找到matplotlib中绘制的两条线图的交点

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='^')

我尝试引用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，可以理解API从文档中做了什么，但没有真正得到背后的概念它，如果有人能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，那将会有很大的帮助。

Answer 1

这是一个丑陋的解决方案（改进版本位于底部）。绘图后，我们知道两个折线图在(6, 7)

的范围内形成一个十字

现在，我们使用以下源代码

绘制此交叉点

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横向范围。这是一个通过循环每两个段的改进版本，但它可能是一个时间消费者。

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

我已经扩展了@SparkAndShine的解决方案以使用3D数据，以及使用KD树进行了一些性能增强。完整的解决方案在此处发布：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()