如何将 GEOS MultiLineString 转换为 Polygon?

发布于 2024-09-04 05:16:22 字数 196 浏览 3 评论 0原文

我正在开发一个 GeoDjango 应用程序,用户可以在其中上传地图文件并执行一些基本的地图操作,例如查询多边形内的要素。

我认识到用户有时会上传“MultiLineString”而不是“Polygon”。这会导致期望闭合几何图形的查询失败。

在 Python 中将 MultiLineString 对象转换为 Polygon 的最佳方法是什么?

I am developing a GeoDjango application where users can upload map files and do some basic mapping operations like querying features inside polygons.

I recognized that users happen to upload "MultiLineString"s instead of "Polygon"s sometimes. This causes the queries expecting closed geometries to fail.

What is the best way to convert a MultiLineString object to a Polygon in Python?

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好菇凉咱不稀罕他 2024-09-11 05:16:23

呵呵,一开始我写了这个:

def close_geometry(self, geometry):
   if geometry.empty or geometry[0].empty:
       return geometry # empty

   if(geometry[-1][-1] == geometry[0][0]):
       return geometry  # already closed

   result = None
   for linestring in geom:
      if result is None:
          resultstring = linestring.clone()
      else:
          resultstring.extend(linestring.coords)

   geom = Polygon(resultstring)

   return geom

但后来我发现有一个漂亮的小方法叫做 convex_hull 自动为您进行多边形转换。

>>> s1 = LineString((0, 0), (1, 1), (1, 2), (0, 1))
>>> s1.convex_hull
<Polygon object at ...>
>>> s1.convex_hull.coords
(((0.0, 0.0), (0.0, 1.0), (1.0, 2.0), (1.0, 1.0), (0.0, 0.0)),)

>>> m1=MultiLineString(s1)
>>> m1.convex_hull
<Polygon object at...>
>>> m1.convex_hull.coords
(((0.0, 0.0), (0.0, 1.0), (1.0, 2.0), (1.0, 1.0), (0.0, 0.0)),)

Hehe, at first I wrote this:

def close_geometry(self, geometry):
   if geometry.empty or geometry[0].empty:
       return geometry # empty

   if(geometry[-1][-1] == geometry[0][0]):
       return geometry  # already closed

   result = None
   for linestring in geom:
      if result is None:
          resultstring = linestring.clone()
      else:
          resultstring.extend(linestring.coords)

   geom = Polygon(resultstring)

   return geom

but then I discovered that there is a nifty little method called convex_hull that does the polygon conversion for you automatically.

>>> s1 = LineString((0, 0), (1, 1), (1, 2), (0, 1))
>>> s1.convex_hull
<Polygon object at ...>
>>> s1.convex_hull.coords
(((0.0, 0.0), (0.0, 1.0), (1.0, 2.0), (1.0, 1.0), (0.0, 0.0)),)

>>> m1=MultiLineString(s1)
>>> m1.convex_hull
<Polygon object at...>
>>> m1.convex_hull.coords
(((0.0, 0.0), (0.0, 1.0), (1.0, 2.0), (1.0, 1.0), (0.0, 0.0)),)
娇俏 2024-09-11 05:16:23

这是 Carlos 答案的修改,它更简单,不仅返回一个元素,还返回源文件中的所有行

import geopandas as gpd
from shapely.geometry import Polygon, mapping

def linestring_to_polygon(fili_shps):
    gdf = gpd.read_file(fili_shps) #LINESTRING
    gdf['geometry'] = [Polygon(mapping(x)['coordinates']) for x in gdf.geometry]
    return gdf

Here is a modification of Carlos answer, it is simpler and returns not just one element, but all the rows in the source file

import geopandas as gpd
from shapely.geometry import Polygon, mapping

def linestring_to_polygon(fili_shps):
    gdf = gpd.read_file(fili_shps) #LINESTRING
    gdf['geometry'] = [Polygon(mapping(x)['coordinates']) for x in gdf.geometry]
    return gdf
甜中书 2024-09-11 05:16:23

这个小代码可以节省大量时间,也许稍后会合并 geopandas 中更短的形式。

import geopandas as gpd
from shapely.geometry import Polygon, mapping

def linestring_to_polygon(fili_shps):
    gdf = gpd.read_file(fili_shps) #LINESTRING
    geom = [x for x in gdf.geometry]
    all_coords = mapping(geom[0])['coordinates']
    lats = [x[1] for x in all_coords]
    lons = [x[0] for x in all_coords]
    polyg = Polygon(zip(lons, lats))
    return gpd.GeoDataFrame(index=[0], crs=gdf.crs, geometry=[polyg])

This small code can save a lot of time, maybe later a shorter form in geopandas will be incorporated.

import geopandas as gpd
from shapely.geometry import Polygon, mapping

def linestring_to_polygon(fili_shps):
    gdf = gpd.read_file(fili_shps) #LINESTRING
    geom = [x for x in gdf.geometry]
    all_coords = mapping(geom[0])['coordinates']
    lats = [x[1] for x in all_coords]
    lons = [x[0] for x in all_coords]
    polyg = Polygon(zip(lons, lats))
    return gpd.GeoDataFrame(index=[0], crs=gdf.crs, geometry=[polyg])
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