如何在matplotlib中找到两条绘制曲线之间的交点？

Question

我想找到两条曲线的交点。例如下面的示例。 可以有多个交点。现在，我正在通过找到x,y坐标之间的距离来找到交点。但是这种方法有时无法在交点位于（17-18 x轴）之间时给出准确的点，如在图中所示。

我需要从曲线中获取所有点才能解决此问题。有什么方法可以让所有人都得到吗？

Answer 1

曲线只是连接每个点的一系列直线。因此，如果要增加点数，可以简单地在每对点之间进行线性外推：

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Answer 2

这是我的方法。我首先仅使用12个采样点创建了两条测试曲线，以说明这一概念。创建带有采样点的阵列后，曲线的精确方程式丢失。

然后，搜索两条曲线之间的交点。通过逐点遍历数组，并检查一条曲线何时从另一条曲线的下方到另一条曲线的上方（或相反），可以通过求解线性方程来计算交点。

然后绘制交点以直观地检查结果。

import numpy as np
from matplotlib import pyplot as plt

N = 12
t = np.linspace(0, 50, N)
curve1 = np.sin(t*.08+1.4)*np.random.uniform(0.5, 0.9) + 1
curve2 = -np.cos(t*.07+.1)*np.random.uniform(0.7, 1.0) + 1
# note that from now on, we don't have the exact formula of the curves, as we didn't save the random numbers
# we only have the points correspondent to the given t values

fig, ax = plt.subplots()
ax.plot(t, curve1,'b-')
ax.plot(t, curve1,'bo')
ax.plot(t, curve2,'r-')
ax.plot(t, curve2,'ro')

intersections = []
prev_dif = 0
t0, prev_c1, prev_c2 = None, None, None
for t1, c1, c2 in zip(t, curve1, curve2):
    new_dif = c2 - c1
    if np.abs(new_dif) < 1e-12: # found an exact zero, this is very unprobable
        intersections.append((t1, c1))
    elif new_dif * prev_dif < 0:  # the function changed signs between this point and the previous
        # do a linear interpolation to find the t between t0 and t1 where the curves would be equal
        # this is the intersection between the line [(t0, prev_c1), (t1, c1)] and the line [(t0, prev_c2), (t1, c2)]
        # because of the sign change, we know that there is an intersection between t0 and t1
        denom = prev_dif - new_dif
        intersections.append(((-new_dif*t0  + prev_dif*t1) / denom, (c1*prev_c2 - c2*prev_c1) / denom))
    t0, prev_c1, prev_c2, prev_dif = t1, c1, c2, new_dif
print(intersections)

ax.plot(*zip(*intersections), 'go', alpha=0.7, ms=10)
plt.show()