Nodes taken from pandas Dataframe is not fully connected by NetworkX - python-3.x

Question

I have a dataframe of that I use to calculate the weight of distribution between OS and limit.
| customer_id | limit_reference | OS | limit |
| ----------- | ------------------ | ------ | ------- |
| 1111 | 1111.A.1::1111.B.1 | 0.1 | 5 |
| 1111 | 1111.A.1 | .08 | 5 |
| 9012 | 1111.B.1::9012.B.1 | .15 | 5 |
The value in limit_reference is of the form: ID.contract_reference.
I need to match all value in customer_id and limit_reference, and group them into groups. So if I created another group_id column, then I will have:
| customer_id | limit_reference | OS | limit | group_id |
| ----------- | ------------------ | ------ | ------- | ---------|
| 1111 | 1111.A.1::1111.B.1 | 0.1 | 5 | 1 |
| 1111 | 1111.A.1 | .08 | 5 | 1 |
| 9012 | 1111.B.1::9012.B.1 | .15 | 5 | 1 |
The problem I have is networkx recognize 1111.A.1::1111.B.1 and 1111.B.1::9012.B.1 as 2 different nodes even though they have the same element 1111.B.1.
I have tried to split limit_reference, but it is unhashable. Here's the code I tried:
import pandas as pd
import networkx as nx
df_ = pd.read_excel('sample2.xlsx')
G = nx.Graph()
G = nx.from_pandas_edgelist(df_, 'customer_id', 'limit_reference')
cnc = nx.connected_components(G)
pos = nx.spring_layout(G, scale=20, k=2/np.sqrt(G.order()))
df_['group_id'] = [label for node in df_.customer_id for label, component in lookup.items() if node in component]
nx.draw(G, pos, node_color='lightgreen', node_size=1000, with_labels=True)
lookup = {i: component for i, component in enumerate(cnc, 1)}

Answer 1

You can split limit_reference with pandas' Series.str.split() method, e.g.
import pandas as pd
import networkx as nx
df_ = pd.DataFrame({'customer_id': [1111, 1111, 9012],
'limit_reference': ['1111.A.1::1111.B.1',
'1111.A.1',
'1111.B.1::9012.B.1']})
G = nx.Graph()
limit_reference_split = [x for sublist in df_['limit_reference'].str.split('::')
for x in sublist]
G.add_nodes_from(limit_reference_split)
list(G.nodes)
['1111.A.1', '1111.B.1', '9012.B.1']