OSMnx：有没有办法找到两个坐标之间准确的最短路径？

Question

我想问是否有办法找到2个坐标之间准确的最短路径。如图所示，这2个坐标是（-33.889606，151.283306），（-33.889927，151.280497）。 黑色路径是理想路径，红色路径使用get_nearest_node。以下是代码：

import folium
import osmnx as ox
import networkx as nx

ox.config(use_cache=True, log_console=True)

G = ox.graph_from_point((-33.889606, 151.283306), dist=3000, network_type='drive')

G = ox.speed.add_edge_speeds(G)
G = ox.speed.add_edge_travel_times(G)

orig = ox.get_nearest_node(G, (-33.889606, 151.283306))
dest = ox.get_nearest_node(G, (-33.889927, 151.280497))
route = nx.shortest_path(G, orig, dest, 'travel_time')

route_map = ox.plot_route_folium(G, route)
route_map.save('test.html')

Answer 1

您可以尝试使用名为Taxicab 的包，它是为此目的而创建的。

import osmnx as ox
import taxicab as tc
import matplotlib.pyplot as plt

G = ox.graph_from_point((-33.889606, 151.283306), dist=3000, network_type='drive')
G = ox.speed.add_edge_speeds(G)
G = ox.speed.add_edge_travel_times(G)

orig = (-33.889606, 151.283306)
dest = (-33.889927, 151.280497)

route = tc.distance.shortest_path(G, orig, dest)

fig, ax = tc.plot.plot_graph_route(G, route, node_size=30, show=False, close=False, figsize=(10,10))
padding = 0.001
ax.scatter(orig[1], orig[0], c='lime', s=200, label='orig', marker='x')
ax.scatter(dest[1], dest[0], c='red', s=200, label='dest', marker='x')
ax.set_ylim([min([orig[0], dest[0]])-padding, max([orig[0], dest[0]])+padding])
ax.set_xlim([min([orig[1], dest[1]])-padding, max([orig[1], dest[1]])+padding])
plt.show()

对于您的情况，这将产生以下结果：

免责声明我创作了出租车模块...