NetworkX expects a square matrix (of nodes and edges), perhaps* you want to pass it:
In [11]: df2 = pd.concat([df, df.T]).fillna(0)
Note: It's important that the index and columns are in the same order!
In [12]: df2 = df2.reindex(df2.columns)
In [13]: df2
Out[13]:
Bar Bat Baz Foo Loc 1 Loc 2 Loc 3 Loc 4 Loc 5 Loc 6 Loc 7 Quux
Bar 0 0 0 0 0 0 1 1 0 1 1 0
Bat 0 0 0 0 0 0 1 0 0 1 0 0
Baz 0 0 0 0 0 0 1 0 0 0 0 0
Foo 0 0 0 0 0 0 1 1 0 0 0 0
Loc 1 0 0 0 0 0 0 0 0 0 0 0 1
Loc 2 0 0 0 0 0 0 0 0 0 0 0 0
Loc 3 1 1 1 1 0 0 0 0 0 0 0 0
Loc 4 1 0 0 1 0 0 0 0 0 0 0 0
Loc 5 0 0 0 0 0 0 0 0 0 0 0 0
Loc 6 1 1 0 0 0 0 0 0 0 0 0 0
Loc 7 1 0 0 0 0 0 0 0 0 0 0 0
Quux 0 0 0 0 1 0 0 0 0 0 0 0
In[14]: graph = nx.from_numpy_matrix(df2.values)
This doesn't pass the column/index names to the graph, if you wanted to do that you could use relabel_nodes
(you may have to be wary of duplicates, which are allowed in pandas' DataFrames):
In [15]: graph = nx.relabel_nodes(graph, dict(enumerate(df2.columns))) # is there nicer way than dict . enumerate ?
*It's unclear exactly what the columns and index represent for the desired graph.
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