如何提取matplotlib axlines交点以用于单应性矩阵？

问题 #

我需要提取matplotlib中绘制的这些axlines的交点，将其用于单应性矩阵（homography matrix）的计算，请问该如何实现？

当前代码实现：

#######################
# Load and Show image #
#######################

import matplotlib.pyplot as plt
from matplotlib.pyplot import figure
import cv2 as cv
from skimage import io
link = "https://i.imgur.com/WKiQZgj.jpeg"
a = io.imread(link)
a = cv.cvtColor(a, cv.COLOR_RGB2GRAY)
figure(figsize=(15,10), dpi = 75)
plt.imshow(a,cmap = "gray")

##############
# Plot Lines #
##############

#red
plt.axline([361, 39], [1051, 15], color = "red")

#blue
plt.axline([1048, 75], [1158, 115], color = "blue")

#green 
plt.axline([361, 229], [1051, 160], color = "green")

#yellow 
plt.axline([123, 75], [151, 115], color = "yellow")

实现步骤 #

1. 编写直线交点计算函数

每条axline由两个点定义，我们可以通过直线的一般式 (Ax + By + C = 0) 求解两条直线的交点。以下是通用计算函数：

def line_intersection(line1, line2):
    # line1、line2格式为 [(x1,y1), (x2,y2)]
    x1, y1 = line1[0]
    x2, y2 = line1[1]
    x3, y3 = line2[0]
    x4, y4 = line2[1]
    
    # 计算直线一般式系数
    A1 = y2 - y1
    B1 = x1 - x2
    C1 = x2*y1 - x1*y2
    
    A2 = y4 - y3
    B2 = x3 - x4
    C2 = x4*y3 - x3*y4
    
    # 解线性方程组，判断是否有唯一交点
    det = A1 * B2 - A2 * B1
    if det == 0:
        return None  # 直线平行/重合，无唯一交点
    x = (B1*C2 - B2*C1) / det
    y = (A2*C1 - A1*C2) / det
    return (x, y)

2. 提取直线点并计算交点

将四条直线的点对存储后，计算所需的四个交点（对应图像中待变换区域的四个角）：

# 存储每条直线的两点
red_line = [(361, 39), (1051, 15)]
blue_line = [(1048, 75), (1158, 115)]
green_line = [(361, 229), (1051, 160)]
yellow_line = [(123, 75), (151, 115)]

# 计算四个角的交点
top_left = line_intersection(red_line, yellow_line)
top_right = line_intersection(red_line, blue_line)
bottom_left = line_intersection(green_line, yellow_line)
bottom_right = line_intersection(green_line, blue_line)

# 输出交点坐标
print("左上交点:", top_left)
print("右上交点:", top_right)
print("左下交点:", bottom_left)
print("右下交点:", bottom_right)

3. 计算单应性矩阵

使用OpenCV的findHomography函数，传入图像中的源交点和目标矩形的顶点，即可得到单应性矩阵：

import numpy as np

# 源点：图像中的四个交点，转为float32格式
src_pts = np.array([top_left, top_right, bottom_left, bottom_right], dtype=np.float32)

# 目标点：自定义的矩形顶点，可根据需求调整尺寸
dst_pts = np.array([[0, 0], [500, 0], [0, 300], [500, 300]], dtype=np.float32)

# 计算单应性矩阵
homography_matrix, _ = cv.findHomography(src_pts, dst_pts)

print("单应性矩阵:\n", homography_matrix)

4. 透视变换验证（可选）

可以用得到的单应性矩阵对原图进行透视变换，验证效果：

# 执行透视变换
height, width = a.shape
warped_img = cv.warpPerspective(a, homography_matrix, (500, 300))

# 显示变换后的图像
plt.figure(figsize=(8,6))
plt.imshow(warped_img, cmap="gray")
plt.show()

完整整合代码 #

import matplotlib.pyplot as plt
from matplotlib.pyplot import figure
import cv2 as cv
from skimage import io
import numpy as np

def line_intersection(line1, line2):
    x1, y1 = line1[0]
    x2, y2 = line1[1]
    x3, y3 = line2[0]
    x4, y4 = line2[1]
    
    A1 = y2 - y1
    B1 = x1 - x2
    C1 = x2*y1 - x1*y2
    
    A2 = y4 - y3
    B2 = x3 - x4
    C2 = x4*y3 - x3*y4
    
    det = A1 * B2 - A2 * B1
    if det == 0:
        return None
    x = (B1*C2 - B2*C1) / det
    y = (A2*C1 - A1*C2) / det
    return (x, y)

# 加载图像
link = "https://i.imgur.com/WKiQZgj.jpeg"
a = io.imread(link)
a = cv.cvtColor(a, cv.COLOR_RGB2GRAY)

# 绘制原图与直线
figure(figsize=(15,10), dpi = 75)
plt.imshow(a,cmap = "gray")

red_line = [(361, 39), (1051, 15)]
blue_line = [(1048, 75), (1158, 115)]
green_line = [(361, 229), (1051, 160)]
yellow_line = [(123, 75), (151, 115)]

plt.axline(*red_line, color = "red")
plt.axline(*blue_line, color = "blue")
plt.axline(*green_line, color = "green")
plt.axline(*yellow_line, color = "yellow")

# 计算并标记交点
top_left = line_intersection(red_line, yellow_line)
top_right = line_intersection(red_line, blue_line)
bottom_left = line_intersection(green_line, yellow_line)
bottom_right = line_intersection(green_line, blue_line)

plt.scatter(*top_left, color='black', s=50)
plt.scatter(*top_right, color='black', s=50)
plt.scatter(*bottom_left, color='black', s=50)
plt.scatter(*bottom_right, color='black', s=50)

plt.show()

# 计算单应性矩阵
src_pts = np.array([top_left, top_right, bottom_left, bottom_right], dtype=np.float32)
dst_pts = np.array([[0, 0], [500, 0], [0, 300], [500, 300]], dtype=np.float32)
homography_matrix, _ = cv.findHomography(src_pts, dst_pts)

print("单应性矩阵:\n", homography_matrix)

# 透视变换验证
height, width = a.shape
warped_img = cv.warpPerspective(a, homography_matrix, (500, 300))
plt.figure(figsize=(8,6))
plt.imshow(warped_img, cmap="gray")
plt.show()

内容的提问来源于stack exchange，提问作者J33T