import sys
import numpy as np
import cv2
import matplotlib 
import matplotlib.pyplot as plt
import PIL
from PIL import Imageprint(f"Python version: {sys.version}")
print(f"Numpy version: {np.__version__}")
print(f"Opencv version: {cv2.__version__}")
print(f"Matplotlib version: {matplotlib.__version__}")
print(f"Pillow version: {PIL.__version__}")

Python version: 3.6.12 |Anaconda, Inc.| (default, Sep  9 2020, 00:29:25) [MSC v.1916 64 bit (AMD64)]
Numpy version: 1.16.6
Opencv version: 3.4.1
Matplotlib version: 3.3.2
Pillow version: 8.0.1

像素间的一些基本关系

像素的相邻像素

• $N_4(p)$
• 坐标$(x,y)$处的像素$p$有2个水平的相信像素和2个垂直的相邻像素，它们的坐标是$(x+1,y),(x-1,y),(x,y+1),(x,y-1)$这组像素称为$p$的4邻域，用$N_4(p)$表示
• $N_D(p)$
• $p$的4个对角相邻像素的坐标是$(x+1,y+1),(x+1,y-1),(x-1,y+1),(x-1,y-1)$用$N_D(p)$表示。
• $N_8(p)$
• 这些相邻像素和4邻域全称为8邻域

如题一个邻域包含$p$，那么称该邻域为闭邻域，否则称该邻域为开邻域。

def n4p(x_0):x = np.zeros(5)y_4 = np.zeros_like(x)x[0] = x_0y_4[0] = x_0x[1] = x[0] + 1y_4[1] = y_4[0]x[2] = x[0] - 1y_4[2] = y_4[0]x[3] = x[0]y_4[3] = y_4[0] + 1x[4] = x[0]y_4[4] = y_4[0] - 1return x, y_4def ndp(x_0):x = np.zeros(5)y_D = np.zeros_like(x)x[0] = x_0y_D[0] = x_0x[1] = x[0] + 1y_D[1] = y_D[0] + 1x[2] = x[0] + 1y_D[2] = y_D[0] - 1x[3] = x[0] - 1y_D[3] = y_D[0] + 1x[4] = x[0] - 1y_D[4] = y_D[0] - 1return x, y_D

x_4, y_4 = n4p(0)
x_d, y_d = ndp(0)
x_8 = np.concatenate((x_4, x_d))
y_8 = np.concatenate((y_4, y_d))plt.figure(figsize=(15, 5))
plt.subplot(1,3,1), plt.scatter(x_4, y_4), plt.title("N_4"),# plt.xticks([]), plt.yticks([])
plt.subplot(1,3,2), plt.scatter(x_d, y_d), plt.title("N_D"),# plt.xticks([]), plt.yticks([])
plt.subplot(1,3,3), plt.scatter(x_8, y_8), plt.title("N_8"),# plt.xticks([]), plt.yticks([])
plt.tight_layout()
plt.show()

# 图像的邻域
height, width = 3, 3
img_ori = np.zeros([height, width], dtype=np.float)img_4 = img_ori.copy()
img_4[0, 1] = 1.0
img_4[1, 0] = 1.0
img_4[1, 1] = 1.0
img_4[1, 2] = 1.0
img_4[2, 1] = 1.0img_d = img_ori.copy()
img_d[0, 0] = 1.0
img_d[0, 2] = 1.0
img_d[1, 1] = 1.0
img_d[2, 0] = 1.0
img_d[2, 2] = 1.0img_8 = (img_d + img_4)
img_8[1, 1] = 0.5plt.figure(figsize=(15, 5))
plt.subplot(1,3,1), plt.imshow(img_4, 'gray'), plt.title("N_4"),# plt.xticks([]), plt.yticks([])
plt.subplot(1,3,2), plt.imshow(img_d, 'gray'), plt.title("N_D"),# plt.xticks([]), plt.yticks([])
plt.subplot(1,3,3), plt.imshow(img_8, 'gray'), plt.title("N_8"),# plt.xticks([]), plt.yticks([])
plt.tight_layout()
plt.show()

距离测试

对于坐标分别为$(x,y),(u,v)和(w,z)的像素p,q和s$，如果
$(a)D(p,q)\ge 0[D(p,q)=0,当且仅当p=q]$，
$(b)D(p,q)==D(q,p)且$
$(c)D(p,s)\le D(p,q)+D(q,s)$，
则D是一个距离函数或距离测度。$p和q$之间的区几里得（欧氏）距离定义为:
$$D_e(p,q)=[(x-u{)}^2+(y-v{)}^2{]}^{\frac{1}{2}}\text{\hspace{1em}(2.19)}$$

对于这个距离测度，到点(x, y)的距离小于等于r的像素，是中心在(x, y)、半径为r的圆盘。

$p和q之间的距离D_4$(称为街区距离-曼哈顿距离)
$$D_4(p,q)=|x-u|+|y-v|\text{\hspace{1em}(2.20)}$$

$p和q之间的距离D_8$(称为棋盘距离)
$$D_4(p,q)=max(|x-u|,|y-v|)\text{\hspace{1em}(2.21)}$$