這是一個固定的網格,每個網格點認爲是存在的樣本數量版本。
然後可以將搜索縮小到相關點周圍的空間。
from random import randint
import math
N = 100
N_SAMPLES = 1000
# create the grid
grd = [[0 for _ in range(N)] for __ in range(N)]
# set the number of points at a given gridpoint
for _ in range(N_SAMPLES):
grd[randint(0, 99)][randint(0, 99)] += 1
def find_neighbours(grid, point, distance):
# this will be: (x, y): number of points there
points = {}
for x in range(point[0]-distance, point[0]+distance):
if x < 0 or x > N-1:
continue
for y in range(point[1]-distance, point[1]+distance):
if y < 0 or y > N-1:
continue
dst = math.hypot(point[0]-x, point[1]-y)
if dst > distance:
continue
if grd[x][y] > 0:
points[(x, y)] = grd[x][y]
return points
print(find_neighbours(grid=grd, point=(45, 36), distance=5))
# -> {(44, 37): 1, (45, 33): 1, ...}
# meadning: there is one neighbour at (44, 37) etc...
進一步optimzation:用於x
和y
測試可以預先計算對於給定gridsize - 在math.hypot(point[0]-x, point[1]-y)
就不必再爲完成每個點。
並且用numpy
陣列替換網格可能是個好主意。
UPDATE
如果你的觀點是float
是你還可以創建一個int
電網以減少搜索空間:
from random import uniform
from collections import defaultdict
import math
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
@property
def x_int(self):
return int(self.x)
@property
def y_int(self):
return int(self.y)
def __str__(self):
fmt = '''{0.__class__.__name__}(x={0.x:5.2f}, y={0.y:5.2f})'''
return fmt.format(self)
N = 100
MIN = 0
MAX = N-1
N_SAMPLES = 1000
# create the grid
grd = [[[] for _ in range(N)] for __ in range(N)]
# set the number of points at a given gridpoint
for _ in range(N_SAMPLES):
p = Point(x=uniform(MIN, MAX), y=uniform(MIN, MAX))
grd[p.x_int][p.y_int].append(p)
def find_neighbours(grid, point, distance):
# this will be: (x_int, y_int): list of points
points = defaultdict(list)
# need to cast a slightly bigger net on the upper end of the range;
# int() rounds down
for x in range(point[0]-distance, point[0]+distance+1):
if x < 0 or x > N-1:
continue
for y in range(point[1]-distance, point[1]+distance+1):
if y < 0 or y > N-1:
continue
dst = math.hypot(point[0]-x, point[1]-y)
if dst > distance + 1: # account for rounding... is +1 enough?
continue
for pt in grd[x][y]:
if math.hypot(pt.x-x, pt.y-y) <= distance:
points[(x, y)].append(pt)
return points
res = find_neighbours(grid=grd, point=(45, 36), distance=5)
for int_point, points in res.items():
print(int_point)
for point in points:
print(' ', point)
輸出看起來是這樣的:
(44, 36)
Point(x=44.03, y=36.93)
(41, 36)
Point(x=41.91, y=36.55)
Point(x=41.73, y=36.53)
Point(x=41.56, y=36.88)
...
爲了方便Points
現在是一類。可能沒有必要,但...這取決於你如何密集或稀疏點
你也可以代表網格爲指向列表或Points
字典...
也find_neighbours
函數接受一個開始僅在該版本中由int
組成的point
。這也可能會被改進。
還有很大的改進空間:y
軸的範圍可以使用三角法進行限制。而對於圈內的分數方式,則不需要單獨檢查;詳細的檢查只需要靠近圓圈的外緣完成。
如果你的網格只有100 * 100,你可以在網格中排列你的點。這樣你可以大大減少搜索空間。 –
http://gis.stackexchange.com/questions/22082/how-can-i-use-r-tree-to-find-points-within-a-distance-in-spatialite –