我想找到两条曲线的交点。例如下面的例子。可以有多个交点。现在,我通过找到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,如图所示。
我需要从曲线中获取所有点才能解决此问题。是否有任何方法可以获取所有这些信息?
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。创建带有采样点的数组后,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(或反向),可以通过求解线性方程来计算交点。
之后绘制相交点以目视检查结果。
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()