检查多边形/多多边形中是否包含一个点-对于许多点

Question

我有一个二维地图，分为矩形网格（其中约45,000个）和许多从shapefile派生的多边形/多面体（我目前使用shapefile库pyshp读取它们）。不幸的是，这些多边形中的一些非常复杂，并且由大量点组成（其中一个点具有640,000个点），并且其中可能有孔。

我要尝试的是针对每个这些多边形检查哪些单元格中心（网格的单元格）落在该特定多边形内。但是，拥有约45,000个像元中心和150个多边形需要花费相当长的时间才能使用形状检查所有内容。这是我正在或多或少要做的事情：

# nx and ny are the number of cells in the x and y direction respectively
# x, y are the coordinates of the cell centers in the grid - numpy arrays
# shapes is a list of shapely shapes - either Polygon or MultiPolygon
# Point is a shapely.geometry.Point class
for j in range(ny):
    for i in range(nx):
        p = Point([x[i], y[j]])
        for s in shapes:
            if s.contains(p):
                # The shape s contains the point p - remove the cell

根据所涉及的多边形，每次检查要花费200毫秒到80毫秒之间的时间，并且要进行600万次以上的检查，运行时间很容易就花了几个小时。

我想必须有一个更明智的方式来解决这个问题-如果可能的话，我宁愿保持身材，但任何建议都将受到赞赏。

谢谢。

Answer 1

在接受的答案中，使用 rtrees in shapely 仅支持多边形的扩展 - 或边界框。正如匀称的文档中所述：

https://shapely.readthedocs.io/en/stable/manual.html#str-packed-r-tree

从 1.7.1 开始，rtree.query 需要检查多边形是否实际包含该点。这意味着额外的检查，但如果多边形不是非常混乱的配置，仍然可以加速算法

也就是说，代码应该是这样的

from shapely.strtree import STRtree
from shapely.geometry import Polygon, Point    

# this is the grid. It includes a point for each rectangle center. 
points = [Point(i, j) for i in range(10) for j in range(10)]
tree = STRtree(points) # create a 'database' of points

# this is one of the polygons. 
p = Polygon([[0.5, 0], [2, 0.2], [3, 4], [0.8, 0.5], [0.5, 0]])

res = [o for o in tree.query(p) if p.contains(o)] # run the query (a single shot) - and test if the found points are actually inside the polygon. 

# do something with the results
print([o.wkt for o in res])

确实是现在的结果

['POINT (2 1)', 'POINT (2 2)']

可以看出这里是多边形内部的点：

-- 奖金编辑 ---

通过在以下意义上定位问题，我观察到了更好的速度改进：

在该区域上制作一个方形网格，STRtree 结构由点和多边形组成。在每个单独的网格多边形上查询点和多边形 - 在每个网格多边形内，检查查询点是否包含在查询多边形内。从而减少对单个多边形的检查量。

from shapely.strtree import STRtree
from shapely.geometry import Polygon, Point
import random    

# build a grid spread over the area. 
grid = [Polygon([[i, j],[i+1,j],[i,j+1],[i+1,j+1]]) for i in range(10) for j in range(10)]

pointtree = STRtree([Point(random.uniform(1,10),random.uniform(1,10)) for q in range(50)]) # create a 'database' of 50 random points - and build a STRtree structure over it

# take the same polygon as above, but put in an STRtree 
polytree = STRtree([Polygon([[0.5, 0], [2, 0.2], [3, 4], [0.8, 0.5], [0.5, 0]])])
res = []
#do nested queries inside the grid
for g in grid:
    polyquery = polytree.query(g)
    pointquery = pointtree.query(g)
    for point in pointquery:
        for poly in polyquery:
            if poly.contains(point):
                 res.append(point)

# do something with the results
print([o.wkt for o in res])

Answer 2

如果您希望执行较少的比较操作，则可以尝试使用匀称的str-tree功能。考虑以下代码：

    npm install -g browser-sync
npm ERR! Linux 4.15.0-101-generic
npm ERR! argv "/usr/bin/nodejs" "/usr/bin/npm" "install" "-g" "browser-sync"
npm ERR! node v4.2.6
npm ERR! npm  v3.5.2
npm ERR! code ENOTFOUND
npm ERR! errno ENOTFOUND
npm ERR! syscall getaddrinfo
npm ERR! network getaddrinfo ENOTFOUND registry.npmjs.org registry.npmjs.org:443
npm ERR! network This is most likely not a problem with npm itself
npm ERR! network and is related to network connectivity.
npm ERR! network In most cases you are behind a proxy or have bad network settings.
npm ERR! network 
npm ERR! network If you are behind a proxy, please make sure that the
npm ERR! network 'proxy' config is set properly.  See: 'npm help config'
npm ERR! Please include the following file with any support request:
npm ERR!     /home/krishna/npm-debug.log`

此代码的结果是：

from shapely.strtree import STRtree
from shapely.geometry import Polygon, Point

# this is the grid. It includes a point for each rectangle center. 
points = [Point(i, j) for i in range(10) for j in range(10)]
tree = STRtree(points). # create a 'database' of points

# this is one of the polygons. 
p = Polygon([[0.5, 0], [2, 0.2], [3, 4], [0.8, 0.5], [0.5, 0]])

res = tree.query(p). # run the query (a single shot). 

# do something with the results
print([o.wkt for o in res])