时间戳

首页 > 解决方案 > 多个源和目标节点上的 Networkx 最短路径分析

python-3.x - 多个源和目标节点上的 Networkx 最短路径分析

问题描述

我有的：

我想要的是：

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

在学校/医院阅读后，拉出并投影 osmnx 街道图

我能够定义一个函数来获取源点和目标点的最近的 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))

而不是返回 486 条路线（每个源和目标一条），我只得到 18 个点的列表（基本上，它只是根据源点和目标点的相应索引计算路线，而不是计算 27 条最短路线每所学校（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

标签： python-3.xroutesnetworkxgeopandasosmnx

解决方案

您正在寻找起点和终点的笛卡尔积，而不是将它们拉在一起。例子：

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

推荐阅读