如果选择“分组依据”，可以在“选择”中使用多个聚合功能吗？

Question

这是表格：

这是我的查询

import netifaces as ni
ni.ifaddresses('eth0')
ip = ni.ifaddresses('eth0')[ni.AF_INET][0]['addr']

allowed_hosts = [ip, '127.0.0.1']

然后我得到了错误

select sum(if(customer_pref_delivery_date = min(order_date), 1, 0)) immidiate_percentage
from Delivery
group by customer_id;

忽略我要在逻辑上进行的操作，我想知道为什么会收到此错误？当我从Invalid use of group function 中删除sum时，它起作用了，所以我想也许在select中，只能在其中使用一个聚合函数（例如sql）做min时select？是真的吗？

Answer 1

您不能将聚合嵌套在选择列表中。
如果要获取与每个客户的订单在同一天需要交付的订单百分比，请执行以下操作：

test[test$id_name %in% keep,]

请参见demo。
结果：

select customer_id, 
  100.0 * avg(customer_pref_delivery_date = order_date) immediate_percentage
from Delivery
group by customer_id;

Answer 2

您不能嵌套聚合功能。这根本没有道理。每个聚合函数每组产生一个值-无法将该值与同一select中组成该组的行进行比较。

我认为您无需再进行任何汇总：

select customer_id, sum( customer_pref_delivery_date = order_date ) as immediate_percentage
from (select d.*,
             min(order_date) over (partition by customer_id) as min_order_date
      from Delivery
     ) d
group by customer_id;

您将总和称为“百分比”似乎很奇怪，但这就是查询中的逻辑。

如果您确实想与每个客户的最短订购日期进行比较，则可以使用窗口功能：

select customer_id, sum( customer_pref_delivery_date = min_order_date ) as immediate_percentage
from (select d.*,
             min(order_date) over (partition by customer_id) as min_order_date
      from Delivery
     ) d
group by customer_id;

Answer 3

您不能嵌套聚合函数。

一种解决此问题的方法是，在子查询中计算每个客户的最短订购日期，然后将其与原始表联系起来，如下所示：

select 
    d.customer_id
    sum(if(d.customer_pref_delivery_date = dmin.min_order_date), 1, 0)) immidiate_percentage
from 
    delivery d
    inner join (
        select customer_id, min(order_date) min_order_date
        from delivery
        group by customer_id
    ) dmin on dmin.customer_id = d.customer_id
group by d.customer_id;

在MySQL 8.0中，您可以使用窗口函数：

select
    d.customer_id,
    sum(if(customer_pref_delivery_date = min_order_date), 1, 0)) immidiate_percentage   
from (
    select 
        customer_id, 
        customer_pref_delivery_date,
        min(order_date) over(partition by customer_id) min_order_date
    from delivery
) t
group by customer_id