python中的加速地理定位算法

问题描述

我有一组100k的地理位置(纬度/经度)和一个六边形网格(4k多边形).我的目标是计算每个多边形内的点总数.

I have a set 100k of of geo locations (lat/lon) and a hexogonal grid (4k polygons). My goal is to calculate the total number of points which are located within each polygon.

我当前的算法使用2个for循环遍历所有地理点和所有多边形，如果我增加多边形的数量，这的确很慢...您将如何加快算法的速度?我上传了一个最小的示例，该示例创建了10万个随机地理位置，并在网格中使用了561个单元格...

My current algorithm uses 2 for loops to loop over all geo points and all polygons, which is really slow if I increase the number of polygons... How would you speedup the algorithm? I have uploaded a minimal example which creates 100k random geo points and uses 561 cells in the grid...

我还看到读取geo json文件(带有4k多边形)需要一些时间，也许我应该将多边形导出到csv中?

I also saw that reading the geo json file (with 4k polygons) takes some time, maybe i should export the polygons into a csv?

hexagon_grid.geojson文件: https://gist.github.com/Arnold1/9e41454e6eea910a4f6cd68ff1901db1

hexagon_grid.geojson file:https://gist.github.com/Arnold1/9e41454e6eea910a4f6cd68ff1901db1

最小的python示例: https://gist.github.com/Arnold1/ee37a2e4b2dfbfdca9bfae7c7c3a3755

minimal python example:https://gist.github.com/Arnold1/ee37a2e4b2dfbfdca9bfae7c7c3a3755

推荐答案

您无需显式测试每个六边形即可查看给定点是否位于其中.

You don't need to explicitly test each hexagon to see whether a given point is located inside it.

目前，让我们假设您所有的点都位于六角形网格的范围内.由于您的六边形形成规则的晶格，因此您只需要真正知道哪个六边形中心最靠近每个点.

Let's assume, for the moment, that all of your points fall somewhere within the bounds of your hexagonal grid. Because your hexagons form a regular lattice, you only really need to know which of the hexagon centers is closest to each point.

使用 scipy.spatial.cKDTree :

import numpy as np
from scipy.spatial import cKDTree
import json

with open('/tmp/grid.geojson', 'r') as f:
    data = json.load(f)

verts = []
centroids = []

for hexagon in data['features']:

    # a (7, 2) array of xy coordinates specifying the vertices of the hexagon.
    # we ignore the last vertex since it's equal to the first
    xy = np.array(hexagon['geometry']['coordinates'][0][:6])
    verts.append(xy)

    # compute the centroid by taking the average of the vertex coordinates
    centroids.append(xy.mean(0))

verts = np.array(verts)
centroids = np.array(centroids)

# construct a k-D tree from the centroid coordinates of the hexagons
tree = cKDTree(centroids)

# generate 10000 normally distributed xy coordinates
sigma = 0.5 * centroids.std(0, keepdims=True)
mu = centroids.mean(0, keepdims=True)
gen = np.random.RandomState(0)
xy = (gen.randn(10000, 2) * sigma) + mu

# query the k-D tree to find which hexagon centroid is nearest to each point
distance, idx = tree.query(xy, 1)

# count the number of points that are closest to each hexagon centroid
counts = np.bincount(idx, minlength=centroids.shape[0])

绘制输出:

from matplotlib import pyplot as plt

fig, ax = plt.subplots(1, 1, subplot_kw={'aspect': 'equal'})
ax.hold(True)
ax.scatter(xy[:, 0], xy[:, 1], 10, c='b', alpha=0.25, edgecolors='none')
ax.scatter(centroids[:, 0], centroids[:, 1], marker='h', s=(counts + 5),
           c=counts, cmap='Reds')
ax.margins(0.01)

我可以考虑几种不同的方式来处理点，这些点取决于所需的精度:

I can think of several different ways you could handle points that fall outside your grid depending on how much accuracy you need:

• 您可以排除位于六边形顶点的外边界矩形之外的点(即x < xmin，x > xmax等).但是，这将无法排除沿着网格边缘位于间隙"内的点.

• You could exclude points that fall outside the outer bounding rectangle of your hexagon vertices (i.e. x < xmin, x > xmax etc.). However, this will fail to exclude points that fall within the 'gaps' along the edges of your grid.

另一个简单的选择是根据您的六边形中心的间距在distance上设置一个截止点，这相当于对您的外部六边形使用圆形近似.

Another straightforward option would be to set a cut-off on distance according to the spacing of your hexagon centers, which is equivalent to using a circular approximation for your outer hexagons.

如果准确性至关重要，那么您可以定义一个与六边形网格的外部顶点对应的matplotlib.path.Path，然后使用其 .contains_points()方法来测试您的点是否包含在其中.与其他两种方法相比，这可能会更慢并且更易于编写代码.

If accuracy is crucial then you could define a matplotlib.path.Path corresponding to the outer vertices of your hexagonal grid, then use its .contains_points() method to test whether your points are contained within it. Compared to the other two methods, this would probably be slower and more fiddly to code.

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