Convolution of two three dimensional arrays with padding on one side too slow

Question

In my current project I need to \"convolve\" two three dimensional arrays in a slightly unusual way:

Assume we have two three dimensional arrays A and B

Answer 1

Have you tried using Numba? It's a package that allows you to wrap Python code that is usually slow with a JIT compiler. I took a quick stab at your problem using Numba and got a significant speed up. Using IPython's magic timeit magic function, the custom_convolution function took ~18 s, while Numba's optimized function took 10.4 ms. That's a speed up of more than 1700.

Here's how Numba is implemented.

import numpy as np
from numba import jit, double

s = 15
array_a = np.random.rand(s ** 3).reshape(s, s, s)
array_b = np.random.rand(s ** 3).reshape(s, s, s)

# Original code
def custom_convolution(A, B):

    dimA = A.shape[0]
    dimB = B.shape[0]
    dimC = dimA + dimB

    C = np.zeros((dimC, dimC, dimC))
    for x1 in range(dimA):
        for x2 in range(dimB):
            for y1 in range(dimA):
                for y2 in range(dimB):
                    for z1 in range(dimA):
                        for z2 in range(dimB):
                            x = x1 + x2
                            y = y1 + y2
                            z = z1 + z2
                            C[x, y, z] += A[x1, y1, z1] * B[x2, y2, z2]
    return C

# Numba'ing the function with the JIT compiler
fast_convolution = jit(double[:, :, :](double[:, :, :],
                        double[:, :, :]))(custom_convolution)

If you compute the residual between the results of both functions you will get zeros. This means that the JIT implementation is working without any problems.

slow_result = custom_convolution(array_a, array_b) 
fast_result = fast_convolution(array_a, array_b)

print np.max(np.abs(slow_result - fast_result))

The output I get for this is 0.0.

You can either install Numba into your current Python setup or try it quickly with the AnacondaCE package from continuum.io.

Last but not least, Numba's function is faster than the scipy.signal.fftconvolve function by a factor of a few.

Note: I'm using Anaconda and not AnacondaCE. There are some differences between the two packages for Numba's performance, but I don't think it will differ too much.

Answer 2

The general rule is to use the correct algorithm for the job, which, unless the convolution kernel is short compared to the data, is an FFT based convolution (short roughly means less than log2(n) where n is the length of the data).

In your case, since you're convolving two equal sizes datasets, you probably want to be considering an FFT based convolution.

Clearly, scipy.signal.fftconvolve is being a touch deficient in this regard. Using a faster FFT algorithm, you can do much better by rolling your own convolution routine (it doesn't help that fftconvolve forces the transform size to powers of two, otherwise it could be monkey patched).

The following code uses pyfftw, my wrappers around FFTW and creates a custom convolution class, CustomFFTConvolution:

class CustomFFTConvolution(object):

    def __init__(self, A, B, threads=1):

        shape = (np.array(A.shape) + np.array(B.shape))-1

        if np.iscomplexobj(A) and np.iscomplexobj(B):
            self.fft_A_obj = pyfftw.builders.fftn(
                    A, s=shape, threads=threads)
            self.fft_B_obj = pyfftw.builders.fftn(
                    B, s=shape, threads=threads)
            self.ifft_obj = pyfftw.builders.ifftn(
                    self.fft_A_obj.get_output_array(), s=shape,
                    threads=threads)

        else:
            self.fft_A_obj = pyfftw.builders.rfftn(
                    A, s=shape, threads=threads)
            self.fft_B_obj = pyfftw.builders.rfftn(
                    B, s=shape, threads=threads)
            self.ifft_obj = pyfftw.builders.irfftn(
                    self.fft_A_obj.get_output_array(), s=shape,
                    threads=threads)

    def __call__(self, A, B):

        fft_padded_A = self.fft_A_obj(A)
        fft_padded_B = self.fft_B_obj(B)

        return self.ifft_obj(fft_padded_A * fft_padded_B)

This is used as:

custom_fft_conv = CustomFFTConvolution(A, B)
C = custom_fft_conv(A, B) # This can contain different values to during construction

with an optional threads argument when constructing the class. The purpose of creating a class is to benefit from the ability of FFTW to plan the transform in advance.

The full demo code below simply extends @Kelsey's answer for the timing and so on.

