how to have access to lists generated through networkx shortest path function

Question

consider the code below :

G2 = nx.DiGraph()

G2.add_edges_from([('A','B'),('A','C'),('F','G'),('A','F'),('F','H'),('A','I')])

nx.draw(G2, with_labels=1)


paths = (nx.all_pairs_shortest_path(G2))

for path in paths :
    print(path)

the output of this code is:

('A', {'A': ['A'], 'B': ['A', 'B'], 'C': ['A', 'C'], 'F': ['A', 'F'], 'I': ['A', 'I'], 'G': ['A', 'F', 'G'], 'H': ['A', 'F', 'H']})
('B', {'B': ['B']})
('C', {'C': ['C']})
('F', {'F': ['F'], 'G': ['F', 'G'], 'H': ['F', 'H']})
('G', {'G': ['G']})
('H', {'H': ['H']})
('I', {'I': ['I']})

I want to know how I can have access to the all lists such as ['A'], ['A', 'B'], ['A', 'C'], ['A', 'F'] , ... and store them in a list.

Thank you.

Answer 1

paths is a generator. It doesn't support item assignment. You might like it to convert to a data structure that does it:

d = dict(paths)  
>>> print(d)
{'A': {'A': ['A'], 'B': ['A', 'B'], 'C': ['A', 'C'], 'F': ['A', 'F'], 'I': ['A', 'I'], 'G': ['A', 'F', 'G'], 'H': ['A', 'F', 'H']}, 'B': {'B': ['B']}, 'C': {'C': ['C']}, 'F': {'F': ['F'], 'G': ['F', 'G'], 'H': ['F', 'H']}, 'G': {'G': ['G']}, 'H': {'H': ['H']}, 'I': {'I': ['I']}}

That's not very presentative. But still, you're able to check specific edges, like:

>>> d['A']['H']
['A', 'F', 'H']

We can also tidy this a little bit to have all possible path traversals:

from itertools import product
>>> print(list(product(d.keys(), repeat=2)))
[('A', 'A'), ('A', 'B'), ('A', 'C'), ..., ('I', 'G'), ('I', 'H'), ('I', 'I')]
dd = {(a, b): d[a][b] for a, b in product(d.keys(), repeat=2) if b in d[a]}
>>> dd
{('A', 'A'): ['A'],
 ('A', 'B'): ['A', 'B'],
 ('A', 'C'): ['A', 'C'],
 ('A', 'F'): ['A', 'F'],
 ('A', 'G'): ['A', 'F', 'G'],
 ('A', 'H'): ['A', 'F', 'H'],
 ('A', 'I'): ['A', 'I'],
 ('B', 'B'): ['B'],
 ('C', 'C'): ['C'],
 ('F', 'F'): ['F'],
 ('F', 'G'): ['F', 'G'],
 ('F', 'H'): ['F', 'H'],
 ('G', 'G'): ['G'],
 ('H', 'H'): ['H'],
 ('I', 'I'): ['I']}

It can be used in a little different way:

>>> dd['A', 'H']
['A', 'F', 'H']

Alternatively, with no itertools:

dd = {(a, b): d[a][b] for a in d.keys() for b in d.keys() if b in d[a]}

Output should be the same.