OSMNx : get coordinates of nodes using OSM id

2019-04-10 02:38发布

I used the Python library OSMNx to draw an optimal route between several steps of a city trip. The final variable is a list of OSM ids.

Now, I'm trying to save this route as a shp or json files. The problem is that I need for that the latitude/longitude of each node, but I didn't found an OSMNx function to do that.

I tried get_route_edge_attributes (but coordinates are not a valid attribute for this function). There is any way to get coordinates of an OSM node with this single id ?

Thanks in advance.

2条回答
We Are One
2楼-- · 2019-04-10 03:00

G.node[38862848]['y'] for latitude and G.node[38862848]['x'] for longitude

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兄弟一词,经得起流年.
3楼-- · 2019-04-10 03:01

you have all the attribute of each node and edge in the Graph. you can get node attributes using:

G.node[38862848]
#out: {'highway': nan,
# 'lat': 45.3210533,
# 'lon': -122.9790558,
# 'osmid': '38862848',
# 'ref': nan,
# 'x': 501641.47862882155,
# 'y': 5018616.5723966481}

G.node[38862848]['lat']
# out: 45.3210533

and to get edge attributes you can use G[u][v]:

G[5035130880][4963510289]
# out: 
#{0: {'bridge': 'yes',
#  'geometry': <shapely.geometry.linestring.LineString at 0x7f90ad7d5860>,
#  'highway': 'secondary',
#  'length': 671.332597496,
#  'name': 'Northwest 185th Avenue',
#  'oneway': False,
#  'osmid': [124454683, 24446714, 124454682]}}

All attributes are also in GeoDataFrame's of the graph. If you have list of nodes, the easiest way to get the geometry of all nodes is:

import osmnx as ox
import networkx as nx

gdf_nodes, gdf_edges = ox.graph_to_gdfs()
path = nx.shortest_path(G, G.nodes()[0], G.nodes()[1])
gdf_nodes.loc[path]
#out: 
#        highway    lat lon    osmid    ref x   y   geometry    traffic_signals
#5035130880 NaN 45.5637 -122.868    5035130880  NaN 510334  5.04558e+06 POINT (510334.0390091945 5045583.999886028) 0
#4963510289 NaN 45.5698 -122.868    4963510289  NaN 510329  5.04625e+06 POINT (510329.3114555664 5046254.728223645) 0
# ... 

the output is a GeoDataFrame.

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