在python中从一组经度/纬度坐标构建一个多边形

Question

我试图用python中的一组long/lat元组构建一个多边形。多边形，我的意思是我需要定义一个包含点的区域，类似于一个凹壳。我使用的python代码：

from shapely.geometry import Polygon
import geopandas as gpd
import geoplot

crs = {'init': 'epsg:4326'}
z=[(-88.09614, 42.21828), (-88.09605, 42.21903), (-88.09558, 42.21758), (-88.09466, 42.21857), (-88.09448, 42.2176), (-88.09425999999999, 42.2191), (-88.09406, 42.2186), (-88.09352, 42.21763), (-88.09329, 42.21859), (-88.09317, 42.21907), (-88.09226, 42.218779999999995), (-88.09185, 42.217659999999995), (-88.09176, 42.218779999999995), (-88.09138, 42.217659999999995), (-88.09127, 42.218779999999995), (-88.09094, 42.217620000000004), (-88.0907, 42.2188), (-88.09052, 42.21753), (-88.09005, 42.218709999999994), (-88.08998000000001, 42.2174), (-88.08957, 42.218309999999995), (-88.08889, 42.217290000000006), (-88.08830999999999, 42.21763)]
poly = Polygon(z)
pg=gpd.GeoDataFrame(index=[0], crs=crs, geometry=[poly])
geoplot.polyplot(pg)

结果是：view plot 这些点是按经度排序的，函数会考虑这种排序，但只要绘制的结果显然不是多边形，它就无关紧要了。在

Answer 1

多边形可以是，但不一定是凸壳。在你的例子中，你有一个自交的多边形，但一个多边形仍然存在。如果您的目标是计算凸壳，可以使用scipy.spatial.ConvexHull，它使用qhull来计算凸壳

来自the documentation：

from scipy.spatial import ConvexHull, convex_hull_plot_2d
points = np.random.rand(30, 2)   # 30 random points in 2-D
hull = ConvexHull(points)

以及

^{pr2}$

策划。在

赫尔的概念是凹的。一个可能的定义是alpha shapes。这些可以用alphashape package生成。事实上，文档中包括使用geopandas的示例：

import os
import geopandas
data = os.path.join(os.getcwd(), 'data', 'Public_Airports_March2018.shp')
gdf = geopandas.read_file(data)

import cartopy.crs as ccrs
gdf_proj = gdf.to_crs(ccrs.AlbersEqualArea().proj4_init)
gdf_proj.plot()

import alphashape
alpha_shape = alphashape.alphashape(gdf_proj)
alpha_shape.plot()

import matplotlib.pyplot as plt
ax = plt.axes(projection=ccrs.PlateCarree())
ax.scatter([p.x for p in gdf_proj['geometry']],
           [p.y for p in gdf_proj['geometry']],
           transform=ccrs.AlbersEqualArea())
ax.add_geometries(
    alpha_shape['geometry'],
    crs=ccrs.AlbersEqualArea(), alpha=.2)
plt.show()