Matching employees from different school and hometown - database

Question

I am new to coding. Now I have a employee table looked like below:
Name
Hometown
School
Jeff
Illinois
Loyola University Chicago
Alice
California
New York University
William
Michigan
University of Illinois at Chicago
Fiona
California
Loyola University Chicago
Charles
Michigan
New York University
Linda
Indiana
Loyola University Chicago
I am trying to get those employees in pairs where two employees come from different state and different university. Each person can only be in one pair. The expected table should look like
employee1
employee2
Jeff
Alice
William
Fiona
Charles
Linda
The real table is over 3,000 rows. I am trying to do it with SQL or Python, but I don't know where to start.

Answer 1

A straightforward approach is to pick employees one by one and search the table after the one for an appropriate peer; found peers are flagged in order to not be paired repeatedly. Since in your case a peer should be found after a few steps, this iteration will likely be faster than operations which construct whole data sets at once.
from io import StringIO
import pandas as pd
# read example employee table
df = pd.read_table(StringIO("""Name Hometown School
Jeff Illinois Loyola University Chicago
Alice California New York University
William Michigan University of Illinois at Chicago
Fiona California Loyola University Chicago
Charles Michigan New York University
Linda Indiana Loyola University Chicago
"""))
# create expected table; its length is half that of the above
ef = pd.DataFrame(index=pd.RangeIndex(len(df)/2), columns=['employee1', 'employee2'])
k = 0 # number of found pairs, index into expected table
# array of flags for already paired employees
paired = pd.Series(False, pd.RangeIndex(len(df)))
# go through the employee table and collect pairs
for i in range(len(df)):
if paired[i]: continue
for j in range(i+1, len(df)):
if not paired[j] \
and df.iloc[j]['Hometown'] != df.iloc[i]['Hometown'] \
and df.iloc[j]['School'] != df.iloc[i]['School']:
# we found a pair - store it, mark employee j paired
ef.iloc[k] = df.iloc[[i, j]]['Name']
k += 1
paired[j] = True
break
else:
print("no peer for", df.iloc[i]['Name'])
print(ef)
output:
employee1 employee2
0 Jeff Alice
1 William Fiona
2 Charles Linda