如何在Python中找到一条线与多条曲线的交点? [英] How to find the points of intersection of a line and multiple curves in Python?
问题描述
我有图中表示的数据.
曲线是外推的，我有一条方程式已知的直线.曲线方程未知.现在，如何找到这条线与每条曲线的交点?
The curves were extrapolated and I have a line whose equation is known. The equation of curves are unknown. Now, how do I find the points of intersection of this line with each of the curves?
可复制的代码:
import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate
x = np.array([[0.12, 0.11, 0.1, 0.09, 0.08],
[0.13, 0.12, 0.11, 0.1, 0.09],
[0.15, 0.14, 0.12, 0.11, 0.1],
[0.17, 0.15, 0.14, 0.12, 0.11],
[0.19, 0.17, 0.16, 0.14, 0.12],
[0.22, 0.19, 0.17, 0.15, 0.13],
[0.24, 0.22, 0.19, 0.16, 0.14],
[0.27, 0.24, 0.21, 0.18, 0.15],
[0.29, 0.26, 0.22, 0.19, 0.16]])
y = np.array([[71.64, 78.52, 84.91, 89.35, 97.58],
[66.28, 73.67, 79.87, 85.36, 93.24],
[61.48, 69.31, 75.36, 81.87, 89.35],
[57.61, 65.75, 71.7, 79.1, 86.13],
[55.12, 63.34, 69.32, 77.29, 83.88],
[54.58, 62.54, 68.7, 76.72, 82.92],
[56.58, 63.87, 70.3, 77.69, 83.53],
[61.67, 67.79, 74.41, 80.43, 85.86],
[70.08, 74.62, 80.93, 85.06, 89.84]])
x1 = np.linspace(0, 0.4, 100)
y1 = -100 * x1 + 100
plt.figure(figsize = (5.15,5.15))
plt.subplot(111)
for i in range(5):
x_val = np.linspace(x[0, i] - 0.05, x[-1, i] + 0.05, 100)
x_int = np.interp(x_val, x[:, i], y[:, i])
poly = np.polyfit(x[:, i], y[:, i], deg=2)
y_int = np.polyval(poly, x_val)
plt.plot(x[:, i], y[:, i], linestyle = '', marker = 'o')
plt.plot(x_val, y_int, linestyle = ':', linewidth = 0.25, color = 'black')
plt.plot(x1, y1, linestyle = '-.', linewidth = 0.5, color = 'black')
plt.xlabel('X')
plt.ylabel('Y')
plt.show()
推荐答案
我们知道.它们的格式为a*x**2 + b*x + c，其中a，b和c是np.polyfit返回的向量的元素.然后，我们只需要找到二次方程的根即可找到交点:
We do know the equations of the curves. They are of the form a*x**2 + b*x + c, where a,b, and c are the elements of the vector returned by np.polyfit. Then we just need to find the roots of a quadratic equation in order to find the intersections:
def quadratic_intersections(p, q):
"""Given two quadratics p and q, determines the points of intersection"""
x = np.roots(np.asarray(p) - np.asarray(q))
y = np.polyval(p, x)
return x, y
上面的函数并不是超级健壮:不需要是真实的根，并且也没有真正检查它.您可以自由地做得更好.
The above function isn't super robust: there doesn't need to be a real root, and it doesn't really check for that. You're free to make it better.
无论如何，我们给quadratic_intersections两个二次方，它返回相交的两个点.将其放入您的代码中，我们可以:
Anyways, we give quadratic_intersections two quadratics, and it returns the two points of intersection. Putting it into your code, we have:
x1 = np.linspace(0, 0.4, 100)
y1 = -100 * x1 + 100
plt.figure(figsize = (7,7))
plt.subplot(111)
plt.plot(x1, y1, linestyle = '-.', linewidth = 0.5, color = 'black')
for i in range(5):
x_val = np.linspace(x[0, i] - 0.05, x[-1, i] + 0.05, 100)
poly = np.polyfit(x[:, i], y[:, i], deg=2)
y_int = np.polyval(poly, x_val)
plt.plot(x[:, i], y[:, i], linestyle = '', marker = 'o')
plt.plot(x_val, y_int, linestyle = ':', linewidth = 0.25, color = 'black')
ix = quadratic_intersections(poly, [0, -100, 100])
plt.scatter(*ix, marker='x', color='black', s=40, linewidth=2)
plt.xlabel('X')
plt.ylabel('Y')
plt.xlim([0,.35])
plt.ylim([40,110])
plt.show()
如下图所示:
Now, if you don't know that the functions you are dealing with are polynomials, you can use the optimization tools in scipy.optimize to find the root. For example:
import scipy.optimize
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.style
matplotlib.style.use('fivethirtyeight')
%matplotlib inline
f = lambda x: np.cos(x) - x
g = np.sin
h = lambda x: f(x) - g(x)
x = np.linspace(0,3,100)
plt.plot(x, f(x), zorder=1)
plt.plot(x, g(x), zorder=1)
x_int = scipy.optimize.fsolve(h, 1.0)
y_int = f(x_int)
plt.scatter(x_int, y_int, marker='x', s=150, zorder=2,
linewidth=2, color='black')
plt.xlim([0,3])
plt.ylim([-4,2])
哪些情节:
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