'AttributeError: module 'networkx' has no attribute 'connected_component_subgraphs'
B = nx.Graph()
B.add_nodes_from(data['movie'].unique(), bipartite=0, label='movie')
B.add_nodes_from(data['actor'].unique(), bipartite=1, label='actor')
B.add_edges_from(edges, label='acted')
A = list(nx.connected_component_subgraphs(B))[0]
I am getting the below given error when am trying to use nx.connected_component_subgraphs(G)
.
In the dataset there are two coumns(movie and actor), and it's in the form bipartite graph.
I want to get connected components for the movie nodes.
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-16-efff4e6fafc4> in <module>
----> 1 A = list(nx.connected_component_subgraphs(B))[0]
AttributeError: module 'networkx' has no attribute 'connected_component_subgraphs'
Solution 1:[1]
This was deprecated with version 2.1, and finally removed with version 2.4.
Use
(G.subgraph(c) for c in connected_components(G))
Or
(G.subgraph(c).copy() for c in connected_components(G))
Solution 2:[2]
connected_component_subgraphs
has been removed from the networkx library.
You can use the alternative described in the deprecation notice.
For your example, refer to the code below:
A = (B.subgraph(c) for c in nx.connected_components(B))
A = list(A)[0]
Solution 3:[3]
Use the following code for single line alternative
A=list(B.subgraph(c) for c in nx.connected_components(B))[0]
Or you can install the previous version of networkx
pip install networkx==2.3
Solution 4:[4]
First I got
AttributeError: module 'matplotlib.cbook' has no attribute 'iterable'.
To fix the above error, I upgraded networkx using
pip install --upgrade --force-reinstall network
It installed unetworkx-2.6.3, I got the error
AttributeError: module networkx has no attribute connected_component_subgraphs.
I used the below code as mentioned by ABHISHEK D, it resolved. Thanks.
A=list(B.subgraph(c) for c in nx.connected_components(B))[0]
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | |
Solution 2 | Thayne |
Solution 3 | |
Solution 4 | m4n0 |