Question

我有两个数据集：（x，y1）和（x，y2）。我想找到这两条曲线相互交叉的位置。目标与此问题类似：Intersection of two graphs in Python, find the x value:

但是，那里描述的方法只找到与最近的数据点的交点。我想找到曲线的交点，其精度高于原始数据间距。一种选择是简单地重新插入到更精细的网格。这是有效的，但是精度取决于我为重新插值选择的点数，这是任意的，需要在精度和效率之间进行权衡。

或者，我可以使用scipy.optimize.fsolve来查找数据集的两个样条插值的精确交集。这很好用，但它不能轻易找到多个交叉点，要求我提供合理的交叉点猜测，并且可能不能很好地扩展。（最终，我想找到数千套（x，y1，y2）的交集，所以一个有效的算法会很好。）

这是我到目前为止所拥有的。有任何改进的想法吗？

import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate, scipy.optimize

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')

interp1 = scipy.interpolate.InterpolatedUnivariateSpline(x, y1)
interp2 = scipy.interpolate.InterpolatedUnivariateSpline(x, y2)

new_x = np.linspace(x.min(), x.max(), 100)
new_y1 = interp1(new_x)
new_y2 = interp2(new_x)
idx = np.argwhere(np.diff(np.sign(new_y1 - new_y2)) != 0)
plt.plot(new_x[idx], new_y1[idx], 'ro', ms=7, label='Nearest data-point method, with re-interpolated data')

def difference(x):
    return np.abs(interp1(x) - interp2(x))

x_at_crossing = scipy.optimize.fsolve(difference, x0=3.0)
plt.plot(x_at_crossing, interp1(x_at_crossing), 'cd', ms=7, label='fsolve method')

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

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

Answer 1

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）。

（我借用“How do I compute the intersection point of two lines in Python?”的代码进行线性插值。）

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()

更新2018-12-13 ：如果有必要找到几个截取，这里是修改版本的代码：

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"""

    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

    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()

```

在Python中以高精度查找由（x，y）数据给出的两条曲线的交集

1 个答案: