使用 Pandas python 将 2 个 groupby 输出与 lambda 结合

Question

Table(df):

 customer_id Order_date 
     1       2015-01-16      
     1       2015-01-19      
     2       2014-12-21      
     2       2015-01-10      
     1       2015-01-10
     3       2018-01-18
     3       2017-03-04
     4       2019-11-05
     4       2010-01-01
     3       2019-02-03
     3       2019-01-01
     3       2018-01-01

Output I want:

Code to extract number of order_dates (where there were at least 3 transactions by a person) using groupby for each customer id and also I need say the most recent transaction dates.

        Customer_id     No_order_date    Most recent order date
            1                3          2015-01-19

            3                5          2019-02-03

Code tried so far:

 freq = 3 
 df.groupby('customer_id')['order_date'].nunique().loc[lambda x: 
 x>=freq].reset_index().rename(columns={'order_date':'No_Order_Dates'})

        Customer_id     No_Order_Dates    
            1                3                              
            3                5          


 df.groupby('customer_id')['order_date'].max().reset_index().rename(columns= 
 {'order_date':'Most recent order Date'})

       Customer_id              Most recent order date
            1                          2015-01-19
            3                          2019-02-03

How can I combine the two groupby outputs? I need both in a single table (is there a way to join without using concatenate or merge or will i have to use concatenate/merge only)

Answer 1

You can use the same named aggregation with .loc[] after the groupby:

(df.groupby('customer_id').agg(No_transactions=('Order_date','nunique'),
           Most_recent_order_date = ('Order_date', 'max'))
            .loc[lambda x: x['No_transactions']>=3])

Or query:

(df.groupby('customer_id').agg(No_transactions=('Order_date','nunique'),
       Most_recent_order_date = ('Order_date', 'max'))
        .query("No_transactions>=3"))

---

             No_transactions Most_recent_order_date
customer_id                                        
1                          3             2015-01-19
3                          5             2019-02-03