如何实现rtree来计算相交面积?

时间:2019-05-07 11:08:19

标签: python-3.x geopandas r-tree

以下是一些示例数据:

import pandas as pd
import geopandas as gp
import shapely.geometry
from shapely.geometry import Polygon
from shapely.geometry import Point
import shapely.affinity
import matplotlib.pyplot as plt

df = gp.GeoDataFrame([['a', Polygon([(0,1), (1,1), (2,2), (1,2)])],
                     ['b', Polygon([(1.5,0.75), (2, 1.25), (3,0.25)])]],
                    columns = ['name', 'geometry'])

df = gp.GeoDataFrame(df, geometry = 'geometry')
df['area'] = df.area

points = gp.GeoDataFrame([['box', Point(1.2, 1.115), 4],
                         ['triangle', Point(2.5, 1.25), 8]],
                        columns = ['name', 'geometry', 'value'],
                        geometry = 'geometry')

buf = points.buffer(0.5, cap_style = 3)
points['buffer'] = buf
points = points.drop(['geometry'], axis = 1)
points = points.rename(columns = {'buffer': 'geometry'})

它看起来像这样:

enter image description here

基本上,我正在尝试查找这些对象之间的相交区域。

到目前为止,我已经使用以下代码完成了此操作:

data = []
for index, geo in df.iterrows():
    for index2, poin in points.iterrows():
        if geo['geometry'].intersects(poin['geometry']):
          data.append({'geometry':geo['geometry'].intersection(poin['geometry']), 'area': geo['geometry'].intersection(poin['geometry']).area})

df2 = gp.GeoDataFrame(data, columns = ['geometry', 'area'])

但是,我将要使用的实际数据具有100,000个多边形,因此此代码将非常耗时。我知道我可以通过使用r树来加快速度。但是,我似乎无法正确实现它。

我尝试过这样的事情:

spatial_index = df.sindex
results_list = []
for index, row in points.iterrows():
    buffer = row['geometry']
    possible_matches_index = list(spatial_index.intersection(buffer.bounds)) 
    possible_matches = df.iloc[possible_matches_index]
    results_list.append({'geometry':possible_matches['geometry'].intersection(row['geometry']), 'area': possible_matches['geometry'].intersection(row['geometry']).area})

df = gp.GeoDataFrame(results_list, columns = ['geometry', 'area'])

但是,这会将每个正方形的所有交点放在一条直线上。

    geometry                                        area
0   name a POLYGON ((1.615 1.615, 1 1, 0.7 1, 0...  name a 0.000037 b 0.000003 dtype: float64
1                                    name a ...     name a 0.000000 b 0.000013 dtype: float64

如何获取此数据以生成一个数据框,该数据框的每个相交点都有一条直线,而其几何形状和面积则由列组成?

1 个答案:

答案 0 :(得分:1)

要计算两个GeoDataFrame之间的交点,可以使用geopandas.overlay函数:

geopandas.overlay(df, points, how='intersection')

这将利用引擎盖下的rtree空间索引(因此应该比蛮力double for循环更为有效),并将两个数据集的所有几何组合的交集作为新的交集返回GeoDataFrame(然后您可以为其计算面积)。

有关此文档,请参见https://geopandas.readthedocs.io/en/latest/set_operations.html