在具有最小最近邻距离和最大密度的 3D space 中随机采样给定点

Question

我在 3D space 中有 n 个点。我想随机采样所有最近邻距离大于 r 的点子集。子集的大小 m 未知，但我希望采样点尽可能密集，即最大化 m.

类似的题也有，但都是关于生成点，而不是从给定的点采样。
Generate random points in 3D space with minimum nearest-neighbor distance

Generate 3-d random points with minimum distance between each of them?

假设我有 300 个随机 3D 点，

import numpy as np
n = 300
points = np.random.uniform(0, 10, size=(n, 3))

在最大化 m 的同时获得具有最小最近邻距离 r = 1 的 m 点子集的最快方法是什么？

Answer 1

这可能不会太快，但是迭代 3D 距离公式，附加到字典，排序，然后获取 id。

3D距离公式给定点(x, y, z)和(x1, y1, z1)：

m = ((x-x1)^2 + (y-y1)^2)^0.5
d = (m^2 + (z-z1)^2)^0.5

（d是总距离，^0.5是平方。）

Answer 2

可能有一个有效的双准则近似方案，但整数规划平均来说这么快，为什么还要费心呢？

import numpy as np

n = 300
points = np.random.uniform(0, 10, size=(n, 3))

from ortools.linear_solver import pywraplp

solver = pywraplp.Solver.CreateSolver("SCIP")
variables = [solver.BoolVar("x[{}]".format(i)) for i in range(n)]
solver.Maximize(sum(variables))
for j, q in enumerate(points):
    for i, p in enumerate(points[:j]):
        if np.linalg.norm(p - q) <= 1:
            solver.Add(variables[i] + variables[j] <= 1)
solver.EnableOutput()
solver.Solve()
print(len([i for (i, variable) in enumerate(variables) if variable.SolutionValue()]))

Answer 3

这不是最佳的大子集，但应该很接近而不需要很长时间，使用 KDTree 优化距离计算：

from scipy.spatial import KDTree
import numpy as np

def space_sample(n = 300, low = 0, high = 10, dist = 1):
    points = np.random.uniform(low, high, size=(n, 3))
    k = KDTree(points)
    pairs = np.array(list(k.query_pairs(dist)))
    
    def reduce_pairs(pairs, remove = []):  #iteratively remove the most connected node
        p = pairs[~np.isin(pairs, remove).any(1)]
        if p.size >0:
            count = np.bincount(p.flatten(), minlength = n)
            r = remove + [count.argmax()]
            return reduce_pairs(p, r)
        else:
            return remove
    
    return np.array([p for i, p in enumerate(points) if not(i in reduce_pairs(pairs))])

subset = space_sample()

迭代删除连接最多的节点不是最优（保留 300 个点中的大约 200 个），但可能是最快的算法 close 到最佳（唯一更快的是随机删除）。您可能 @njit reduce_pairs 使该部分更快（如果我以后有时间会尝试）。

Answer 4

用 30 个给定点测试@David Eisenstat 的答案：

from ortools.linear_solver import pywraplp
import numpy as np

def subset_David_Eisenstat(points, r):
    solver = pywraplp.Solver.CreateSolver("SCIP")
    variables = [solver.BoolVar("x[{}]".format(i)) for i in range(len(points))]
    solver.Maximize(sum(variables))
    for j, q in enumerate(points):
        for i, p in enumerate(points[:j]):
            if np.linalg.norm(p - q) <= r:
                solver.Add(variables[i] + variables[j] <= 1)
    solver.EnableOutput()
    solver.Solve()
    indices = [i for (i, variable) in enumerate(variables) if variable.SolutionValue()]
    return points[indices]

points = np.array(
[[ 7.32837882, 12.12765786, 15.01412241],
 [ 8.25164031, 11.14830379, 15.01412241],
 [ 8.21790113, 13.05647987, 13.05647987],
 [ 7.30031002, 13.08276009, 14.05452502],
 [ 9.18056467, 12.0800735 , 13.05183844],
 [ 9.17929647, 11.11270337, 14.03027534],
 [ 7.64737905, 11.48906945, 15.34274827],
 [ 7.01315886, 12.77870699, 14.70301668],
 [ 8.88132414, 10.81243313, 14.68685022],
 [ 7.60617372, 13.39792166, 13.39792166],
 [ 8.85967682, 12.72946394, 12.72946394],
 [ 9.50060705, 11.43361294, 13.37866092],
 [ 8.21790113, 12.08471494, 14.02824481],
 [ 7.32837882, 12.12765786, 16.98587759],
 [ 8.25164031, 11.14830379, 16.98587759],
 [ 7.30031002, 13.08276009, 17.94547498],
 [ 8.21790113, 13.05647987, 18.94352013],
 [ 9.17929647, 11.11270337, 17.96972466],
 [ 9.18056467, 12.0800735 , 18.94816156],
 [ 7.64737905, 11.48906945, 16.65725173],
 [ 7.01315886, 12.77870699, 17.29698332],
 [ 8.88132414, 10.81243313, 17.31314978],
 [ 7.60617372, 13.39792166, 18.60207834],
 [ 8.85967682, 12.72946394, 19.27053606],
 [ 9.50060705, 11.43361294, 18.62133908],
 [ 8.21790113, 12.08471494, 17.97175519],
 [ 7.32837882, 15.01412241, 12.12765786],
 [ 8.25164031, 15.01412241, 11.14830379],
 [ 7.30031002, 14.05452502, 13.08276009],
 [ 9.18056467, 13.05183844, 12.0800735 ],])

当期望的最小距离为1时：

subset1 = subset_David_Eisenstat(points, r=1.)
print(len(subset1))
# Output: 18

检查最小距离：

from scipy.spatial.distance import cdist
dist = cdist(subset1, subset1, metric='euclidean')
# Delete diagonal
res = dist[~np.eye(dist.shape[0],dtype=bool)].reshape(dist.shape[0],-1)
print(np.min(res))
# Output: 1.3285513450926985

将预期的最小距离更改为 2:

subset2 = subset_David_Eisenstat(points, r=2.)
print(len(subset2))
# Output: 10

检查最小距离：

from scipy.spatial.distance import cdist
dist = cdist(subset2, subset2, metric='euclidean')
# Delete diagonal
res = dist[~np.eye(dist.shape[0],dtype=bool)].reshape(dist.shape[0],-1)
print(np.min(res))
# Output: 2.0612041004376223

在具有最小最近邻距离和最大密度的 3D space 中随机采样给定点

Sample given points stochastically in a 3D space with minimum nearest-neighbor distance and maximum density

python

random

algorithm

numpy

nearest-neighbor