python - 如何使用分组依据并返回具有空值的行

Question

我有一个关于电子邮件和购买的数据集，如下所示。

Email          Purchaser    order_id    amount 
a@gmail.com    a@gmail.com    1         5
b@gmail.com         
c@gmail.com    c@gmail.com    2         10
c@gmail.com    c@gmail.com    3         5

我想求出数据集中的总人数、购买人数和总订单数以及总收入金额。我知道如何使用 left join 和聚合函数通过 SQL 完成此操作，但我不知道如何使用 Python/ 复制它 Pandas 。

对于 Python，我尝试使用 pandas 和 numpy:

table1 = table.groupby(['Email', 'Purchaser']).agg({'amount': np.sum, 'order_id': 'count'})

table1.agg({'Email': 'count', 'Purchaser': 'count', 'amount': np.sum, 'order_id': 'count'})

问题是 - 它只返回有顺序的行(第一行和第三行)而不是其他行(第二行)

Email          Purchaser      order_id    amount 
a@gmail.com    a@gmail.com    1           5
c@gmail.com    c@gmail.com    2           15

SQL 查询应如下所示:

SELECT count(Email) as num_ind, count(Purchaser) as num_purchasers, sum(order) as orders , sum(amount) as revenue
    FROM
        (SELECT Email, Purchaser, count(order_id) as order, sum(amount) as amount
        FROM table 1 
        GROUP BY Email, Purchaser) x 

如何在 Python 中复制它？

Answer 1

它现在没有在 pandas 中实现 - see .

所以一个糟糕的解决方案是将 NaN 替换为某个字符串，然后在 agg 替换回 NaN 之后:

table['Purchaser'] = table['Purchaser'].replace(np.nan, 'dummy')

print table
         Email    Purchaser  order_id  amount
0  a@gmail.com  a@gmail.com         1       5
1  b@gmail.com          NaN       NaN     NaN
2  c@gmail.com  c@gmail.com         2      10
3  c@gmail.com  c@gmail.com         3       5

table['Purchaser'] = table['Purchaser'].replace(np.nan, 'dummy')
print table
         Email    Purchaser  order_id  amount
0  a@gmail.com  a@gmail.com         1       5
1  b@gmail.com        dummy       NaN     NaN
2  c@gmail.com  c@gmail.com         2      10
3  c@gmail.com  c@gmail.com         3       5

table1 = table.groupby(['Email', 'Purchaser']).agg({'amount': np.sum, 'order_id': 'count'})
print table1
                         order_id  amount
Email       Purchaser                    
a@gmail.com a@gmail.com         1       5
b@gmail.com dummy               0     NaN
c@gmail.com c@gmail.com         2      15

table1 = table1.reset_index()
table1['Purchaser'] = table1['Purchaser'].replace('dummy', np.nan)
print table1
         Email    Purchaser  order_id  amount
0  a@gmail.com  a@gmail.com         1       5
1  b@gmail.com          NaN         0     NaN
2  c@gmail.com  c@gmail.com         2      15