scipy.ndimage.filters.convolve与傅立叶变换相乘得出不同的结果 [英] scipy.ndimage.filters.convolve and multiplying Fourier Transforms give different results

问题描述

这是我的代码:

from scipy.ndimage import filters
import numpy

a = numpy.array([[2,43,42,123,461],[453,12,111,123,55] ,[123,112,233,12,255]])
b = numpy.array([[0,2,2,3,0],[0,15,12,100,0],[0,45,32,22,0]])

ab = filters.convolve(a,b, mode='constant', cval=0)

af = numpy.fft.fftn(a)
bf = numpy.fft.fftn(b)

abf = af*bf

abif = numpy.fft.ifftn(abf)

print numpy.around(ab)
print numpy.around(abif)

结果是:

[[ 1599  2951  7153 13280 18311]
 [ 8085 51478 13028 40239 30964]
 [18192 32484 23527 36122  8726]]

[[ 37416.+0.j  32251.+0.j  46375.+0.j  32660.+0.j  23986.+0.j]
 [ 30265.+0.j  33206.+0.j  62450.+0.j  19726.+0.j  17613.+0.j]
 [ 40239.+0.j  38095.+0.j  24492.+0.j  51478.+0.j  13028.+0.j]]

我该如何修正使用FFT进行卷积的方式，以确保其产生与scipy.ndimage.filters.convolve相同的结果?

How can I fix my way of doing convolution using FFT so to guarantee that it gives the same result as scipy.ndimage.filters.convolve?

谢谢.

推荐答案

正如@senderle所指出的那样，当您使用FFT实现卷积时，会得到 circular 卷积. @senderle的答案显示了如何调整filters.convolve的参数以进行循环卷积.要修改FFT计算以生成与最初使用filters.convolve相同的结果，可以将参数填充为0，然后提取结果的相应部分:

As @senderle points out, when you use the FFT to implement the convolution, you get the circular convolution. @senderle's answer shows how to adjust the arguments of filters.convolve to do a circular convolution. To modify the FFT calculation to generate the same result as your original use of filters.convolve, you can pad the arguments with 0, and then extract the appropriate part of the result:

from scipy.ndimage import filters
import numpy

a = numpy.array([[2.0,43,42,123,461], [453,12,111,123,55], [123,112,233,12,255]])
b = numpy.array([[0.0,2,2,3,0], [0,15,12,100,0], [0,45,32,22,0]])

ab = filters.convolve(a,b, mode='constant', cval=0)

print numpy.around(ab)
print

nrows, ncols = a.shape
# Assume b has the same shape as a.
# Pad the bottom and right side of a and b with zeros.
pa = numpy.pad(a, ((0, nrows-1), (0, ncols-1)), mode='constant')
pb = numpy.pad(b, ((0, nrows-1), (0, ncols-1)), mode='constant')
paf = numpy.fft.fftn(pa)
pbf = numpy.fft.fftn(pb)
pabf = paf*pbf
p0 = nrows // 2
p1 = ncols // 2
pabif = numpy.fft.ifftn(pabf).real[p0:p0+nrows, p1:p1+ncols]

print pabif

输出:

[[  1599.   2951.   7153.  13280.  18311.]
 [  8085.  51478.  13028.  40239.  30964.]
 [ 18192.  32484.  23527.  36122.   8726.]]

[[  1599.   2951.   7153.  13280.  18311.]
 [  8085.  51478.  13028.  40239.  30964.]
 [ 18192.  32484.  23527.  36122.   8726.]]

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