python - 在没有 opencv 的情况下使用 python 实现 Sobel 运算符

Question

给定一个灰度 8 位图像（像素强度值从 0 到 255 的二维数组），我想在图像上实现 Sobel 运算符（掩码）。下面的 Sobel 函数基本上围绕给定像素循环，对像素应用以下权重：

然后应用给定的公式：

我正在尝试实现此链接中的公式： http://homepages.inf.ed.ac.uk/rbf/HIPR2/sobel.htm

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.image as mpimg
import Image


def Sobel(arr,rstart, cstart,masksize, divisor):
  sum = 0;
  x = 0
  y = 0

  for i in range(rstart, rstart+masksize, 1):
    x = 0
    for j in range(cstart, cstart+masksize, 1):
        if x == 0 and y == 0:
            p1 = arr[i][j]
        if x == 0 and y == 1:
            p2 = arr[i][j]
        if x == 0 and y == 2:
            p3 = arr[i][j]
        if x == 1 and y == 0:
            p4 = arr[i][j]
        if x == 1 and y == 1:
            p5 = arr[i][j]           
        if x == 1 and y == 2:
            p6 = arr[i][j]
        if x == 2 and y == 0:
            p7 = arr[i][j]
        if x == 2 and y == 1:
            p8 = arr[i][j]
        if x == 2 and y == 2:
            p9 = arr[i][j]
        x +=1
    y +=1
  return np.abs((p1 + 2*p2 + p3) - (p7 + 2*p8+p9)) + np.abs((p3 + 2*p6 + p9) - (p1 + 2*p4 +p7)) 


def padwithzeros(vector, pad_width, iaxis, kwargs):
   vector[:pad_width[0]] = 0
   vector[-pad_width[1]:] = 0
   return vector

im = Image.open('charlie.jpg')
im.show()
img = np.asarray(im)
img.flags.writeable = True
p = 1
k = 2
m = img.shape[0]
n = img.shape[1]
masksize = 3
img = np.lib.pad(img, p, padwithzeros) #this function padds image with zeros to cater for pixels on the border.
x = 0
y = 0
for row in img:
  y = 0
  for col in row:
    if not (x < p or y < p or y > (n-k) or x > (m-k)):
        img[x][y] = Sobel(img, x-p,y-p,masksize,masksize*masksize)
    y = y + 1
  x = x + 1

img2 = Image.fromarray(img)
img2.show()

鉴于此灰度 8 位图像

我在应用函数时得到这个：

但应该得到这个：

我已经用python实现了其他高斯过滤器，我不确定我哪里出错了？

Answer 1

如果使用 NumPy 和 SciPy 没有问题，那么一个简单的解决方案是使用 SciPy 的 convolve2d()。

import numpy as np
from scipy.signal import convolve2d
from scipy.ndimage import imread

# Load sample data
with np.DataSource().open("http://i.stack.imgur.com/8zINU.gif", "rb") as f:
    img = imread(f, mode="I")

# Prepare the kernels
a1 = np.matrix([1, 2, 1])
a2 = np.matrix([-1, 0, 1])
Kx = a1.T * a2
Ky = a2.T * a1

# Apply the Sobel operator
Gx = convolve2d(img, Kx, "same", "symm")
Gy = convolve2d(img, Ky, "same", "symm")
G = np.sqrt(Gx**2 + Gy**2)
# or using the absolute values
G = np.abs(Gx) + np.abs(Gy)

Answer 2

紧贴您的代码正在执行的操作，一种优雅的解决方案是使用scipy.ndimage.filters.generic_filter() 和上面提供的公式。

import numpy as np
from scipy.ndimage.filters import generic_filter
from scipy.ndimage import imread

# Load sample data
with np.DataSource().open("http://i.stack.imgur.com/8zINU.gif", "rb") as f:
    img = imread(f, mode="I")

# Apply the Sobel operator
def sobel_filter(P):
    return (np.abs((P[0] + 2 * P[1] + P[2]) - (P[6] + 2 * P[7] + P[8])) +
            np.abs((P[2] + 2 * P[6] + P[7]) - (P[0] + 2 * P[3] + P[6])))
G = generic_filter(img, sobel_filter, (3, 3))

在示例图像上运行此操作大约需要 400 毫秒。相比之下，convolve2d 的性能约为 6.5 毫秒。

Answer 3

我遇到了和你一样的问题。我通过读取“灰色”格式的图像来修复它，你可以在下面看到

import PIL.Image
img = PIL.Image.open('image.gif').convert('L')