CODEWITHSUNDEEP
pythonnetwork-programminggraphnetworkx
BrnCPrz
BrnCPrz

Reputation: 35

Python NetworkX edge lists out of range

I'm currently working with NetworkX in Python2 and just faced a weird situation here. I wanted to discover what nodes are connected to each node in a graph, so I wrote this for loop. (I know that each node is only connected to two other nodes, but I need the exact labels) 

    for i in sorted(graph.nodes(), key=int):
        b = [int(edge_tuple[0]) for edge_tuple in PedigNetwork.in_edges(i)]
        print i, b

So, for every element i, b is a list of length 2, containing the identification of the two nodes that connect to node i.

But when I try to access the FIRST (b[0]) element of that list, it returns an error pointing that the index is out of range. 

    print i, b[0]
    IndexError: list index out of range

But if the list has two elements, [0] is clearly not out of range.

Any ideia how could I solve it? Am I missing something really basic here?

I want to access the first and second elements individually and store them into lists after that.

Thank you very much in advance.

Upvotes: 1

Views: 1156

Answers (1)

unutbu
unutbu

Reputation: 879421

The error implies b is the empty list, []:

In [143]: b = []

In [147]: b[0]
IndexError: list index out of range

That could happen if there is a node i for which PedigNetwork.in_edges(i) returns an empty list. To test this theory, you could run

for i in sorted(graph.nodes(), key=int):
    b = [int(edge_tuple[0]) for edge_tuple in PedigNetwork.in_edges(i)]
    if not b:
        print('Empty list: {}, {}'.format(i, b))
        break

If it prints Empty list: ... then you have your culprit.

If you wish to skip nodes where PedigNetwork.in_edges(i) is empty, you could use

for i in sorted(graph.nodes(), key=int):
    b = [int(edge_tuple[0]) for edge_tuple in PedigNetwork.in_edges(i)]
    if len(b) < 2: continue
    print(i, b[0], b[1])

or

for i in sorted(graph.nodes(), key=int):
    b = [int(edge_tuple[0]) for edge_tuple in PedigNetwork.in_edges(i)]
    if len(b) >= 2:
        print(i, b[0], b[1])

The advantage of using the first option, if len(b) < 2: continue is that it saves you one level of indentation.

Upvotes: 2

Related Questions