I have a graph called testg

testg = nx.Graph()
testg.add_nodes_from([1,2,3,4])
testg.add_edges_from([(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)])

It looks like this:

I want to remove each node, using a loop, and see if the resulting graph is connected. I know it will be, but in my real problem that’s not the case. I think it makes sense to recreate the graph as a different object.

testg2 = nx.Graph()
testg2.add_nodes_from([1,2,3,4])
testg2.add_edges_from([(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)])

If I want to remove node 3 from the graph, and then put the graph back together, the below works:

testg2.remove_node(3)
testg2.add_node(3)
testg2.add_edges_from(testg.edges(3))

I’m wondering how to do this in a loop, and store the boolean results of is_connected in a list? In other words: remove edge –> is resulting graph connected? store response for each node –> put graph back together –> repeat.

nodez = list(range(1,5,1))
for i in nodez:
  c = testg2.remove_node(i)
  stor = nx.is_connected(c)
  back = c.add_node(i)
  beck = c.add_edges_from(testg.edges(i))
  print (stor[i])

I don’t think I’m "putting the graph back together" correctly in the loop. I’m new to python, so nay help would be great. Thank you.

>Solution :

I would use a subgraph to avoid the deletion/insertion.

You can easily do this with set operations:

out = {n: nx.is_connected(testg.subgraph(testg.nodes-{n})) for n in testg.nodes}

Output:

{1: True, 2: True, 3: True, 4: True}

Example with another graph:

testg = nx.Graph()
testg.add_nodes_from([1,2,3,4])
testg.add_edges_from([(1,2),(1,3),(1,4),(2,3)])

{n: nx.is_connected(testg.subgraph(testg.nodes-{n})) for n in testg.nodes}
# {1: False, 2: True, 3: True, 4: True}

graph:

generalization

If you want, you can also easily generalize to test "removal" of combinations of nodes (here n=2 nodes):

from itertools import combinations

n = 2

out = {c: nx.is_connected(testg.subgraph(testg.nodes-set(c)))
       for c in combinations(testg.nodes, n)}

Output:

{(1, 2): True,
 (1, 3): True,
 (1, 4): True,
 (2, 3): True,
 (2, 4): True,
 (3, 4): True}