Numpy 点云中的局部极大值

我有一个点云C，其中每个点都有一个关联的值。假设点在二维空间中，所以每个点都可以用三元组（x，y，v）表示

我想找到局部极大值点的子集。也就是说，对于一些半径R，我想找到C中点S的子集，这样对于S中的任何点Pi（值为vi），在Pi的R距离内C中没有点Pj，其值vj大于vi

我知道如何在O（N^2）时间内完成这项工作，但这似乎是浪费。有没有一种有效的方法可以做到这一点

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旁注：

• 这个问题的根源是我试图在稀疏矩阵中找到局部极大值，所以在我的例子中，x，y是有序整数索引-如果这简化了问题，请告诉我
• 如果解决方案只针对曼哈顿距离或其他地方，我会非常高兴
• 我是python的，所以如果有一种很好的矢量化numpy方法来实现这一点，那就太好了

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我找到了这个解决方案，但它可能是O（N^2）：

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使用二维树（图形的二维实例）。经过N.Log（N）时间预处理后，它将允许您在大约Log（N）+K时间（平均找到K个邻居）内围绕所有点执行固定半径的近邻搜索，总共N.Log（N）+K.N。它将与曼哈顿距离完美匹配。

根据Yves的建议，下面是一个使用scipy的答案：

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谢谢但问题更为普遍。这解决了点在矩形网格中排列时的问题，并且由于掩蔽（尽管使用切片可能会更快），时间为O（N^2）。在我的问题中，这些点位于任意（x，y）位置。编辑速度比O（N^2）快，可以处理任意位置。谢谢，但仍然是N^2，因为对于每个点，您都会检查每个其他点是否在形成遮罩的半径内。看起来亚N^2解决方案需要使用K-D树或一些巧妙的排序。好的，我明白了。。。我还没有进入

kdtree

，所以我现在已经完成了！祝你好运你似乎自相矛盾。如果一个点在一个规则的网格上，完全改变了问题，但在commemt中你说它们是任意的，我不知道怎么做。在第一个解决方案中，将点表示为密集阵列，即矩形网格，这只是任意位置更一般情况下的一个特定实例。在特定矩阵情况下，您只需要搜索附近的索引。这简化了事情的顺序。注意——如果你想提高速度，不要使用SciPy的KDTree——而是使用SKLearn的，它快了300倍。这个答案仍然是金黄色的！非常适合：“from sklearn.KDTree；kdtree=kdtree（坐标）；邻居=kdtree.query_半径（坐标，r=5）'

import numpy as np

# generate test data
n = 10
foo = np.random.rand(n,n)

# fixed test data for visual back-checking
# foo = np.array([[ 0.12439309,  0.88878825,  0.21675684,  0.21422532,  0.7016789 ],
#                 [ 0.14486462,  0.40642871,  0.4898418 ,  0.41611303,  0.12764404],
#                 [ 0.41853585,  0.22216484,  0.36113181,  0.5708699 ,  0.3874901 ],
#                 [ 0.24314391,  0.22488507,  0.22054467,  0.25387521,  0.46272496],
#                 [ 0.99097341,  0.76083447,  0.37941783,  0.932519  ,  0.9668254 ]])

# list to collect local maxima
local_maxima = []

# distance in x / y to define region of interest around current center coordinate
# roi = 1 corresponds to a region of interest of 3x3 (except at borders)
roi = 1

# give pseudo-coordinates
x,y = np.meshgrid(range(foo.shape[0]), range(foo.shape[1]))

for i in range(foo.shape[0]):
    for j in range(foo.shape[1]):
        x0 = x[i,j]
        y0 = y[i,j]
        z0 = foo[i,j]
        # index calculation to avoid out-of-bounds error when taking sub-matrix
        mask_x = abs(x - x0) <= roi
        mask_y = abs(y - y0) <= roi
        mask = mask_x & mask_y
        if np.max(foo[mask]) == z0:
            local_maxima.append((i, j))

print local_maxima

import numpy as np

# generate test data
# points = np.random.rand(10,3)

points = np.array([[ 0.08198248,  0.25999721,  0.07041999],
                   [ 0.19091977,  0.05404123,  0.25826508],
                   [ 0.8842875 ,  0.90132467,  0.50512316],
                   [ 0.33320528,  0.74069399,  0.36643752],
                   [ 0.27789568,  0.14381512,  0.13405309],
                   [ 0.73586202,  0.4406952 ,  0.52345838],
                   [ 0.76639731,  0.70796547,  0.70692905],
                   [ 0.09164532,  0.53234394,  0.88298593],
                   [ 0.96164975,  0.60700481,  0.22605181],
                   [ 0.53892635,  0.95173308,  0.22371167]])

# list to collect local maxima
local_maxima = []

# distance in x / y to define region of interest around current center coordinate
radius = 0.25

for i in range(points.shape[0]):
        # radial mask with radius radius, could be beautified via numpy.linalg
        mask = np.sqrt((points[:,0] - points[i,0])**2 + (points[:,1] - points[i,1])**2) <= radius
        # if current z value equals z_max in current region of interest, append to result list
        if points[i,2] == np.max(points[mask], axis = 0)[2]:
            local_maxima.append(tuple(points[i]))

local_maxima = [
 (0.19091976999999999, 0.054041230000000003, 0.25826507999999998), 
 (0.33320527999999999, 0.74069399000000002, 0.36643752000000002), 
 (0.73586202000000001, 0.44069520000000001, 0.52345838), 
 (0.76639731, 0.70796546999999999, 0.70692904999999995), 
 (0.091645320000000002, 0.53234393999999996, 0.88298593000000003), 
 (0.53892635, 0.95173308000000001, 0.22371167)
]

from scipy.spatial.kdtree import KDTree
import numpy as np

def locally_extreme_points(coords, data, neighbourhood, lookfor = 'max', p_norm = 2.):
    '''
    Find local maxima of points in a pointcloud.  Ties result in both points passing through the filter.

    Not to be used for high-dimensional data.  It will be slow.

    coords: A shape (n_points, n_dims) array of point locations
    data: A shape (n_points, ) vector of point values
    neighbourhood: The (scalar) size of the neighbourhood in which to search.
    lookfor: Either 'max', or 'min', depending on whether you want local maxima or minima
    p_norm: The p-norm to use for measuring distance (e.g. 1=Manhattan, 2=Euclidian)

    returns
        filtered_coords: The coordinates of locally extreme points
        filtered_data: The values of these points
    '''
    assert coords.shape[0] == data.shape[0], 'You must have one coordinate per data point'
    extreme_fcn = {'min': np.min, 'max': np.max}[lookfor]
    kdtree = KDTree(coords)
    neighbours = kdtree.query_ball_tree(kdtree, r=neighbourhood, p = p_norm)
    i_am_extreme = [data[i]==extreme_fcn(data[n]) for i, n in enumerate(neighbours)]
    extrema, = np.nonzero(i_am_extreme)  # This line just saves time on indexing
    return coords[extrema], data[extrema]