Pandas Group By and Score

Question

A pandas dataframe df has 3 columns:

user_id, session, income

Now I want to make the df group a unique user_id and get 2 new columns - one of which is called number_sessions (counts the number of sessions associated with a specific user_id), and the other is number_transactions (counts the number of rows under the revenue column which has a value> 0 for each user_id) . How should I do it?

I tried to do something like this:

df.groupby('user_id')['session', 'revenue'].agg({'number sessions': lambda x: len(x.session), 
'number_transactions': lambda x: len(x[x.revenue>0])})

Answer 1

I think you can use:

df = pd.DataFrame({'user_id':['a','a','s','s','s'],
                   'session':[4,5,4,5,5],
                   'revenue':[-1,0,1,2,1]})

print (df)
   revenue  session user_id
0       -1        4       a
1        0        5       a
2        1        4       s
3        2        5       s
4        1        5       s

a = df.groupby('user_id') \
      .agg({'session': len, 'revenue': lambda x: len(x[x>0])}) \
      .rename(columns={'session':'number sessions','revenue':'number_transactions'})
print (a)
         number sessions  number_transactions
user_id                                      
a                      2                    0
s                      3                    3

---

a = df.groupby('user_id') \
      .agg({'session':{'number sessions': len}, 
            'revenue':{'number_transactions': lambda x: len(x[x>0])}}) 
a.columns = a.columns.droplevel()

print (a)
         number sessions  number_transactions
user_id                                      
a                      2                    0
s                      3                    3

Answer 2

nunique session,

funcs = dict(session={'number sesssions': 'nunique'},
             revenue={'number transactions': lambda x: x.gt(0).sum()})
df.groupby('user_id').agg(funcs)

df = pd.DataFrame({'user_id':['a','a','s','s','s'],
                   'session':[4,5,4,5,5],
                   'revenue':[-1,0,1,2,1]})