Efficient 2D convolution/cross-correlation along only one axis (1D output)

Question

I wish to convolve/cross-correlate two images

and

but, only horizontally, yielding 1D output.

So, first output point would be sum(x * h), second sum(x * h_shift1), where h_shift1 is h horizontally shifted by 1 sample, in Python np.roll(h, axis=1). Basically, pass images through each other horizontally.

I know that 2D cross-correlation is efficiently done as

$$ \texttt{iFFT}_{2d}\bigg( \texttt{FFT}_{2d}\big(x\big) \cdot \overline{\texttt{FFT}_{2d}\big(\texttt{iFFTSHIFT}_{2d}(h)\big)} \bigg) $$

but this yields a 2D output which I don't need. Can my case be handled efficiently?

Convention

For sake of this post, assume a "backward" (which I think should be standard) definition, where $\overline{f(\cdot)}$ is complex conjugation:

$$ (x \star h)(\tau) = \int_{-\infty}^{\infty} x(t) \overline{h(t - \tau)} dt $$

which just flips the order of arguments relative to standard. An answer with the standard definition is also acceptable.

Clarification

This is not "batched 1D" / "many 1D". The sum and product for every shift is 2D:

$$ (\mathbf{a} \star \mathbf{b})(\tau) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathbf{a}(y, x) \overline{\mathbf{b}(y, x - \tau)} dxdy $$

Again, out[i] = sum(a * shift(b, i)), where a and b are 2D and sum yields a scalar.

Brute force example

MATLAB -- Python -- same outputs

Answer 1

We observe that "2D along 1D" is equivalently: first do 1D horizontally (each row independently), then sum vertically. The complete operation for an output point, except for a shift, is sum(a * b), which is a 2D product and 2D sum.

1. 1D convolution for row m does sum(a[m, :] * b[m, :]) for all shifts of b by i, b_i.
2. Summing vertically for a given i is hence summing sum(a[m, :] * b_i[m, :]) for all m.
3. (2) is same as sum(a * b_i), i.e. sum(a[:, :] * b_i[:, :]).

So, if we let hf = ifftshift(conj(fft(h, axis=1)), axes=1), and prod = fft(x, axis=1) * hf, then it's just sum(ifft(prod, axis=1), axis=0). But we observe, by linearity, we can move sum inside ifft for a great speedup. All together,

$$ \texttt{CC}_{2d1d}(x, h) = \texttt{iFFT}_{1d} \left( \sum_{m=0}^{M - 1} \left( \texttt{FFT}_{1d}\big(x\big) \cdot \overline{\texttt{FFT}_{1d}\big(\texttt{iFFTSHIFT}_{1d}(h)\big)} \right)[m] \right) $$

where 2D indexing is $x[m, n]$, and $\texttt{op}_{1d}$ denotes 1D operation along $n$'s axis.

Thanks to @CrisLuengo and @Royi for pointers.

Example in question

Applying in code (extending the code at bottom)...

import matplotlib.pyplot as plt
from PIL import Image
# load image as greyscale
x = np.array(Image.open("cim0.png").convert("L")) / 255.
h = np.array(Image.open("cim1.png").convert("L")) / 255.
# blank regions default to `1`, undo that
x[x==1] = 0
h[h==1] = 0
# compute
out = cc2d1d(x, h)[0].real
# plot
plt.plot(out); plt.show()

the peak is near center, as expected:

Applications

I used it to identify abrupt changes in audio, by cross-correlating CWT's impulse response with non-linearly filtered version of SSQ_CWT. So one major use is 2D template matching upon underlying 1D structures. Surely there's plenty others.

(Note for those curious in the linked post) But! I by no means did this with images like in this post. An "image" involves up to three major modifications - compression, color-mapping, and clipping (vmin, vmax args in plt.imshow) - which change its numeric representation once loaded from image into array. Instead I operated on the original arrays, and it's clear from the worse results in this post.

Convolution? Vertical?

• Convolution: remove np.conj
• Vertical: ifftshift(, axes=1) -> ifftshift(, axes=0), and mean(axis=0) -> mean(axis=1)
• Boundary effects / "time aliasing": pad and unpad exactly same as with 1D convolutions. But note, if $h$ isn't reusable, it's faster to adjust unpad indices instead of doing ifftshift, as shown in Royi's answer on conv2 (ignore vertical unpad).

Benchmarks (CPU)

For reusable $h$:

def cc2d1d_hf(x, hf):
    return ifft((fft(x) * hf).sum(axis=0))
    
shapes = [(8192, 8192), (256, 262144), (262144, 256)]
for shape in shapes:
    x = np.random.randn(*shape)
    hf = np.conj(fft(ifftshift(np.random.randn(*shape), axes=1)))
    %timeit cc2d1d_hf(x, hf)

3.01 s ± 122 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
4.21 s ± 138 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
2.3 s ± 78.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Code

import numpy as np
from numpy.fft import fft, ifft, ifftshift

def cc2d1d(x, h):
    prod = fft(x) * np.conj(fft(ifftshift(h, axes=1)))
    return ifft(prod.sum(axis=0))

def cc2d1d_brute(x, h):
    out = np.zeros(x.shape[-1], dtype=x.dtype)
    h = ifftshift(np.conj(h), axes=1)
    for i in range(len(out)):
        out[i] = np.sum(x * np.roll(h, i, axis=1))
    return out

for M in (128, 129):
    for N in (128, 129):
        x = np.random.randn(M, N) + 1j*np.random.randn(M, N)
        h = np.random.randn(M, N) + 1j*np.random.randn(M, N)
        out0 = cc2d1d(x, h)
        out1 = cc2d1d_brute(x, h)
        assert np.allclose(out0, out1)

Answer 2

OLD (& slower) ANSWER - see other for better solution

We observe that this "2D along 1D" is simply a subset of the full 2D operation - namely, we obtain the 1D vector of interest by slicing the full 2D output along horizontal, i.e. indexing a row. Equivalently, it's maximum subsampling along vertical - so applying "Subsampling in time $\Leftrightarrow$ Folding in Fourier", we let

$$ \texttt{prod} = \texttt{FFT}_{2d}\big(x\big) \cdot \overline{\texttt{FFT}_{2d}\big(\texttt{iFFTSHIFT}_{2d}(h)\big)} $$

and then the desired output is out = ifft2(prod.reshape(sub, 1, N).mean(axis=0)), where M, N = x.shape and sub = M. Except, this is equivalent to out = out_full[0], which assumes that the 1D vector of interest is the very first row. That's in fact not the case with $\texttt{iFFTSHIFT}_\text{2d}$.

Following the linked post, we could use the offset relation to instead make it like out_full[offset:][0], with an appropriately chosen offset - but the more efficient option is just to shift h the right way. Turns out to do this, we simply don't shift it along vertical: in turn, the FFTs will vertical-center it during cross-correlation.

Lastly, we realize that fft(x) == x for len(x) == 1, so ifft2 along vertical & horizontal simplifies to ifft along horizontal. In full-ish math...

$$ \texttt{CC}_{2d1d}(x, h) = \texttt{iFFT}_{1d} \left( \frac{1}{M}\sum_{m=0}^{M - 1} \left( \texttt{FFT}_{2d}\big(x\big) \cdot \overline{\texttt{FFT}_{2d}\big(\texttt{iFFTSHIFT}_{1d;n}(h)\big)} \right)[m] \right) $$

where 2D indexing is $x[m, n]$, and $\texttt{iFFTSHIFT}_{1d;n}$ denotes 1D ifftshift along $n$'s axis. (anyone got better notation? and I don't mean a "strictly accurate" complete mess)

Benchmarks (CPU)

For reusable $h$:

def cc2d1d_hf(x, hf):
    return ifft((fft2(x) * hf).mean(axis=0))

shapes = [(8192, 8192), (256, 262144), (262144, 256)]
for shape in shapes:
    x = np.random.randn(*shape)
    hf = np.conj(fft2(ifftshift(np.random.randn(*shape), axes=1)))
    %timeit cc2d1d_hf(x, hf)

4.65 s ± 270 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
4.9 s ± 133 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
5.52 s ± 404 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Code

import numpy as np
from numpy.fft import fft2, ifft2, ifftshift

def cc2d1d(x, h):
    prod = fft2(x) * np.conj(fft2(ifftshift(h, axes=1)))
    return ifft(prod.mean(axis=0))

def cc2d1d_brute(x, h):
    out = np.zeros(x.shape[-1], dtype=x.dtype)
    h = ifftshift(np.conj(h), axes=1)
    for i in range(len(out)):
        out[i] = np.sum(x * np.roll(h, i, axis=1))
    return out

for M in (128, 129):
    for N in (128, 129):
        x = np.random.randn(M, N) + 1j*np.random.randn(M, N)
        h = np.random.randn(M, N) + 1j*np.random.randn(M, N)
        out0 = cc2d1d(x, h)
        out1 = cc2d1d_brute(x, h)
        assert np.allclose(out0, out1)