftx)
  ffty_mean = np.mean(ffty)
  ffty_std = np.std(ffty)

  result = ifft((fftx - fftx_mean) * (ffty - ffty_mean))
  result = fftshift(result)
  return [i/(fftx_std * ffty_std) for i in result.real]

我已经运行了一些测试数据，看起来它确实应该做到了应该但是我并不完全确定我没有搞砸了什么，只是偶然得到了一些正确的结果;）

Answer 1

Maple的AutoCorrelation函数似乎使用了定义

def AutoCorrelation(x):
    x = np.asarray(x)
    y = x-x.mean()
    result = np.correlate(y, y, mode='full')
    result = result[len(result)//2:]
    result /= result[0]
    return result 

In [189]: AutoCorrelation([1,2,1,2,1,2,1,2])
Out[189]: array([ 1.   , -0.875,  0.75 , -0.625,  0.5  , -0.375,  0.25 , -0.125])

现在，看看我们是否可以使用FFT重现这个结果会很有趣。 NumPy的np.fft.fft是periodic convolution，而np.correlate是线性卷积。要使用np.fft.fft，我们需要添加足够的零填充以使计算基本上是线性的：

def autocorrelation(x):
    """
    Compute autocorrelation using FFT
    """
    x = np.asarray(x)
    N = len(x)
    x = x-x.mean()
    s = fft.fft(x, N*2-1)
    result = np.real(fft.ifft(s * np.conjugate(s), N*2-1))
    result = result[:N]
    result /= result[0]
    return result

以下是一些测试，确认AutoCorrelation和autocorrelation一致并返回与Maple的AutoCorrelation函数返回的值相同的值 - 至少对于我所知道的有限示例。

import numpy as np
fft = np.fft

def autocorrelation(x):
    """
    Compute autocorrelation using FFT
    The idea comes from 
    http://dsp.stackexchange.com/a/1923/4363 (Hilmar)
    """
    x = np.asarray(x)
    N = len(x)
    x = x-x.mean()
    s = fft.fft(x, N*2-1)
    result = np.real(fft.ifft(s * np.conjugate(s), N*2-1))
    result = result[:N]
    result /= result[0]
    return result

def AutoCorrelation(x):
    x = np.asarray(x)
    y = x-x.mean()
    result = np.correlate(y, y, mode='full')
    result = result[len(result)//2:]
    result /= result[0]
    return result 

def autocorrelate(x):
    fftx = fft.fft(x)
    fftx_mean = np.mean(fftx)
    fftx_std = np.std(fftx)

    ffty = np.conjugate(fftx)
    ffty_mean = np.mean(ffty)
    ffty_std = np.std(ffty)

    result = fft.ifft((fftx - fftx_mean) * (ffty - ffty_mean))
    result = fft.fftshift(result)
    return [i / (fftx_std * ffty_std) for i in result.real]


np.set_printoptions(precision=3, suppress=True)

"""
These tests come from
http://www.maplesoft.com/support/help/Maple/view.aspx?path=Statistics/AutoCorrelation
http://www.maplesoft.com/support/help/Maple/view.aspx?path=updates%2fMaple15%2fcomputation
"""
tests = [
    ([1,2,1,2,1,2,1,2], [1,-0.875,0.75,-0.625,0.5,-0.375,0.25,-0.125]),
    ([1,-1,1,-1], [1, -0.75, 0.5, -0.25]),
    ]

for x, answer in tests:
    x = np.array(x)
    answer = np.array(answer)
    # print(autocorrelate(x)) 
    print(autocorrelation(x))
    print(AutoCorrelation(x))
    assert np.allclose(AutoCorrelation(x), answer)
    print

"""
Test that autocorrelation() agrees with AutoCorrelation()
"""
for i in range(1000):
    x = np.random.random(np.random.randint(2,100))*100
    assert np.allclose(autocorrelation(x), AutoCorrelation(x))

使用numpy.fft进行自相关的统计缩放

1 个答案: