如何在matplotlib中找到axvline和axline之间交点的数值？

Question

我有下面的代码，我需要找到axline和axvline之间的交点的数值，我不知道如何简单地解决这个问题，有人知道如何解决吗？提前无限感谢！ :)

!pip install matplotlib==3.4

%matplotlib inline  
import matplotlib.pyplot as plt 

x = [0,2,4,6,8,10,12,14.3,16.2,18,20.5,22.2,25.1,
     26.1,28,30,33.3,34.5,36,38,40]
y = [13.4,23.7,35.1,48.3,62.7,76.4,91.3,106.5,119.6,131.3,
     146.9,157.3,173.8,180.1,189.4,199.5,215.2,220.6,227,234.7,242.2]
slope = (131.3-119.6)/(18-16.2)
plt.figure(figsize=(10, 5))
plt.axline((16.2,119.6), slope = slope, linestyle = '--', color = 'r')
plt.grid()
plt.minorticks_on()
plt.axvline(30,linestyle = '--', color = 'black')
plt.plot(x,y, linewidth = 2.5)
plt.show()

Plot Result

Answer 1

首先，您需要找到斜线的方程，y=slope*x+q：这很容易做到，因为您知道斜线的斜率和点。

接下来求解方程组：y=slope*x+q、x=x0（垂直线）。

这里我用绿色标记绘制了交点。

import matplotlib.pyplot as plt 

x = [0,2,4,6,8,10,12,14.3,16.2,18,20.5,22.2,25.1,
     26.1,28,30,33.3,34.5,36,38,40]
y = [13.4,23.7,35.1,48.3,62.7,76.4,91.3,106.5,119.6,131.3,
     146.9,157.3,173.8,180.1,189.4,199.5,215.2,220.6,227,234.7,242.2]
slope = (131.3-119.6)/(18-16.2)
plt.figure(figsize=(10, 5))

point_1 = (16.2, 119.6)
q = point_1[1] - slope * point_1[0]
x2 = 30
point_2 = (x2, slope * x2 + q)
plt.axline(point_1, slope = slope, linestyle = '--', color = 'r')
plt.grid()
plt.minorticks_on()
plt.axvline(x2,linestyle = '--', color = 'black')
plt.plot(x,y, linewidth = 2.5)
plt.scatter(*point_2, color="g", s=100)
plt.show()

如何在matplotlib中找到axvline和axline之间交点的数值？

How to find the numeric value of the intersection point between an axvline and axline in matplotlib?

python

matplotlib

intersection

axvline