Pandas: Split a specified datasets into groups on customer id and calculate the number of customers starting with 'C', the list of all products

Pandas Grouping and Aggregating: Split-Apply-Combine Exercise-23 with Solution

Write a Pandas program to split the following datasets into groups on customer id and calculate the number of customers starting with 'C', the list of all products and the difference of maximum purchase amount and minimum purchase amount.

Test Data:

    ord_no  purch_amt    ord_date customer_id  salesman_id
0    70001     150.50  05-10-2012       C3001         5002
1    70009     270.65  09-10-2012       C3001         5005
2    70002      65.26  05-10-2012       D3005         5001
3    70004     110.50  08-17-2012       D3001         5003
4    70007     948.50  10-09-2012       C3005         5002
5    70005    2400.60  07-27-2012       D3001         5001
6    70008    5760.00  10-09-2012       C3005         5001
7    70010    1983.43  10-10-2012       D3001         5006
8    70003    2480.40  10-10-2012       D3005         5003
9    70012     250.45  06-17-2012       C3001         5002
10   70011      75.29  07-08-2012       D3005         5007
11   70013    3045.60  04-25-2012       D3005         5001

Sample Solution:

Python Code :

import pandas as pd
pd.set_option('display.max_rows', None)
#pd.set_option('display.max_columns', None)
df = pd.DataFrame({
'ord_no':[70001,70009,70002,70004,70007,70005,70008,70010,70003,70012,70011,70013],
'purch_amt':[150.5, 270.65, 65.26, 110.5, 948.5, 2400.6, 5760, 1983.43, 2480.4, 250.45, 75.29, 3045.6],
'ord_date': ['05-10-2012','09-10-2012','05-10-2012','08-17-2012','10-09-2012','07-27-2012','10-09-2012','10-10-2012','10-10-2012','06-17-2012','07-08-2012','04-25-2012'],
'customer_id':['C3001','C3001','D3005','D3001','C3005','D3001','C3005','D3001','D3005','C3001','D3005','D3005'],
'salesman_id': [5002,5005,5001,5003,5002,5001,5001,5006,5003,5002,5007,5001]})
print("Original Orders DataFrame:")
print(df)
def customer_id_C(x):
    return (x.str[0] == 'C').sum()
result = df.groupby(['salesman_id'])\
  .agg(customer_id_start_C = ('customer_id', customer_id_C),
       customer_id_list = ('customer_id', lambda x: ', '.join(x)),
       purchase_amt_gap   = ('purch_amt', lambda x: x.max()-x.min())
      )
print("\nNumber of customers  starting with ‘C’, the list of all products and the difference of maximum purchase amount and minimum purchase amount:")
print(result)

Sample Output:

Original Orders DataFrame:
    ord_no  purch_amt    ord_date customer_id  salesman_id
0    70001     150.50  05-10-2012       C3001         5002
1    70009     270.65  09-10-2012       C3001         5005
2    70002      65.26  05-10-2012       D3005         5001
3    70004     110.50  08-17-2012       D3001         5003
4    70007     948.50  10-09-2012       C3005         5002
5    70005    2400.60  07-27-2012       D3001         5001
6    70008    5760.00  10-09-2012       C3005         5001
7    70010    1983.43  10-10-2012       D3001         5006
8    70003    2480.40  10-10-2012       D3005         5003
9    70012     250.45  06-17-2012       C3001         5002
10   70011      75.29  07-08-2012       D3005         5007
11   70013    3045.60  04-25-2012       D3005         5001

Number of customers  starting with ‘C’, the list of all products and the difference of maximum purchase amount and minimum purchase amount:
             customer_id_start_C            customer_id_list  purchase_amt_gap
salesman_id                                                                   
5001                           1  D3005, D3001, C3005, D3005           5694.74
5002                           3         C3001, C3005, C3001            798.00
5003                           0                D3001, D3005           2369.90
5005                           1                       C3001              0.00
5006                           0                       D3001              0.00
5007                           0                       D3005              0.00

Note: Run on Spyder Python 3.7.1

Python Code Editor:

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