我有一个二维数组,即一个也是数组的序列数组。对于每个序列,我想计算自相关,因此对于(5,4)数组,我将得到5个结果,或维数(5,7)。
我知道我可以绕过第一个维度,但这很慢,我的最后一招。还有另一种方式吗?
谢谢!
编辑:
根据所选答案加上mtrw的评论,我有以下功能:
def xcorr(x):
"""FFT based autocorrelation function, which is faster than numpy.correlate"""
# x is supposed to be an array of sequences, of shape (totalelements, length)
fftx = fft(x, n=(length*2-1), axis=1)
ret = ifft(fftx * np.conjugate(fftx), axis=1)
ret = fftshift(ret, axes=1)
return ret
请注意,length是我的代码中的全局变量,因此请务必声明它。我也没有将结果限制为实数,因为我也需要考虑复数。
答案 0 :(得分:8)
import numpy
from numpy.fft import fft, ifft
data = numpy.arange(5*4).reshape(5, 4)
print data
##[[ 0 1 2 3]
## [ 4 5 6 7]
## [ 8 9 10 11]
## [12 13 14 15]
## [16 17 18 19]]
dataFT = fft(data, axis=1)
dataAC = ifft(dataFT * numpy.conjugate(dataFT), axis=1).real
print dataAC
##[[ 14. 8. 6. 8.]
## [ 126. 120. 118. 120.]
## [ 366. 360. 358. 360.]
## [ 734. 728. 726. 728.]
## [ 1230. 1224. 1222. 1224.]]
我对你对维度(5,7)的答案的陈述感到有些困惑,所以也许有些重要我不理解。
编辑:根据mtrw的建议,填充版本不包围:
import numpy
from numpy.fft import fft, ifft
data = numpy.arange(5*4).reshape(5, 4)
padding = numpy.zeros((5, 3))
dataPadded = numpy.concatenate((data, padding), axis=1)
print dataPadded
##[[ 0. 1. 2. 3. 0. 0. 0. 0.]
## [ 4. 5. 6. 7. 0. 0. 0. 0.]
## [ 8. 9. 10. 11. 0. 0. 0. 0.]
## [ 12. 13. 14. 15. 0. 0. 0. 0.]
## [ 16. 17. 18. 19. 0. 0. 0. 0.]]
dataFT = fft(dataPadded, axis=1)
dataAC = ifft(dataFT * numpy.conjugate(dataFT), axis=1).real
print numpy.round(dataAC, 10)[:, :4]
##[[ 14. 8. 3. 0. 0. 3. 8.]
## [ 126. 92. 59. 28. 28. 59. 92.]
## [ 366. 272. 179. 88. 88. 179. 272.]
## [ 734. 548. 363. 180. 180. 363. 548.]
## [ 1230. 920. 611. 304. 304. 611. 920.]]
必须有一种更有效的方法来做到这一点,特别是因为自相关是对称的,我没有利用它。
答案 1 :(得分:2)
对于非常大的数组,使n = 2 ** p变得很重要,其中p是整数。这将为您节省大量时间。例如:
def xcorr(x):
l = 2 ** int(np.log2(length * 2 - 1))
fftx = fft(x, n = l, axis = 1)
ret = ifft(fftx * np.conjugate(fftx), axis = 1)
ret = fftshift(ret, axes=1)
return ret
这可能会给你环绕错误。但对于大型数组,自然相关在边缘附近应该是无关紧要的。
答案 2 :(得分:1)
也许这只是一种偏好,但我想从定义中得出结论。我个人觉得这样做更容易一些。这是我对任意nd数组的实现。
from itertools import product
from numpy import empty, roll
def autocorrelate(x):
"""
Compute the multidimensional autocorrelation of an nd array.
input: an nd array of floats
output: an nd array of autocorrelations
"""
# used for transposes
t = roll(range(x.ndim), 1)
# pairs of indexes
# the first is for the autocorrelation array
# the second is the shift
ii = [list(enumerate(range(1, s - 1))) for s in x.shape]
# initialize the resulting autocorrelation array
acor = empty(shape=[len(s0) for s0 in ii])
# iterate over all combinations of directional shifts
for i in product(*ii):
# extract the indexes for
# the autocorrelation array
# and original array respectively
i1, i2 = asarray(i).T
x1 = x.copy()
x2 = x.copy()
for i0 in i2:
# clip the unshifted array at the end
x1 = x1[:-i0]
# and the shifted array at the beginning
x2 = x2[i0:]
# prepare to do the same for
# the next axis
x1 = x1.transpose(t)
x2 = x2.transpose(t)
# normalize shifted and unshifted arrays
x1 -= x1.mean()
x1 /= x1.std()
x2 -= x2.mean()
x2 /= x2.std()
# compute the autocorrelation directly
# from the definition
acor[tuple(i1)] = (x1 * x2).mean()
return acor