Show the results of Cypher query with Pandas Dataframe

Question

I'm working with Neo4j from Python and I compare two graphs at the node level in this way:

query = """
  MATCH (n)-[r1]-(), (n)-[r2]-()
  WHERE r1.disease = 'MyLabel1'
  AND r2.disease = 'MyLabel2'
  RETURN DISTINCT n
"""

results = driver.execute_query(query)
results[0]

1. The results of the query are stored in one at the first index of a list, I would like to show the results using a Pandas Dataframe;

2. I would like to plot the count of nodes in each graph and the nodes in common using a Venn diagramm in matplotlib

Answer 1

Example 1: Counting using Cypher, creating 1-row DataFrame

import pandas as pd
from matplotlib_venn import venn2
import matplotlib.pyplot as plt
from neo4j import GraphDatabase

uri = "bolt://localhost:7687"
user = "neo4j"
password = "xxxxxx"

driver = GraphDatabase.driver(uri, auth=(user, password))

query = """
  MATCH (n1)-[r1]-()
  WHERE r1.disease = 'MyLabel1'
  WITH COLLECT(DISTINCT n1) AS n1s
  MATCH (n2)-[r2]-()
  WHERE r2.disease = 'MyLabel2'
  WITH n1s, COLLECT(DISTINCT n2) AS n2s
  RETURN SIZE(n1s) AS count1, SIZE(n2s) AS count2, SIZE([n IN n1s WHERE n IN n2s]) AS countIntersection
"""

with driver.session() as session:
    results = session.run(query)
    data = tuple(record.values() for record in results)

df = pd.DataFrame(data, columns=('count1', 'count2', 'countIntersection'))

if not df.empty:
    venn2(subsets=df.iloc[0], set_labels=('MyLabel1 nodes', 'MyLabel2 nodes'))
    plt.show()
else:
    print("no results")

And here is a sample resulting Venn diagram:

]

Notes:

• This Cypher query just returns a single row of counts, so it is probably not actually worth putting in a DataFrame. Nevertheless, the example shows how to do create a DataFrame from the results and use it to generate the plot.
• The Cypher query uses undirected relationship patterns (like your original query), so the nodes at each end of the same relationship are included in count1 and count2.
• Breaking up your MATCH into 2 MATCHes in the way shown above (where the results from the first MATCH are aggregated before doing the second MATCH) avoids producing a cartesian product.

Example 2: Using DataFrames with all data

This query assumes that every node has a unique id property.

import pandas as pd
from matplotlib_venn import venn2
import matplotlib.pyplot as plt
from neo4j import GraphDatabase

uri = "bolt://localhost:7687"
user = "neo4j"
password = "xxxxxx"

driver = GraphDatabase.driver(uri, auth=(user, password))

query = """
  MATCH (n1)-[r1]-()
  WHERE r1.disease = 'MyLabel1'
  WITH COLLECT(DISTINCT {labels: labels(n1), properties: properties(n1)}) AS n1s
  MATCH (n2)-[r2]-()
  WHERE r2.disease = 'MyLabel2'
  RETURN n1s, COLLECT(DISTINCT {labels: labels(n2), properties: properties(n2)}) AS n2s
"""

with driver.session() as session:
    results = session.run(query)
    data = tuple(record.values() for record in results)

# Create list of dictionaries for every node.
n1s_data = []
for node in data[0][0]:
    node_dict = node['properties']
    node_dict['labels'] = node['labels']
    n1s_data.append(node_dict)

n2s_data = []
for node in data[0][1]:
    node_dict = node['properties']
    node_dict['labels'] = node['labels']
    n2s_data.append(node_dict)

df_n1s = pd.DataFrame(n1s_data)
df_n2s = pd.DataFrame(n2s_data)

n1s_ids = set(node['id'] for node in df_n1s.to_dict('records'))
n2s_ids = set(node['id'] for node in df_n2s.to_dict('records'))

venn2(subsets=(len(n1s_ids), len(n2s_ids), len(n1s_ids & n2s_ids)), set_labels=('MyLabel1 nodes', 'MyLabel2 nodes'))
plt.show()