The speedup is substantial over both the numba solution and the vanilla fftconvolve solution. For n = 33, it's about 40-45x faster than both.

from timeit import Timer
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import fftconvolve
from numba import jit, double
import pyfftw

# Original code
def custom_convolution(A, B):

    dimA = A.shape[0]
    dimB = B.shape[0]
    dimC = dimA + dimB

    C = np.zeros((dimC, dimC, dimC))
    for x1 in range(dimA):
        for x2 in range(dimB):
            for y1 in range(dimA):
                for y2 in range(dimB):
                    for z1 in range(dimA):
                        for z2 in range(dimB):
                            x = x1 + x2
                            y = y1 + y2
                            z = z1 + z2
                            C[x, y, z] += A[x1, y1, z1] * B[x2, y2, z2]
    return C

# Numba'ing the function with the JIT compiler
numba_convolution = jit(double[:, :, :](double[:, :, :],
                        double[:, :, :]))(custom_convolution)

def fft_convolution(A, B):
    return fftconvolve(A, B, mode='full')

class CustomFFTConvolution(object):

    def __init__(self, A, B, threads=1):

        shape = (np.array(A.shape) + np.array(B.shape))-1

        if np.iscomplexobj(A) and np.iscomplexobj(B):
            self.fft_A_obj = pyfftw.builders.fftn(
                    A, s=shape, threads=threads)
            self.fft_B_obj = pyfftw.builders.fftn(
                    B, s=shape, threads=threads)
            self.ifft_obj = pyfftw.builders.ifftn(
                    self.fft_A_obj.get_output_array(), s=shape,
                    threads=threads)

        else:
            self.fft_A_obj = pyfftw.builders.rfftn(
                    A, s=shape, threads=threads)
            self.fft_B_obj = pyfftw.builders.rfftn(
                    B, s=shape, threads=threads)
            self.ifft_obj = pyfftw.builders.irfftn(
                    self.fft_A_obj.get_output_array(), s=shape,
                    threads=threads)

    def __call__(self, A, B):

        fft_padded_A = self.fft_A_obj(A)
        fft_padded_B = self.fft_B_obj(B)

        return self.ifft_obj(fft_padded_A * fft_padded_B)

def run_test():
    reps = 10
    nt, ft, cft, cft2 = [], [], [], []
    x = range(2, 34)

    for N in x:
        print N
        A = np.random.rand(N, N, N)
        B = np.random.rand(N, N, N)

        custom_fft_conv = CustomFFTConvolution(A, B)
        custom_fft_conv_nthreads = CustomFFTConvolution(A, B, threads=2)

        C1 = numba_convolution(A, B)
        C2 = fft_convolution(A, B)
        C3 = custom_fft_conv(A, B)
        C4 = custom_fft_conv_nthreads(A, B)

        assert np.allclose(C1[:-1, :-1, :-1], C2)
        assert np.allclose(C1[:-1, :-1, :-1], C3)
        assert np.allclose(C1[:-1, :-1, :-1], C4)

        t = Timer(lambda: numba_convolution(A, B))
        nt.append(t.timeit(number=reps))
        t = Timer(lambda: fft_convolution(A, B))
        ft.append(t.timeit(number=reps))
        t = Timer(lambda: custom_fft_conv(A, B))
        cft.append(t.timeit(number=reps))
        t = Timer(lambda: custom_fft_conv_nthreads(A, B))
        cft2.append(t.timeit(number=reps))

    plt.plot(x, ft, label='scipy.signal.fftconvolve')
    plt.plot(x, nt, label='custom numba convolve')
    plt.plot(x, cft, label='custom pyfftw convolve')
    plt.plot(x, cft2, label='custom pyfftw convolve with threading')        
    plt.legend()
    plt.show()

if __name__ == '__main__':
    run_test()

EDIT: More recent scipy does a better job of not always padding to powers of 2 length so is closer in output to the pyFFTW case.

Answer 3

The Numba method above is a neat trick, but will only be an advantage for relatively small N. An O(N^6) algorithm will kill you every time, regardless of how fast it is implemented. In my tests, the fftconvolve method crosses over to be faster around N=20, and by N=32 is 10x as fast. Leaving out the definition of custom_convolution above:

from timeit import Timer
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import fftconvolve
from numba import jit, double

# Numba'ing the function with the JIT compiler
numba_convolution = jit(double[:, :, :](double[:, :, :],
                        double[:, :, :]))(custom_convolution)

def fft_convolution(A, B):
    return fftconvolve(A, B, mode='full')

if __name__ == '__main__':
    reps = 3
    nt, ft = [], []
    x = range(2, 34)
    for N in x:
        print N
        A = np.random.rand(N, N, N)
        B = np.random.rand(N, N, N)
        C1 = numba_convolution(A, B)
        C2 = fft_convolution(A, B)
        assert np.allclose(C1[:-1, :-1, :-1], C2)
        t = Timer(lambda: numba_convolution(A, B))
        nt.append(t.timeit(number=reps))
        t = Timer(lambda: fft_convolution(A, B))
        ft.append(t.timeit(number=reps))
    plt.plot(x, ft, x, nt)
    plt.show()

It is also very dependent on N, since the FFT will be a lot faster for powers of 2. The time for the FFT version is essentially constant for N=17 to N=32, and it is still faster at N=33, where it starts diverging quickly again.

You could try wrapping an FFT implementation in Numba, but you can't do that directly with the scipy version.

(Sorry to create a new answer, but I don't have the points to reply directly.)