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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