在Python中需要循环FFT卷积

Question

I need a faster analog of

scipy.signal.convolve2d(data, filter, boundary="wrap", mode="same")

Cannot you advice me how to replace it?

PS scipy.signal.fftconvolve is fast enough, but it does not have boundary option and I cannot make it work in circular convolution mode.

Answer 1

If you compute the following:

from scipy.fftpack import fft2, ifft2

f2 = ifft2(fft2(data, shape=data.shape) * fft2(filter, shape=data.shape)).real

then f2 contains the same values as convolve2d(data, filt, boundary='wrap', mode='same') , but the values are shifted ("rolled", in numpy terminology) in each axis. (This is an application of the convolution theorem .)

Here's a short function that rolls the result to the give same result as the convolve2d function call:

def fftconvolve2d(x, y):
    # This assumes y is "smaller" than x.
    f2 = ifft2(fft2(x, shape=x.shape) * fft2(y, shape=x.shape)).real
    f2 = np.roll(f2, (-((y.shape[0] - 1)//2), -((y.shape[1] - 1)//2)), axis=(0, 1))
    return f2

For example,

In [91]: data = np.random.rand(256, 256)

In [92]: filt = np.random.rand(16, 16)

In [93]: c2d = convolve2d(data, filt, boundary='wrap', mode='same')

In [94]: f2 = fftconvolve2d(data, filt)

Verify that the results are the same:

In [95]: np.allclose(c2d, f2)
Out[95]: True

Check the performance:

In [96]: %timeit c2d = convolve2d(data, filt, boundary='wrap', mode='same')
44.9 ms ± 77.3 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [97]: %timeit f2 = fftconvolve2d(data, filt)
5.23 ms ± 11.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

The FFT version is much faster (but note that I chose the dimensions of data to be a power of 2).