Convolution computations in Numpy/Scipy

Question

Profiling some computational work I'm doing showed me that one bottleneck in my program was a function that basically did this (np is numpy, sp is scipy):

def mix1(signal1, signal2):
    spec1 = np.fft.fft(signal1, axis=1)
    spec2 = np.fft.fft(signal2, axis=1)
    return np.fft.ifft(spec1*spec2, axis=1)

Both signals have shape (C, N) where C is the number of sets of data (usually less than 20) and N is the number of samples in each set (around 5000). The computation for each set (row) is completely independent of any other set.

I figured that this was just a simple convolution, so I tried to replace it with:

def mix2(signal1, signal2):
    outputs = np.empty_like(signal1)

    for idx, row in enumerate(outputs):
        outputs[idx] = sp.signal.convolve(signal1[idx], signal2[idx], mode='same')

    return outputs

...just to see if I got the same results. But I didn't, and my questions are:

Why not?
Is there a better way to compute the equivalent of mix1()?

(I realise that mix2 probably wouldn't have been faster as-is, but it might have been a good starting point for parallelisation.)

Here's the full script I used to quickly check this:

import numpy as np
import scipy as sp
import scipy.signal

N = 4680
C = 6

def mix1(signal1, signal2):
    spec1 = np.fft.fft(signal1, axis=1)
    spec2 = np.fft.fft(signal2, axis=1)
    return np.fft.ifft(spec1*spec2, axis=1)

def mix2(signal1, signal2):
    outputs = np.empty_like(signal1)

    for idx, row in enumerate(outputs):
        outputs[idx] = sp.signal.convolve(signal1[idx], signal2[idx], mode='same')

    return outputs

def test(num, chans):
    sig1 = np.random.randn(chans, num)
    sig2 = np.random.randn(chans, num)
    res1 = mix1(sig1, sig2)
    res2 = mix2(sig1, sig2)

    np.testing.assert_almost_equal(res1, res2)

if __name__ == "__main__":
    np.random.seed(0x1234ABCD)
    test(N, C)

Convolution computations in Numpy/Scipy

3 Answers