python中的3D卷积

Question

I need to wite a code to perform a 3D convolution in python using numpy, with 3x3 kernels. I've done it right for 2D arrays like B&W images but when i try to extend it to 3D arrays like RGB is a mess. I need help to improve my method. Here is the 2D code:

def convolucion_3x3(arreglo, kernel):
  (dim_x, dim_y) = arreglo.shape
  (ker_x, ker_y) = kernel.shape

  matriz_convolucionada = np.zeros((dim_x, dim_y))

  for i in range(dim_x):
    for j in range(dim_y):
      resultado = 0
      for x in range(-1, 2):
        try:
          if i + x not in range(dim_x):
              raise ValueError()
          for y in range(-1, 2):
            try:
              if j + y not in range(dim_y):
                  raise ValueError()

              resultado += arreglo[i + x, j + y] * kernel[x + 1][y + 1]
              '''
              Para el kernel sumo un 1 a cada índice para que lo corra desde 0 hasta 2 y no de -1 a 1
              '''
            except ValueError:
                pass
        except ValueError:
            pass
      matriz_convolucionada[i][j] = resultado
  return matriz_convolucionada

The next one is my attempt to the RGB images:

def convolucion(arreglo, kernel): (dim_x, dim_y, dim_z) = arreglo.shape (ker_x, ker_y) = kernel.shape

matriz_convolucionada = np.zeros((dim_x, dim_y, dim_z))

for k in range(dim_z):
    for i in range(dim_x):
        for j in range(dim_y):
            resultado = 0
            for x in range(-1, 2):
                try:
                    if i + x not in range(dim_x):
                        raise ValueError()

                    for y in range(-1, 2):
                        try:
                            if j + y not in range(dim_y):
                                raise ValueError()

                            resultado += arreglo[i + x, j + y, k] * kernel[x + 1][y + 1]

                            '''
                            Para el kernel sumo un 1 a cada índice para que lo corra desde 0 hasta 2 y no de -1 a 1
                            '''

                        except ValueError:
                            pass

                except ValueError:
                    pass

            matriz_convolucionada[i][j][k] = resultado

return matriz_convolucionada

Answer 1

While looping through would work, it can also be difficult to follow the nested loops. You might consider invoking the convolution theorem to perform the convolution easier. See here .

Using numpy's fft module, you can compute an n-dimensional discrete Fourier transform of the original stack of images and multiply it by the n-dimensional Fourier transform (documentation found here )of a kernel of the same size . Since your 2D kernel is a 3x3 array, it's a 3x3xz square 'pillar.' You can just pad this array with zeros to increase the dimensions accordingly.

Try this:

import numpy as np
import math

radius = 2
r2 = np.arange(-radius, radius+1)**2
sphere = r2[:, None, None] + r2[:, None] + r2
sphere -= np.max(sphere)
sphere = -sphere*2
array_len = 10*radius

array = np.zeros((array_len, array_len, array_len))
center = slice(array_len//2-radius,
               array_len//2+radius+1), slice(array_len//2-radius,
                                             array_len//2+radius+1),slice(array_len//2-radius,
                                                                          array_len//2+radius+1)
array[center] = sphere


k_len = 3
kernel_2D = np.ones((k_len,k_len))
kernel = np.zeros_like(array)

center_k = slice(array_len//2-math.ceil(k_len/2),
           array_len//2+k_len//2), slice(array_len//2-math.ceil(k_len/2),
                                         array_len//2+k_len//2)
for i in range(kernel.shape[2]):
    kernel[center_k+(i,)] = kernel_2D

def fft(array):
  fft = np.fft.ifftshift(np.fft.fftn(np.fft.fftshift(array)))
  return fft

def ifft(array):
  ifft = np.fft.fftshift(np.fft.ifftn(np.fft.ifftshift(array)))
  return ifft

def conv_3D(array, kernel):
  conv = np.abs(ifft(fft(array)*fft(kernel)))
  return conv

conv = conv_3D(array, kernel)

This convolves a sphere of radius 2 with a pillar of side length 3.