如何使用 Spatial.kdTree 的树获取具有 point_id 的对象点的最近邻居

Question

帮帮我。我被困住了。我有一个class 或point。该类的属性是lat、Long、point_id、name。有些点具有相同的Lat 和Long。所以，我需要point_id 作为积分标识。实际上，我想使用kdTree 的树来获得我点的最近邻居。我在构造kdTree 时没有问题，但是当我想知道我的点最近邻居时，结果是最近邻居的列表(Lat,Long) 而我想要point_id 作为我的结果，因为有些点具有相同的@987654334 @ 和 Long。

这是我的代码：

import numpy as np
from scipy import spatial
import psycopg2
import math
import psycopg2.extensions   

class Point:
def __init__(self, id_point, name, Lat, Long):
    self.id_point=id_point
    self.name=name
    self.Lat=Lat
    self.Long=Long

def Struktur_KdTree(points):
    tree= spatial.KDTree(points)
    return tree

def getNearestPoint(tree,point,radius):
    return tree.query_ball_point(point,radius)

try:
   conn = psycopg2.connect("dbname='Chicago_crime' user='postgres' host='localhost' password='1234'")
except :
    print "I'm unable to connect to the database"

cur = conn.cursor()
cur.execute("""SELECT id_objek, primary_descript, lat, long from data_crime""")

data = []
Lat = []
Long = []
for id_jenis, jenis, lat, long in cur.fetchall():
     data.append(Point(id_point,name,Lat,Long))
     Lat.append(lat)
     Long.append(long)

dataPoints = zip (Lat,Long)

tree = Struktur_KdTree(dataPoints)
result=[]
radius=2
for point in dataPoint:
   result.append(getNearestPoint(tree,point,radius))

请给我一些建议？

Answer 1

使用字典（或collections.defaultdict）记录从(lat, lng) 元组到Points 列表的映射。 有了这个字典（我们称之为pointmap），给定一个(lat, lng)，你会 能够在该位置查找所有Points。

pointmap = collections.defaultdict(list)
locations = set()
for id_jenis, jenis, lat, lng in np.random.randint(10, size=(N, 4)):
    point = Point(id_jenis, jenis, lat, lng)
    pointmap[lat, lng].append(point)

pointmap dict 中的键是 (lat, lng) 元组，我们可以将它们形成一个 NumPy 数组传递给spatial.KDTree：

locations = np.array(pointmap.keys())
tree = spatial.KDTree(locations)

现在，我们可以遍历每个位置并找到 最近的点：

for loc in locations:
    indices = getNearestPoint(tree, loc, radius)

请注意，query_ball_point 返回一个索引列表（该索引进入 locations)。实际位置由locations[indices] 给出。我们在这里 利用 NumPy 花式索引——因为locations 是 一个 NumPy 数组。

对于location[indices] 中的每个位置，我们现在可以通过使用我们的字典pointmap 查找关联点来获得Points 的列表：

near_points = [point for near_loc in locations[indices]
               for point in pointmap[tuple(near_loc)]]

---

把它们放在一起：

import numpy as np
from scipy import spatial
import collections

class Point:

    def __init__(self, id_point, name, Lat, Long):
        self.id_point = id_point
        self.name = name
        self.Lat = Lat
        self.Long = Long

    def __repr__(self):
        return 'Point({}, {}, {}, {})'.format(
            self.id_point, self.name, self.Lat, self.Long)

def getNearestPoint(tree, point, radius):
    return tree.query_ball_point(point, radius)

pointmap = collections.defaultdict(list)

N = 10
for id_jenis, jenis, lat, lng in np.random.randint(10, size=(N, 4)):
    point = Point(id_jenis, jenis, lat, lng)
    pointmap[lat, lng].append(point)

locations = np.array(pointmap.keys())
tree = spatial.KDTree(locations)

result = []
radius = 2

for loc in locations:
    indices = getNearestPoint(tree, loc, radius)
    near_points = [point for near_loc in locations[indices]
                   for point in pointmap[tuple(near_loc)]]
    print('{} close to {}'.format(loc, near_points))

产生如下输出：

[8 3] close to [Point(2, 9, 8, 3)]
[7 1] close to [Point(8, 6, 7, 1), Point(2, 1, 6, 1)]
[4 5] close to [Point(7, 1, 4, 5), Point(4, 9, 3, 6)]
[6 1] close to [Point(8, 6, 7, 1), Point(2, 1, 6, 1)]
[2 0] close to [Point(0, 1, 2, 0), Point(4, 3, 4, 0)]
[3 6] close to [Point(7, 1, 4, 5), Point(4, 9, 3, 6)]
[7 9] close to [Point(1, 9, 7, 9)]
[9 6] close to [Point(8, 5, 9, 6)]
[2 4] close to [Point(4, 4, 2, 4)]
[4 0] close to [Point(0, 1, 2, 0), Point(4, 3, 4, 0)]