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Using the networkx library, I define a MultDiGraph. I then calculate the shortest path between two given nodes. As my graph holds parallel edges, I would like to know the keys of the edges that make up the shortest path. Here's an example:

import networkx as nx

G = nx.MultiDiGraph()
G.add_edge('a', 'b', key=0, weight=1)
G.add_edge('a', 'b', key=1, weight=2)
G.add_edge('b', 'c', key=0, weight=1)

shortest_path = nx.shortest_path(G, source='a', target='c')
shortest_path_length = nx.shortest_path_length(G, source='a', target='c')
    
print(shortest_path)
print(shortest_path_length)

The result looks like this:

['a', 'b', 'c']
2

This is only correct if the shortest path between nodes 'a' and 'b' is via the key=0 (weight=1). I couldn't find anything in the docs that would allow me to retrieve the keys of the actual edges involved in the shortest path. How could I get to this information?

Johngoldenboy
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1 Answers1

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Short answer is you cannot. Networkx does not have a functionality that returns the edges that comprise the shortest path. At most it can return the nodes that are in that shortest path.

A work around this is to assume that the edges in the shortest path are the shortest ones, therefore if two nodes in the path can form an edge with a key of either 1 or 0, assume it is the shortest one.

I have faced the same issue as you in: How does osmnx know which edges to plot in shortest path

I used a library called OSMNx and the feature I wanted needed to know which edges make the shortest path. The feature makes the same assumption as I have stated above (you can see the answer to the question)

DPM
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    Thanks for your response. I understand your assumption and on first thought it makes sense, too. I wonder if one can find an example where this assumption does not hold true. Anyhow, for now I go with it and will mark the answer as correct if no one comes around with a proper way of retrieving the information. – Johngoldenboy Feb 25 '22 at 09:20