Strided convolution of 2D in numpy


I tried to implement strided convolution of a 2D array using for loop i.e

arr = np.array([[2,3,7,4,6,2,9],

arr2 = np.array([[3,4,4],

def stride_conv(arr1,arr2,s,p):
    beg = 0
    end = arr2.shape[0]
    final = []
    for i in range(0,arr1.shape[0]-1,s):
        k = []
        for j in range(0,arr1.shape[0]-1,s):
            k.append(np.sum(arr1[beg+i : end+i, beg+j:end+j] * (arr2)))

    return np.array(final)


This results in 3*3 array:

array([[ 91, 100,  88],
       [ 69,  91, 117],
       [ 44,  72,  74]])

Is there a numpy function or scipy function to do the same? My approach is not that good. How can I vectorize this?

Asked By: Bharath



Ignoring the padding argument and trailing windows that won’t have enough lengths for convolution against the second array, here’s one way with np.lib.stride_tricks.as_strided

def strided4D(arr,arr2,s):
    strided = np.lib.stride_tricks.as_strided
    s0,s1 = arr.strides
    m1,n1 = arr.shape
    m2,n2 = arr2.shape    
    out_shp = (1+(m1-m2)//s, m2, 1+(n1-n2)//s, n2)
    return strided(arr, shape=out_shp, strides=(s*s0,s*s1,s0,s1))

def stride_conv_strided(arr,arr2,s):
    arr4D = strided4D(arr,arr2,s=s)
    return np.tensordot(arr4D, arr2, axes=((2,3),(0,1)))

Alternatively, we can use the scikit-image built-in view_as_windows to get those windows elegantly, like so –

from skimage.util.shape import view_as_windows

def strided4D_v2(arr,arr2,s):
    return view_as_windows(arr, arr2.shape, step=s)
Answered By: Divakar

Here is an O(N^d (log N)^d) fft-based approach. The idea is to chop up both operands into strides-spaced grids at all offsets modulo strides, do the conventional fft convolution between grids of corresponding offsets and then pointwise sum the results. It is a bit index-heavy but I’m afraid that can’t be helped:

import numpy as np
from numpy.fft import fftn, ifftn

def strided_conv_2d(x, y, strides):
    s, t = strides
    # consensus dtype
    cdt = (x[0, 0, ...] + y[0, 0, ...]).dtype
    xi, xj = x.shape
    yi, yj = y.shape
    # round up modulo strides
    xk, xl, yk, yl = map(lambda a, b: -a//b * -b, (xi,xj,yi,yj), (s,t,s,t))
    # zero pad to avoid circular convolution
    xp, yp = (np.zeros((xk+yk, xl+yl), dtype=cdt) for i in range(2))
    xp[:xi, :xj] = x
    yp[:yi, :yj] = y
    # fold out strides
    xp = xp.reshape((xk+yk)//s, s, (xl+yl)//t, t)
    yp = yp.reshape((xk+yk)//s, s, (xl+yl)//t, t)
    # do conventional fft convolution
    xf = fftn(xp, axes=(0, 2))
    yf = fftn(yp, axes=(0, 2))
    result = ifftn(xf * yf.conj(), axes=(0, 2)).sum(axis=(1, 3))
    # restore dtype
    if cdt in (int, np.int_, np.int64, np.int32):
        result = result.real.round()
    return result.astype(cdt)

arr = np.array([[2,3,7,4,6,2,9],

arr2 = np.array([[3,4,4],

print(strided_conv_2d(arr, arr2, (2, 2)))


[[ 91 100  88  23   0  29]
 [ 69  91 117  19   0  38]
 [ 44  72  74  17   0  22]
 [ 16  53  26  12   0   0]
 [  0   0   0   0   0   0]
 [ 19  11  21  -9   0   6]]
Answered By: Paul Panzer

I think we can do a “valid” fft convolution and pick out only those results at strided locations, like this:

def strideConv(arr,arr2,s):
    idx=(np.arange(0,cc.shape[1],s), np.arange(0,cc.shape[0],s))
    return cc[yidx,xidx]

This gives same results as other people’s answers.
But I guess this only works if the kernel size is odd numbered.

Also I’ve flipped the kernel in arr2[::-1,::-1] just to stay consistent with others, you may want to omit it depending on context.


We currently have a few different ways of doing 2D or 3D convolution using numpy and scipy alone, and I thought about doing some comparisons to give some idea on which one is faster on data of different sizes. I hope this won’t be regarded as off-topic.

Method 1: FFT convolution (using scipy.signal.fftconvolve):

def padArray(var,pad,method=1):
    if method==1:
    return var_pad

def conv3D(var,kernel,stride=1,pad=0,pad_method=1):
    '''3D convolution using scipy.signal.convolve.

    if var_ndim<2 or var_ndim>3 or kernel_ndim<2 or kernel_ndim>3:
        raise Exception("<var> and <kernel> dimension should be in 2 or 3.")

    if var_ndim==2 and kernel_ndim==3:
        raise Exception("<kernel> dimension > <var>.")

    if var_ndim==3 and kernel_ndim==2:

    if pad>0:


    if stride>1:

    return conv

Method 2: Special conv (see this anwser):

def conv3D2(var,kernel,stride=1,pad=0):
    '''3D convolution by sub-matrix summing.


    if pad>0:

    for ii in range(ky*kx):
        slabii=var_pad[yi:2*pad+ny-ky+yi+1:1, xi:2*pad+nx-kx+xi+1:1,...]*kernel[yi,xi]
        if var_ndim==3:

    if stride>1:

    return result

Method 3: Strided-view conv, as suggested by Divakar:

def asStride(arr,sub_shape,stride):
    '''Get a strided sub-matrices view of an ndarray.

