多个源和目标节点上的 Networkx 最短路径分析

问题描述

我有什么：

学校点的地理数据框（来源 - 总共 18 个）
医院 pts 的地理数据框（目标 - 总共 27 个）
投影的 Osmnx 图（节点 + 边）

我想要的：

一个包含从每所学校到每家医院的最短路线几何图形的地理数据框（表中共有 486 个 [18*27] 个要素，每个要素都有一条路线）即

学校编号	医院编号	路线
xxxxxxxxx	xxxxxxxxxxx	Linestring(x,x,x)

在学校/医院阅读后，绘制并投影osmnx街图

我可以定义一个函数来获取源点和目标点的 Neareset osm 节点

# import neceessary modules
import pandas as pd
import geopandas as gpd
import osmnx as ox
import networkx as nx
import matplotlib.pyplot as plt
from pyproj import CRS
from shapely.geometry import polygon,Point,Linestring
)

# read in file
hosp_fp = r'vtData/hospitals.shp'
school_fp = r'vtData/schools.shp'

# read files
hospitals = gpd.read_file(hosp_fp)
schools = gpd.read_file(school_fp)

#### skip the reading osmnx part (the error isn't here and,yes,everything in same crs) ######

# Create function to find nearest node
def get_nearest_node(points,graph):
    # a function that finds the nearest node
    # params: points (a gdf of points with an x and y column); graph (an osm network graph)
    points['nearest_osm'] = None
    for i in tqdm(points.index,desc='find nearest osm node from input points',position=0):
        points.loc[i,'nearest_osm'] = ox.get_nearest_node(graph,[points.loc[i,'y'],points.loc[i,'x']],method='euclidean') # find the nearest node from hospital location
    return(points)

# use the function to find each destination point nearest node
## returns the original gdfs with corresponding osmid column
source = get_nearest_node(schools,graph)
target = get_nearest_node(hospitals,graph)

# extract osmid's from list
src_list = list(source['nearest_osm'])
trg_list = list(target['nearest_osm'])

### WHERE I AM STUCK ####
# a function meant to construct shortest path routes to each target from each source

def get_routes(graph,src_list,trg_list):
    # a function that constructs a shortest routes to each target from the source
    # params: graph_proj (a projected osmnx graph); src (source pts); trg (target pts)

    # an empty list to append route linestring geometries to
    routes = []

    # a loop to construct all shortest path geometries
    for src,trg in zip(src_list,trg_list):
        sp = nx.shortest_path(graph,source=src,target=trg,weight='length')
        routes.append(sp)
        print(len(routes))

我没有返回 486 条路由（每个源和目标一个），而是只得到一个列表 18点（基本上，它只是根据相应的索引计算路线源点和目标点，而不是计算每所学校的 27 条最短路线（医院总数）

从这里开始，我会将列表附加到一个名为“路线”的新地理数据框中，但我的 486 条路线中只有 18 条

解决方法

您正在寻找起点和目的地的笛卡尔积，而不是将它们压缩在一起。示例：

import numpy as np
import osmnx as ox
from itertools import product
ox.config(log_console=True)

# get a graph and add edge travel times
G = ox.graph_from_place('Piedmont,CA,USA',network_type='drive')
G = ox.add_edge_travel_times(ox.add_edge_speeds(G))

# randomly choose 10 origins and 10 destinations
n = 10
origs = np.random.choice(G.nodes,size=n)
dests = np.random.choice(G.nodes,size=n)

# calculate 100 (10 origins x 10 destinations) shortest paths
paths = []
for o,d in product(origs,dests):
    path = ox.shortest_path(G,o,d,weight='travel_time')
    paths.append(path)
    
len(paths) #100

geopandas networkx osmnx python-3.x routes routes