Python中两个图的交集，找到x值

Question

让0

plt.plot(x, f, '-')
plt.plot(x, g, '*')

我想找到曲线相交的点“x”。我不想找到f和g的交点。我可以通过以下方法简单地找到：

set(f) & set(g)

Answer 1

以下是一个解决方案：

• 适用于N维数据
• 使用欧几里德距离，而不仅仅是在y轴上查找交叉点
• 对于大量数据更有效（它查询KD树，KD树应该以对数时间而不是线性时间进行查询）
• 您可以在KD树查询中更改距离上限，以定义距离是否足够近
• 如果需要，您可以同时查询具有多个点的KD树。注意：如果您需要一次查询数千个点，您可以通过使用另一个KD树查询KD树来显著提高性能

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()

Answer 2

对于那些正在使用或开放使用Shapely库进行几何体相关计算的人来说，获取交点会容易得多。您只需从每条线构造LineString，并获得它们的交点，如下所示：

import numpy as np
import matplotlib.pyplot as plt
from shapely.geometry import LineString

x = np.arange(0, 1000)
f = np.arange(0, 1000)
g = np.sin(np.arange(0, 10, 0.01) * 2) * 1000

plt.plot(x, f)
plt.plot(x, g)

first_line = LineString(np.column_stack((x, f)))
second_line = LineString(np.column_stack((x, g)))
intersection = first_line.intersection(second_line)

if intersection.geom_type == 'MultiPoint':
    plt.plot(*LineString(intersection).xy, 'o')
elif intersection.geom_type == 'Point':
    plt.plot(*intersection.xy, 'o')

要获得x和y值作为NumPy数组，您只需编写：

x, y = LineString(intersection).xy
# x: array('d', [0.0, 149.5724669847373, 331.02906176584617, 448.01182730277833, 664.6733061190541, 743.4822641140581])
# y: array('d', [0.0, 149.5724669847373, 331.02906176584617, 448.01182730277833, 664.6733061190541, 743.4822641140581])

或者，如果交点仅为一个点：

x, y = intersection.xy

Answer 3

您可以使用np。与np组合使用符号。差异和np。argwhere获取线交叉点的索引（在本例中，点是[0、149、331、448、664、743]）：

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(0, 1000)
f = np.arange(0, 1000)
g = np.sin(np.arange(0, 10, 0.01) * 2) * 1000

plt.plot(x, f, '-')
plt.plot(x, g, '-')

idx = np.argwhere(np.diff(np.sign(f - g))).flatten()
plt.plot(x[idx], f[idx], 'ro')
plt.show()

首先，它使用np计算f-g和相应的符号。签名。应用np。diff显示符号变化的所有位置（例如，线交叉）。使用np。argwhere为我们提供了准确的索引。