查找在 matplotlib 中绘制的两个折线图的交点

Question

有没有办法在matplotlib中找到两个折线图的交点？

考虑代码

import numpy as np
import matplotlib.pyplot as plt


fig = plt.figure()


ax = fig.add_subplot(111)

ax.plot([1,2,3,4,5,6,7,8],[20,100,50,120,55,240,50,25],color='lightblue',linewidth=3)
ax.plot([3,4,5,6,7,8,9], [25,35,14,67,88,44,120], color='darkgreen', marker='^')

我尝试参考 Python - matplotlib: find intersection of lineplots ，但该方法似乎过于复杂 - 它涉及高级数学概念，如分段多项式插值，可以从文档中理解 API 在做什么，但并没有真正了解其背后的概念，如果有人可以提供更简单的解决方案或解释分段多项式解决方案中发生的事情，那将有很大帮助。

Answer 1

这是一个丑陋的解决方案（改进版本在底部）。绘制后，我们知道两个折线图在(6, 7)的范围内交叉

现在，我们用下面的源代码绘制这个交叉点，

import numpy as np
import matplotlib.pyplot as plt

fig = plt.figure()
ax = fig.add_subplot(111)

x1 = [1,2,3,4,5,6,7,8]
y1 = [20,100,50,120,55,240,50,25]
x2 = [3,4,5,6,7,8,9]
y2 = [25,35,14,67,88,44,120]

ax.plot(x1, y1, color='lightblue',linewidth=3)
ax.plot(x2, y2, color='darkgreen', marker='^')


# Plot the cross point

x3 = np.linspace(6, 7, 1000)        # (6, 7) intersection range
y1_new = np.linspace(240, 50, 1000) # (6, 7) corresponding to (240, 50) in y1
y2_new = np.linspace(67, 88, 1000)  # (6, 7) corresponding to (67, 88) in y2

idx = np.argwhere(np.isclose(y1_new, y2_new, atol=0.1)).reshape(-1)
ax.plot(x3[idx], y2_new[idx], 'ro')

plt.show()

---

最终用户不乐意手动输入交叉范围。这是一个改进的版本，每两个片段循环一次，但它可能是一个时间消耗者。

import numpy as np
import matplotlib.pyplot as plt

fig = plt.figure()
ax = fig.add_subplot(111)

x1 = [1,2,3,4,5,6,7,8]
y1 = [20,100,50,120,55,240,50,25]
x2 = [3,4,5,6,7,8,9]
y2 = [25,35,14,67,88,44,120]

ax.plot(x1, y1, color='lightblue',linewidth=3)
ax.plot(x2, y2, color='darkgreen', marker='^')

# Get the common range, from `max(x1[0], x2[0])` to `min(x1[-1], x2[-1])`   
x_begin = max(x1[0], x2[0])     # 3
x_end = min(x1[-1], x2[-1])     # 8

points1 = [t for t in zip(x1, y1) if x_begin<=t[0]<=x_end]  # [(3, 50), (4, 120), (5, 55), (6, 240), (7, 50), (8, 25)]
points2 = [t for t in zip(x2, y2) if x_begin<=t[0]<=x_end]  # [(3, 25), (4, 35), (5, 14), (6, 67), (7, 88), (8, 44)]

idx = 0
nrof_points = len(points1)
while idx < nrof_points-1:
    # Iterate over two line segments
    y_min = min(points1[idx][1], points1[idx+1][1]) 
    y_max = max(points1[idx+1][1], points2[idx+1][1]) 

    x3 = np.linspace(points1[idx][0], points1[idx+1][0], 1000)      # e.g., (6, 7) intersection range
    y1_new = np.linspace(points1[idx][1], points1[idx+1][1], 1000)  # e.g., (6, 7) corresponding to (240, 50) in y1
    y2_new = np.linspace(points2[idx][1], points2[idx+1][1], 1000)  # e.g., (6, 7) corresponding to (67, 88) in y2

    tmp_idx = np.argwhere(np.isclose(y1_new, y2_new, atol=0.1)).reshape(-1)
    if tmp_idx:
        ax.plot(x3[tmp_idx], y2_new[tmp_idx], 'ro')                 # Plot the cross point

    idx += 1

plt.show()

Answer 2

我扩展了@SparkAndShine 的解决方案以处理 3D 数据，并使用 KD-tree 进行了一些性能增强。完整的解决方案在这里发布：https://stackoverflow.com/a/51145981/4212158

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.spatial import cKDTree
from scipy import interpolate

fig = plt.figure()
ax = fig.add_axes([0, 0, 1, 1], projection='3d')
ax.axis('off')

def upsample_coords(coord_list):
    # s is smoothness, set to zero
    # k is degree of the spline. setting to 1 for linear spline
    tck, u = interpolate.splprep(coord_list, k=1, s=0.0)
    upsampled_coords = interpolate.splev(np.linspace(0, 1, 100), tck)
    return upsampled_coords

# target line
x_targ = [1, 2, 3, 4, 5, 6, 7, 8]
y_targ = [20, 100, 50, 120, 55, 240, 50, 25]
z_targ = [20, 100, 50, 120, 55, 240, 50, 25]
targ_upsampled = upsample_coords([x_targ, y_targ, z_targ])
targ_coords = np.column_stack(targ_upsampled)

# KD-tree for nearest neighbor search
targ_kdtree = cKDTree(targ_coords)

# line two
x2 = [3,4,5,6,7,8,9]
y2 = [25,35,14,67,88,44,120]
z2 = [25,35,14,67,88,44,120]
l2_upsampled = upsample_coords([x2, y2, z2])
l2_coords = np.column_stack(l2_upsampled)

# plot both lines
ax.plot(x_targ, y_targ, z_targ, color='black', linewidth=0.5)
ax.plot(x2, y2, z2, color='darkgreen', linewidth=0.5)

# find intersections
for i in range(len(l2_coords)):
    if i == 0:  # skip first, there is no previous point
        continue

    distance, close_index = targ_kdtree.query(l2_coords[i], distance_upper_bound=.5)

    # strangely, points infinitely far away are somehow within the upper bound
    if np.isinf(distance):
        continue

    # plot ground truth that was activated
    _x, _y, _z = targ_kdtree.data[close_index]
    ax.scatter(_x, _y, _z, 'gx')
    _x2, _y2, _z2 = l2_coords[i]
    ax.scatter(_x2, _y2, _z2, 'rx')  # Plot the cross point


plt.show()