這裏有幾個功能,自動計算,並與有限的互相關滯後。選擇乘法(和複合情況下的共軛)的順序以匹配numpy.correlate
的相應行爲。
import numpy as np
from numpy.lib.stride_tricks import as_strided
def _check_arg(x, xname):
x = np.asarray(x)
if x.ndim != 1:
raise ValueError('%s must be one-dimensional.' % xname)
return x
def autocorrelation(x, maxlag):
"""
Autocorrelation with a maximum number of lags.
`x` must be a one-dimensional numpy array.
This computes the same result as
numpy.correlate(x, x, mode='full')[len(x)-1:len(x)+maxlag]
The return value has length maxlag + 1.
"""
x = _check_arg(x, 'x')
p = np.pad(x.conj(), maxlag, mode='constant')
T = as_strided(p[maxlag:], shape=(maxlag+1, len(x) + maxlag),
strides=(-p.strides[0], p.strides[0]))
return T.dot(p[maxlag:].conj())
def crosscorrelation(x, y, maxlag):
"""
Cross correlation with a maximum number of lags.
`x` and `y` must be one-dimensional numpy arrays with the same length.
This computes the same result as
numpy.correlate(x, y, mode='full')[len(a)-maxlag-1:len(a)+maxlag]
The return vaue has length 2*maxlag + 1.
"""
x = _check_arg(x, 'x')
y = _check_arg(y, 'y')
py = np.pad(y.conj(), 2*maxlag, mode='constant')
T = as_strided(py[2*maxlag:], shape=(2*maxlag+1, len(y) + 2*maxlag),
strides=(-py.strides[0], py.strides[0]))
px = np.pad(x, maxlag, mode='constant')
return T.dot(px)
例如,
In [367]: x = np.array([2, 1.5, 0, 0, -1, 3, 2, -0.5])
In [368]: autocorrelation(x, 3)
Out[368]: array([ 20.5, 5. , -3.5, -1. ])
In [369]: np.correlate(x, x, mode='full')[7:11]
Out[369]: array([ 20.5, 5. , -3.5, -1. ])
In [370]: y = np.arange(8)
In [371]: crosscorrelation(x, y, 3)
Out[371]: array([ 5. , 23.5, 32. , 21. , 16. , 12.5, 9. ])
In [372]: np.correlate(x, y, mode='full')[4:11]
Out[372]: array([ 5. , 23.5, 32. , 21. , 16. , 12.5, 9. ])
(這將是很好,有這樣的功能在numpy的本身。)
該函數只是numpy.correlate的一個包裝。不幸的是,雖然它返回適當的長度向量,但它沒有任何性能節省,因爲它實際上計算了完整的互相關,然後拋出額外的條目。 – honi