如何根据 osmnx 上的一些“手动选择”节点绘制路径？

Question

我想在 osmnx 地图上绘制一条路线。当我尝试手动创建路线时，它不起作用。

import networkx as nx

import osmnx as ox

import requests 

import matplotlib.cm as cm

import matplotlib.colors as colors

location_point = (50.345559621674596, 7.576892382877531)

G = ox.graph_from_point(location_point, distance=500, 

distance_type='network', network_type='walk')

origin_node = ox.get_nearest_node(G, location_point)

destination_node = list(G.nodes())[-1]

fig, ax = ox.plot_graph(G)

G=nx.convert_node_labels_to_integers(G)

route = [3 ,1 ,41 ,40 ,98 ,62 ,67 ,107 ,94 ,83 ,39 ]

ox.plot_route_folium(G,route)

我希望有一张带有彩色路线的地图，但我却收到了这个错误。

    ---------------------------------------------------------------------------

    IndexError                                Traceback (most recent call last)

    <ipython-input-9-11235dc07ef4> in <module>()

    ----> 1 ox.plot_route_folium(G,route)

    ~\Anaconda3\lib\site-packages\osmnx\plot.py in plot_route_folium(G, route, route_map, popup_attribute, tiles, zoom, fit_bounds, route_color, route_width, route_opacity)

        907     gdf_edges = graph_to_gdfs(G, nodes=False, fill_edge_geometry=True)

        908     route_nodes = list(zip(route[:-1], route[1:]))

    --> 909     index = [gdf_edges[(gdf_edges['u']==u) & (gdf_edges['v']==v)].index[0] for u, v in route_nodes]

        910     gdf_route_edges = gdf_edges.loc[index]

        911 

    ~\Anaconda3\lib\site-packages\osmnx\plot.py in <listcomp>(.0)

        907     gdf_edges = graph_to_gdfs(G, nodes=False, fill_edge_geometry=True)

        908     route_nodes = list(zip(route[:-1], route[1:]))

    --> 909     index = [gdf_edges[(gdf_edges['u']==u) & (gdf_edges['v']==v)].index[0] for u, v in route_nodes]

        910     gdf_route_edges = gdf_edges.loc[index]

        911 

如果我尝试通过最短路径算法找到路线并将其在地图上可视化，则效果很好。但我想绘制这条特定路线

Answer 1

OSMnx 使用节点的 OSMID。当您运行时G=nx.convert_node_labels_to_integers(G)，您会覆盖这些 OSMID。但是，您也没有在图边u和v属性中覆盖它们，这些属性标识了边的端点（使用 OSMID）。