Recursive SQL CTE query in Pandas?
pandas, python
Solution
First, you need to correct the typo in python code `MgrID` list: `[0,1,1,2,0,0,5,6]`
Second, if this job is done recursively in SQL, why do you expect Python/Pandas can do it without recursive method? It isn't too hard:
def nlevel(id, mgr_dict=df.MgrID, _cache={0:0}):
if id in _cache:
return _cache[id]
return 1+nlevel(mgr_dict[id],mgr_dict)
df['nLevel'] = df.ID.map(nlevel)
print df[['nLevel','ID','Name']]
Then the output(`nLevel`) is what you need (except the order, which I don't understand from your SQL):
nLevel ID Name
ID
1 1 1 Keith
2 2 2 Josh
3 2 3 Robin
4 3 4 Raja
5 1 5 Tridip
6 1 6 Arijit
7 2 7 Amit
8 2 8 Dev
[8 rows x 3 columns]
Problem
how do I elegantly port the below recursive SQL query to Pandas python code? Somehow I don't see a straightforward way without writing own recursive function? Python sample code: ``` import datetime import numpy as np import pandas as pd import pandas.io.data from pandas import Series, DataFrame data = { 'ID': [1,2,3,4,5,6,7,8], 'Name': ['Keith','Josh','Robin','Raja','Tridip','Arijit','Amit','Dev'], 'MgrID': [0,1,1,2,0,5,5,6] } df = pd.DataFrame.from_dict(data) df.set_index('ID', inplace=True, drop=False, append=False) df.ix[df.query('MgrID >0')['MgrID']] ``` Trying to get this: ``` nLevel ID Name ================================ 1 6 Arijit 2 8 Dev 1 1 Keith 2 2 Josh 2 3 Robin 3 4 Raja 1 5 Tridip 2 7 Amit ``` Recursive SQL Query: ``` ;WITH Employee (ID, Name, MgrID) AS ( SELECT 1, 'Keith', NULL UNION ALL SELECT 2, 'Josh', 1 UNION ALL SELECT 3, 'Robin', 1 UNION ALL SELECT 4, 'Raja', 2 UNION ALL SELECT 5, 'Tridip', NULL UNION ALL SELECT 6, 'Arijit', NULL UNION ALL SELECT 7, 'Amit', 5 UNION ALL SELECT 8, 'Dev', 6 ) ,Hierarchy AS ( -- Anchor SELECT ID ,Name ,MgrID ,nLevel = 1 ,Family = ROW_NUMBER() OVER (ORDER BY Name) FROM Employee WHERE MgrID IS NULL UNION ALL -- Recursive query SELECT E.ID ,E.Name ,E.MgrID ,H.nLevel+1 ,Family FROM Employee E JOIN Hierarchy H ON E.MgrID = H.ID ) SELECT nLevel ,ID,space(nLevel+(CASE WHEN nLevel > 1 THEN nLevel ELSE 0 END))+Name Name FROM Hierarchy ORDER BY Family, nLevel ```