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，我们可以通过原帖中显示的“最近数据点”方法找到交叉点的大致位置。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位置。
这种方法非常快，并且适用于 2D numpy 数组，以防您想一次计算多条曲线的交叉点（就像我想在我的应用程序中做的那样）。
（我借用了“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()
2018 年 12 月 13 日更新：
如果需要找到多个拦截，这里是一个修改版本的代码：
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()
```