    <arr>: ndarray of rank 2.
    <sub_shape>: tuple of length 2, window size: (ny, nx).
    <stride>: int, stride of windows.

    Return <subs>: strided window view.

    See also skimage.util.shape.view_as_windows()


    return subs

def conv3D3(var,kernel,stride=1,pad=0):
    '''3D convolution by strided view.

    if var_ndim<2 or var_ndim>3 or kernel_ndim<2 or kernel_ndim>3:
        raise Exception("<var> and <kernel> dimension should be in 2 or 3.")

    if var_ndim==2 and kernel_ndim==3:
        raise Exception("<kernel> dimension > <var>.")

    if var_ndim==3 and kernel_ndim==2:

    if pad>0:

    #return numpy.tensordot(aa,kernel,axes=((2,3),(0,1)))
    if numpy.ndim(kernel)==2:

    return conv

I did 3 sets of comparisons:

  1. convolution on 2D data, with different input size and different kernel size, stride=1, pad=0. Results below (color as time used for convolution repeated for 10 times):

enter image description here

So “FFT conv” is in general the fastest. “Special conv” and “Stride-view conv” get slow as kernel size increases, but decreases again as it approaches the size of input data. The last subplot shows the fastest method, so the big triangle of purple indicates FFT being the winner, but note there is a thin green column on the left side (probably too small to see, but it’s there), suggesting that “Special conv” has advantage for very small kernels (smaller than about 5×5). And when kernel size approaches input, “stride-view conv” is fastest (see the diagonal line).

Comparison 2: convolution on 3D data.

Setup: pad=0, stride=2, input dimension=nxnx5, kernel shape=fxfx5.

I skipped computations of “Special Conv” and “Stride-view conv” when kernel size is in the mid of input. Basically “Special Conv” shows no advantage now, and “Stride-view” is faster than FFT for both small and large kernels.

enter image description here

One additional note: when sizes goes above 350, I notice considerable memory usage peaks for the “Stride-view conv”.

Comparison 3: convolution on 3D data with larger stride.

Setup: pad=0, stride=5, input dimension=nxnx10, kernel shape=fxfx10.

This time I omitted the “Special Conv”. For a larger area “Stride-view conv” surpasses FFT, and last subplots shows that the difference approaches 100 %.
Probably because as the stride goes up, the FFT approach will have more wasted numbers so the “stride-view” gains more advantages for small and large kernels.

enter image description here

Answered By: Jason

How about using signal.convolve2d from scipy?

My approach is similar to Jason’s one but using indexing.

def strideConv(arr, arr2, s):
    return signal.convolve2d(arr, arr2[::-1, ::-1], mode='valid')[::s, ::s]

Note that the kernal has to be reversed. For details, please see discussion here and here. Otherwise use signal.correlate2d.


 >>> strideConv(arr, arr2, 1)
 array([[ 91,  80, 100,  84,  88],
        [ 99, 106, 126,  92,  77],
        [ 69,  98,  91,  93, 117],
        [ 80,  79,  87,  93,  61],
        [ 44,  72,  72,  63,  74]])
 >>> strideConv(arr, arr2, 2)
 array([[ 91, 100,  88],
        [ 69,  91, 117],
        [ 44,  72,  74]])
Answered By: pe-perry

As far as I know, there is no direct implementation of convolution filter in numpy or scipy that supports stride and padding so I think it’s better to use a DL package such as torch or tensorflow, then cast the final result to numpy. a torch implementation might be:

import torch
import torch.nn.functional as F

arr = torch.tensor(np.expand_dims(arr, axis=(0,1))
arr2 = torch.tensor(np.expand_dims(arr2, axis=(0,1))
output = F.conv2d(arr, arr2, stride=2, padding=0)
output = output.numpy().squeeze()

array([[ 91, 100,  88],
       [ 69,  91, 117],
       [ 44,  72,  74]])
Answered By: Sajad.sni

Convolution which supports strides and dilation. numpy.lib.stride_tricks.as_strided is used.

import numpy as np
from numpy.lib.stride_tricks import as_strided

def conv_view(X, F_s, dr, std):
    X_s = np.array(X.shape)
    F_s = np.array(F_s)
    dr = np.array(dr)
    Fd_s = (F_s - 1) * dr + 1
    if np.any(Fd_s > X_s):
        raise ValueError('(Dilated) filter size must be smaller than X')
    std = np.array(std)
    X_ss = np.array(X.strides)
    Xn_s = (X_s - Fd_s) // std + 1
    Xv_s = np.append(Xn_s, F_s)
    Xv_ss = np.tile(X_ss, 2) * np.append(std, dr)
    return as_strided(X, Xv_s, Xv_ss, writeable=False)

def convolve_stride(X, F, dr=None, std=None):
    if dr is None:
        dr = np.ones(X.ndim, dtype=int)
    if std is None:
        std = np.ones(X.ndim, dtype=int)
    if not (X.ndim == F.ndim == len(dr) == len(std)):
        raise ValueError('X.ndim, F.ndim, len(dr), len(std) must be the same')
    Xv = conv_view(X, F.shape, dr, std)
    return np.tensordot(Xv, F, axes=X.ndim)
Answered By: lovetl2002
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