For example, given:
G = nx.DiGraph()
G.add_path([0, 1])
G.add_path([0, 2])
G.add_path([0, 3])
G.add_path([1, 11])
G.add_path([1, 12])
G.add_path([2, 21])
G.add_path([2, 22])
G.add_path([3, 31])
G.add_path([3, 32])
I want this:
0, 1, 2, 3, 11, 12, 21, 22, 31, 32
Why does the networkx documentation of bfs_tree actually uses bfs_edges in its example? And not bfs_tree? The key line from the example given in the bfs_tree documentation:
print(list(nx.bfs_edges(G,0)))
Using bfs_edge instead, e.g. via print(list(nx.algorithms.bfs_tree(G, 0).edges())) seems to result in the same list as returned by the example code but is obviously more complicated. Can I use bfs_tree() in a simpler way to obtain a list of _nodes_ of a directed graph in BFS order?
Of course, I could iterate the list returned from bfs_tree or bfs_edges and only take the second element. But isn't there a simpler way?
Why not flatten the list of edge tuples (each 2 nodes), and then take the set of nodes to ensure uniqueness?
list(set(sum(list(nx.algorithms.bfs_tree(G, 0).edges()), ())))
In your hypothetical solution, you would ignore the 0 node, and it would not be included in your output.
Or you could use the bfs_successors() method to get a dict of nodes (passing in the 0 node) and take the values.
[0].extend(nx.algorithms.bfs_successors(G, 0).values())
# Get your first, node, and extend with a list of all successor nodes in BFS order