PROSAGA码农传奇-部署模型-在Python中以高精度查找由（x，y）数据给出的两条曲线的交集

<div class =“post-text”itemprop =“text”>
  
    最佳（也是最有效）的答案可能取决于数据集及其采样方式。但是，许多数据集的一个很好的近似是它们在数据点之间几乎是线性的。因此，我们可以通过原始帖子中显示的“最近的数据点”方法找到交点的近似位置。然后，我们可以使用线性插值来细化最近的两个数据点之间的交点的位置。
  
  
    这个方法非常快，并且适用于2D numpy数组，以防你想一次计算多条曲线的交叉（正如我想在我的应用程序中那样）。
  
  
    （我从“借来的代码”
    <a href="https://stackoverflow.com/questions/20677795/how-do-i-compute-the-intersection-point-of-two-lines-in-python">
      如何在Python中计算两条线的交点？
    </A>
    “用于线性插值。）
  
   <pre>
 <code>
 from __future__ import division 
import numpy as np
import matplotlib.pyplot as plt

def interpolated_intercept(x, y1, y2):
    """Find the intercept of two curves, given by the same x data"""

def intercept(point1, point2, point3, point4):
        """find the intersection between two lines
        the first line is defined by the line between point1 and point2
        the first line is defined by the line between point3 and point4
        each point is an (x,y) tuple.

So, for example, you can find the intersection between
        intercept((0,0), (1,1), (0,1), (1,0)) = (0.5, 0.5)

Returns: the intercept, in (x,y) format
        """

def line(p1, p2):
            A = (p1[1] - p2[1])
            B = (p2[0] - p1[0])
            C = (p1[0]*p2[1] - p2[0]*p1[1])
            return A, B, -C

def intersection(L1, L2):
            D  = L1[0] * L2[1] - L1[1] * L2[0]
            Dx = L1[2] * L2[1] - L1[1] * L2[2]
            Dy = L1[0] * L2[2] - L1[2] * L2[0]

x = Dx / D
            y = Dy / D
            return x,y

L1 = line([point1[0],point1[1]], [point2[0],point2[1]])
        L2 = line([point3[0],point3[1]], [point4[0],point4[1]])

R = intersection(L1, L2)

return R

idx = np.argwhere(np.diff(np.sign(y1 - y2)) != 0)
    xc, yc = intercept((x[idx], y1[idx]),((x[idx+1], y1[idx+1])), ((x[idx], y2[idx])), ((x[idx+1], y2[idx+1])))
    return xc,yc

def main():
    x  = np.linspace(1, 4, 20)
    y1 = np.sin(x)
    y2 = 0.05*x

plt.plot(x, y1, marker='o', mec='none', ms=4, lw=1, label='y1')
    plt.plot(x, y2, marker='o', mec='none', ms=4, lw=1, label='y2')

idx = np.argwhere(np.diff(np.sign(y1 - y2)) != 0)

plt.plot(x[idx], y1[idx], 'ms', ms=7, label='Nearest data-point method')

# new method!
    xc, yc = interpolated_intercept(x,y1,y2)
    plt.plot(xc, yc, 'co', ms=5, label='Nearest data-point, with linear interpolation')

plt.legend(frameon=False, fontsize=10, numpoints=1, loc='lower left')

plt.savefig('curve crossing.png', dpi=200)
    plt.show()

if __name__ == '__main__': 
    main()

</code>
 </pre>
  
    <a href="https://i.stack.imgur.com/vzVal.png" rel="nofollow noreferrer">
      <img src =“https://i.stack.imgur.com/vzVal.png”alt =“曲线穿越”/>
    </A>
  
  
    的
      更新2018-12-13
    </强>
    ：
如果有必要找到几个截取，这里是一个修改版本的代码：
  
   <pre>
 <code>
 from __future__ import division 
import numpy as np
import matplotlib.pyplot as plt

def interpolated_intercepts(x, y1, y2):
    """Find the intercepts of two curves, given by the same x data"""

So, for example, you can find the intersection between
        intercept((0,0), (1,1), (0,1), (1,0)) = (0.5, 0.5)

Returns: the intercept, in (x,y) format
        """

def line(p1, p2):
            A = (p1[1] - p2[1])
            B = (p2[0] - p1[0])
            C = (p1[0]*p2[1] - p2[0]*p1[1])
            return A, B, -C

def intersection(L1, L2):
            D  = L1[0] * L2[1] - L1[1] * L2[0]
            Dx = L1[2] * L2[1] - L1[1] * L2[2]
            Dy = L1[0] * L2[2] - L1[2] * L2[0]

x = Dx / D
            y = Dy / D
            return x,y

L1 = line([point1[0],point1[1]], [point2[0],point2[1]])
        L2 = line([point3[0],point3[1]], [point4[0],point4[1]])

R = intersection(L1, L2)

return R

idxs = np.argwhere(np.diff(np.sign(y1 - y2)) != 0)

xcs = []
    ycs = []

for idx in idxs:
        xc, yc = intercept((x[idx], y1[idx]),((x[idx+1], y1[idx+1])), ((x[idx], y2[idx])), ((x[idx+1], y2[idx+1])))
        xcs.append(xc)
        ycs.append(yc)
    return np.array(xcs), np.array(ycs)

def main():
    x  = np.linspace(1, 10, 50)
    y1 = np.sin(x)
    y2 = 0.02*x

plt.plot(x, y1, marker='o', mec='none', ms=4, lw=1, label='y1')
    plt.plot(x, y2, marker='o', mec='none', ms=4, lw=1, label='y2')

idx = np.argwhere(np.diff(np.sign(y1 - y2)) != 0)

plt.plot(x[idx], y1[idx], 'ms', ms=7, label='Nearest data-point method')

# new method!
    xcs, ycs = interpolated_intercepts(x,y1,y2)
    for xc, yc in zip(xcs, ycs):
        plt.plot(xc, yc, 'co', ms=5, label='Nearest data-point, with linear interpolation')

plt.legend(frameon=False, fontsize=10, numpoints=1, loc='lower left')

plt.savefig('curve crossing.png', dpi=200)
    plt.show()

if __name__ == '__main__': 
    main()

</code>
 </pre>
  
    ```