计算3条水平线和三次样条的交点

Question

我有一个问题，类似于那些，我现在发帖。我想计算一个三次样条和 3 条水平线之间的交点。对于所有这些水平线，我都知道 y 值，并且我想找出交点对应的 x 值。我希望你能帮助我。我相信对于更有经验的编码人员来说，解决起来非常容易！

import matplotlib.pyplot as plt
from scipy import interpolate
import numpy as np

x = np.arange(0, 10)
y = np.exp(-x**2.0)

spline = interpolate.interp1d(x, y, kind = "cubic")
xnew = np.arange(0, 9, 0.1)
ynew = spline(xnew)   


x1=np.arange(0,10)
y1=1/10*np.ones(10)

x2=np.arange(0,10)
y2=2/10*np.ones(10)

x3=np.arange(0,10)
y3=3/10*np.ones(10)


plt.plot(x,y,'o', xnew, ynew, '-', x1,y1, '-.', x2,y2, '-.', x3,y3, '-.')
plt.show()


for i in range(1,4):
    idx = np.argwhere(np.diff(np.sign(spline-y_i))).flatten()
    list_idx.append(idx)


print(list_idx)

Answer 1

您可以使用scipy.interpolate.InterpolatedUnivariateSpline的roots()函数求根。所以首先你必须从函数中减去 y-value 并找到根，这给你 x-value 在那个特定的 y-value.

import matplotlib.pyplot as plt
from scipy import interpolate
import numpy as np

x = np.arange(0, 10)
y = np.exp(-x**2.0)

spline = interpolate.interp1d(x, y, kind = "cubic")
xnew = np.arange(0, 9, 0.1)
ynew = spline(xnew)   


x1=np.arange(0,10)
y1=1*np.ones(10)/10

x2=np.arange(0,10)
y2=2*np.ones(10)/10

x3=np.arange(0,10)
y3=3*np.ones(10)/10


plt.plot(x,y,'o', xnew, ynew, '-', x1,y1, '-.', x2,y2, '-.', x3,y3, '-.')
plt.show()


y_val = 0.2
func = np.array(y) - y_val
sub_funct = interpolate.InterpolatedUnivariateSpline(x, func) # to find the roots we need to substract y_val from the function
root = sub_funct.roots() # find roots here
print(root)

当 y_val=0.2 为

时打印 x 值

 [1.36192179]

编辑

您可以绘制输出图如下。

plt.arrow(root, y_val, 0, -y_val, head_width=0.2, head_length=0.06)

计算3条水平线和三次样条的交点

Calculating the intersection points of 3 horizontal lines and a cubic spline

python

scipy

line

intersection

spline