python - 如何使用Python在给定斜率和截距坐标（x，y）的图像上画一条线

Question

我有一条线的斜率，并且我有我希望这条线通过的截距的 x 和 y 坐标。本质上，我希望线条显示在图像本身上。

以下基本上是我迄今为止唯一的代码（只是变量）：

slope = 3
intercepts = [7, 10]

我想使用 Python 来完成这项任务。任何帮助将不胜感激！

标签： pythonimagematplotlib

Answer 1

您可以从 Tommaso Di Noto 自定义此答案：

import matplotlib.pyplot as plt
import random

x_intercept = 7  
y_intercept = 10 
my_slope = 3 

def find_second_point(slope,x0,y0):
    # this function returns two points which belongs to the line that has the slope 
    # inserted by the user and that intercepts the point (x0,y0) inserted by the user
    q = y0 - (slope*x0)  # calculate q
    new_x1 = x0 + random.randint(x0,x0+10)  
    new_y1 = (slope*new_x1) + q  
    new_x2 = x0 - random.randint(x0,x0+10)  
    new_y2 = (slope*new_x2) + q 
    return new_x1, new_y1, new_x2, new_y2   


new_x1, new_y1,new_x2, new_y2 = find_second_point(my_slope , x_intercept, y_intercept )

def slope(x1, y1, x2, y2):
    return (y2-y1)/(x2-x1)
print(slope(x_intercept,y_intercept, new_x1, new_y1))


plt.figure(1)  # create new figure
plt.plot((new_x2, new_x1),(new_y2, new_y1), c='r', label='Segment')
plt.scatter(x_intercept, y_intercept, c='b', linewidths=3, label='Intercept')
plt.scatter(new_x1, new_y1, c='g', linewidths=3, label='New Point 1')
plt.scatter(new_x2, new_y2, c='cyan', linewidths=3, label='New Point 2')
plt.legend()  # add legend to image

plt.show()

输出：

slope
3.0