pandas - Merge nearly duplicate rows based on column value

I was using some code that I didn't think was optimal and eventually found jezrael's answer. But after using it and running a timeit test, I actually went back to what I was doing, which was:

cmnts = {}
for i, row in df.iterrows():
    while True:
        try:
            if row['Use_Case']:
                cmnts[row['Name']].append(row['Use_Case'])

            else:
                cmnts[row['Name']].append('n/a')

            break

        except KeyError:
            cmnts[row['Name']] = []

df.drop_duplicates('Name', inplace=True)
df['Use_Case'] = ['; '.join(v) for v in cmnts.values()]

According to my 100 run timeit test, the iterate and replace method is an order of magnitude faster than the groupby method.

import pandas as pd
from my_stuff import time_something

df = pd.DataFrame({'a': [i / (i % 4 + 1) for i in range(1, 10001)],
                   'b': [i for i in range(1, 10001)]})

runs = 100

interim_dict = 'txt = {}\n' \
               'for i, row in df.iterrows():\n' \
               '    try:\n' \
               "        txt[row['a']].append(row['b'])\n\n" \
               '    except KeyError:\n' \
               "        txt[row['a']] = []\n" \
               "df.drop_duplicates('a', inplace=True)\n" \
               "df['b'] = ['; '.join(v) for v in txt.values()]"

grouping = "new_df = df.groupby('a')['b'].apply(str).apply('; '.join).reset_index()"

print(time_something(interim_dict, runs, beg_string='Interim Dict', glbls=globals()))
print(time_something(grouping, runs, beg_string='Group By', glbls=globals()))

yields:

Interim Dict
  Total: 59.1164s
  Avg: 591163748.5887ns

Group By
  Total: 430.6203s
  Avg: 4306203366.1827ns

where time_something is a function which times a snippet with timeit and returns the result in the above format.

You can groupby and apply the list function:

>>> df['Use_Case'].groupby([df.Name, df.Sid, df.Revenue]).apply(list).reset_index()
    Name    Sid     Revenue     0
0   A   xx01    $10.00  [Voice, SMS]
1   B   xx02    $5.00   [Voice]
2   C   xx03    $15.00  [Voice, SMS, Video]

(In case you are concerned about duplicates, use set instead of list.)

I think you can use groupby with aggregate first and custom function ', '.join:

df = df.groupby('Name').agg({'Sid':'first', 
                             'Use_Case': ', '.join, 
                             'Revenue':'first' }).reset_index()

#change column order                           
print df[['Name','Sid','Use_Case','Revenue']]                              
  Name   Sid           Use_Case Revenue
0    A  xx01         Voice, SMS  $10.00
1    B  xx02              Voice   $5.00
2    C  xx03  Voice, SMS, Video  $15.00

Nice idea from comment, thanks Goyo:

df = df.groupby(['Name','Sid','Revenue'])['Use_Case'].apply(', '.join).reset_index()

#change column order                           
print df[['Name','Sid','Use_Case','Revenue']]                              
  Name   Sid           Use_Case Revenue
0    A  xx01         Voice, SMS  $10.00
1    B  xx02              Voice   $5.00
2    C  xx03  Voice, SMS, Video  $15.00