根据 cmets 更新答案：
你有很多选择。
这里有 3 种不同的方法：
1.使用 scipy.CKDTree：
优点：
缺点：
- 不太准确，因为计算出的距离是欧式的
- 半径将与您的输入相同，所以这里以度为单位
我会使用 cKDTree 和半径查询来查找半径中的所有点，从列表中删除这些点，然后继续剩余点。这不是最佳的，但可以作为一个很好的基础。
from scipy.spatial import cKDTree
points = [(41.81014,-72.550028), (41.995833,-72.581525), (41.377211,-72.150307), (41.710626,-72.763862), (41.55254,-72.815454), (41.415022,-73.401914), (41.0554,-73.54142), (41.660572,-72.725673), (41.350949,-72.871673), (41.280278,-72.987515), (41.23354,-73.151677), (41.235174,-73.038092), (41.58254,-73.034321), (41.89121,-72.6521), (41.340446,-73.078943), (41.81886,-73.0755), (41.228735,-73.225326), (41.839019,-71.883778), (41.585192,-71.99693), (41.611472,-72.901357), (41.783976,-72.748229), (43.634242,-70.347774), (44.842191,-68.74156), (43.934038,-69.985271), (43.474,-70.5141), (44.312403,-69.804993), (42.552616,-70.937616), (41.877743,-71.068577), (41.940344,-71.351931), (42.399035,-71.071855), (42.168221,-72.642232), (42.518609,-71.135461), (42.160827,-71.498868), (42.481583,-71.024154), (42.305328,-71.398387), (42.29247,-71.7751), (41.796058,-71.321145), (42.376695,-71.090028), (42.364178,-71.156462), (41.971125,-70.716858), (42.280435,-71.655929), (42.359487,-71.607159), (42.503468,-70.919421), (42.194395,-71.774687), (42.357311,-72.547241), (42.328872,-71.062845), (42.033714,-71.310581), (42.39976,-71.000326), (42.527193,-71.71374), (42.495264,-73.206116), (41.63729,-71.003268), (42.110519,-70.927683), (42.152383,-71.073541), (42.02714,-71.1438), (42.740784,-71.161323), (41.773672,-70.745562), (42.788072,-71.115959), (42.623622,-71.318304), (42.137401,-70.83883), (42.348748,-71.504967), (41.749066,-71.207427), (42.2045,-71.1553), (42.22142,-71.021844), (42.589718,-71.159895), (42.344172,-71.099961), (42.364561,-71.102575), (42.2882972,-71.1267483), (42.350679,-71.114022), (42.494932,-71.103401), (42.42072,-71.09902), (42.388648,-71.118659), (42.484104,-71.186185), (41.666927,-70.294616), (42.275401,-71.029299), (42.299241,-71.062748), (42.361045,-71.0626), (42.764475,-71.215039), (43.2189,-71.485199), (42.702771,-71.437791), (43.045615,-71.461202), (42.79899,-71.53679), (42.941002,-71.473513), (42.928188,-72.301906), (43.235048,-70.884519), (43.048951,-70.818587), (43.633682,-72.322002), (44.466154,-73.18226)]
# Radius of circle. Note that the unit is the same as in your list (here, degrees.)
radius = 1
num_circles = 0
list_is_no_empty = True
while(list_is_no_empty):
# Take the first point in order to find all points within distance radius
start_point = points[0]
# Create a KDTree
tree = cKDTree(points)
# Find indexes of all points in radius
indexes_of_points_in_radius = tree.query_ball_point(start_point, radius)
# Create the list of points to remove (points that were found within distance radius)
points_to_remove = [points[i] for i in indexes_of_points_in_radius]
# Remove these points
points = list(set(points) - set(points_to_remove))
# Increment the number of circles
num_circles += 1
# If no points remain, exit loop
if points == []:
list_is_no_empty = False
print("Number of circles:", num_circles)
2。使用 sklearn.neighbors.BallTree：
优点：
- 这会更准确，因为此处计算的距离是大圆距离（Haversine 公式）。
缺点：
- 与 cKDTree 一样，半径将与您的输入相同，因此以度为单位。
- 比 scipy.cKDTree 慢一点（我测试时慢了 2 倍）
还要注意，我发现有些人建议将您的输入转换为弧度，因为这对于 Haversine 公式 (https://scikit-learn.org/stable/modules/generated/sklearn.neighbors.DistanceMetric.html) 是必需的。但根据我的测试，使用 scikit-learn 1.0.1，这不是必需的。但以防万一，你会这样做：
from math import radians
points = [tuple(map(radians, point)) for point in points]
start_point = (radians(start_point[0]), radians(start_point[1]))
radius = radians(radius)
使用 BallTree 编写代码：
from sklearn.neighbors import BallTree
import numpy as np
num_circles = 0
list_is_no_empty = True
while(list_is_no_empty):
# Take the first point in order to find all points within distance radius
start_point = np.array([points[0]])
# Create a BallTree, and chose the Haversine formula (great circle distance)
tree = BallTree(points, metric="haversine")
# Find indexes of all points in radius
indexes_of_points_in_radius = tree.query_radius(start_point, r=radius)[0]
# Create the list of points to remove (points that were found within distance radius)
points_to_remove = [points[i] for i in indexes_of_points_in_radius]
# Remove these points
points = list(set(points) - set(points_to_remove))
# Increment the number of circles
num_circles += 1
# If no points remain, exit loop
if points == []:
list_is_no_empty = False
print("Number of circles:", num_circles)
3。使用 sklearn.neighbors.BallTree，使用用户定义的距离函数：
优点：
- 我们将能够在这里使用非常准确的距离
- 我们将能够以英里（或米）为单位指定此距离
缺点：
- 比 cKDTree 慢很多（我测试时是 10 倍）
from pyproj import Geod
from sklearn.neighbors import BallTree
import numpy as np
# Create a WGS84 ellipsoid
geod = Geod(ellps='WGS84')
# User-defined function for BallTree
# We use the "inv" method of pyproj in order to get the distance in meters between 2 points
# It computes the geodesic distance using the wonderful Karney's algorithm
def geodedsic_distance(point_01, point_02):
lat1,lon1 = point_01
lat2,lon2 = point_02
az12,az21,distance_in_meters = geod.inv(lon1,lat1,lon2,lat2)
return distance_in_meters
def miles_to_meters(miles):
return miles * 1609.344
# Radius in miles
radius_in_miles = 50
radius_in_meters = miles_to_meters(50)
num_circles = 0
list_is_no_empty = True
while(list_is_no_empty):
# Take the first point in order to find all points within distance radius
start_point = np.array([points[0]])
# Create a BallTree, and chose our custom function
tree = BallTree(points, metric=geodedsic_distance)
# Find indexes of all points in radius, specified in meters
indexes_of_points_in_radius = tree.query_radius(start_point, r=radius_in_meters)[0]
# Create the list of points to remove (points that were found within distance radius)
points_to_remove = [points[i] for i in indexes_of_points_in_radius]
# Remove these points
points = list(set(points) - set(points_to_remove))
# Increment the number of circles
num_circles += 1
# If no points remain, exit loop
if points == []:
list_is_no_empty = False
print("Number of circles:", num_circles)
如果您想了解更多关于英里到度数的转换（以及为什么，事实上，我们不能）和计算地球上的距离：
Is the Haversine Formula or the Vincenty's Formula better for calculating distance?
https://gis.stackexchange.com/questions/84885/difference-between-vincenty-and-great-circle-distance-calculations