Clarify expected behavior for Leiden on a disconnected graph

Is your feature request related to a problem? Please describe.

Several times I have been confused by output of Leiden only to realize it is because some nodes are disconnected. These nodes then will not show up in the partition dictionary.

Describe the solution you’d like

Just some clarification and/or a warning for what to expect

Provide references (if applicable)

in this example node 0 is disconnected from the graph and thus key 0 is not present in the output dictionary

g = nx.complete_graph(10)
nodelist = sorted(g.nodes)
for node in nodelist[1:]: 
  g.remove_edge(0, node)
leiden(g)

 {3: 0, 4: 0, 7: 0, 2: 0, 9: 0, 6: 0, 5: 0, 1: 0, 8: 0}

cc: @daxpryce

[daxpryce]

Might I suggest a different approach?

if isinstance(graph, nx.Graph):
  # check if graph.number_of_nodes() == len(partitions), if not, we can presume isolates are removed
else if <is matrix form>:
  # check if graph.shape[0] == len(partitions), if not, we can presume isolates are removed

Doing it this way means fewer check flags, which are an evil (though sometimes necessary), and also somewhat keeps the divergent-paths-depending-on-graph-representation-type stuff to a minimum

[daxpryce]

I don’t know that the LCC should matter - it makes sense that your embeddings lose coherency/utility, but > 1 CC is perfectly fine in the leiden case. I don’t think we should lcc check, but we should make sure we warn if we’re returning a partitions dictionary that doesn’t contain a partition for every node 😃