python自编程序实现——robert算子、sobel算子、Laplace算子进行图像边缘提取

实现思路:

  1,将传进来的图片矩阵用算子进行卷积求和(卷积和取绝对值)

  2,用新的矩阵(与原图一样大小)去接收每次的卷积和的值

  3,卷积图片所有的像素点后,把新的矩阵数据类型转化为uint8

注意:

  必须对求得的卷积和的值求绝对值;矩阵数据类型进行转化。

完整代码:

import cv2
import numpy as np

# robert 算子[[-1,-1],[1,1]]
def robert_suanzi(img):
	r, c = img.shape
	r_sunnzi = [[-1,-1],[1,1]]
	for x in range(r):
		for y in range(c):
			if (y + 2 <= c) and (x + 2 <= r):
				imgChild = img[x:x+2, y:y+2]
				list_robert = r_sunnzi*imgChild
				img[x, y] = abs(list_robert.sum())		# 求和加绝对值
	return img
				
# # sobel算子的实现
def sobel_suanzi(img):
	r, c = img.shape
	new_image = np.zeros((r, c))
	new_imageX = np.zeros(img.shape)
	new_imageY = np.zeros(img.shape)
	s_suanziX = np.array([[-1,0,1],[-2,0,2],[-1,0,1]])		# X方向
	s_suanziY = np.array([[-1,-2,-1],[0,0,0],[1,2,1]])		
	for i in range(r-2):
		for j in range(c-2):
			new_imageX[i+1, j+1] = abs(np.sum(img[i:i+3, j:j+3] * s_suanziX))
			new_imageY[i+1, j+1] = abs(np.sum(img[i:i+3, j:j+3] * s_suanziY))
			new_image[i+1, j+1] = (new_imageX[i+1, j+1]*new_imageX[i+1,j+1] + new_imageY[i+1, j+1]*new_imageY[i+1,j+1])**0.5
	# return np.uint8(new_imageX)
	# return np.uint8(new_imageY)
	return np.uint8(new_image)	# 无方向算子处理的图像

# Laplace算子
# 常用的Laplace算子模板  [[0,1,0],[1,-4,1],[0,1,0]]   [[1,1,1],[1,-8,1],[1,1,1]]
def Laplace_suanzi(img):
	r, c = img.shape
	new_image = np.zeros((r, c))
	L_sunnzi = np.array([[0,-1,0],[-1,4,-1],[0,-1,0]])		
	# L_sunnzi = np.array([[1,1,1],[1,-8,1],[1,1,1]])		
	for i in range(r-2):
		for j in range(c-2):
			new_image[i+1, j+1] = abs(np.sum(img[i:i+3, j:j+3] * L_sunnzi))
	return np.uint8(new_image)


img = cv2.imread('1.jpg', cv2.IMREAD_GRAYSCALE)
cv2.imshow('image', img)

# # robers算子
out_robert = robert_suanzi(img)
cv2.imshow('out_robert_image', out_robert)

# sobel 算子
out_sobel = sobel_suanzi(img)
cv2.imshow('out_sobel_image', out_sobel)

# Laplace算子
out_laplace = Laplace_suanzi(img)
cv2.imshow('out_laplace_image', out_laplace)

cv2.waitKey(0)
cv2.destroyAllWindows()

  

 

posted @ 2018-11-03 20:53  成长中的菜鸟zhy  阅读(10630)  评论(2编辑  收藏  举报