使用 LAG 比较今天和 7 天前（不是两者之间）的数据

Question

我目前正在尝试比较今天和恰好 7 天前的汇总数字（不是今天和 7 天前之间，而是简单地比较这两个离散日期）。我已经有办法使用大量子查询来实现它，但是性能很差，我现在正在尝试优化。

这是我到目前为止的想法（示例查询，由于保密原因，没有真正的 table 名称和列）：

Select current_date, previous_date, current_sum, previous_sum, percentage
From   (Select date as current_date, sum(numbers) as current_sum, 
             lag (sum(numbers)) over (partition by date order by date) as previous_sum,
             (Select max(date)-7 From t1 ) as previous_date,
             (current_sum - previous_sum)*100/current_sum as percentage
        From t1 where date>=sysdate-7 group by date,previous_date)

但我肯定做错了什么，因为在输出中 previous_sum 显示为空，当然百分比也是。

对我做错了什么有什么想法吗？我以前没有使用过 LAG，所以它一定是存在的。谢谢！

Answer 1

使用预聚合子查询的联接。

with agg as (
select sum(numbers) as sum_numbers, date from t1 group by date
)

select curr.sum_numbers as current_sum, 
       prev.sum_numbers as prev_sum, 
       curr.date        as curr_date, 
       prev.date        as prev_date
  from agg curr
       left join agg prev on curr.date-7=prev.date

使用延迟：

    with agg as (
    select sum(numbers) as sum_numbers, date from t1 group by date
    )

select sum_numbers       as current_sum, 
       lag(sum_numbers, 7) over(order by date)  as prev_sum,
       a.date            as curr_date,
       lag(a.date,7) over(order by date) as prev_date
  from agg a

如果您只需要 2 个日期（今天和今天 - 7），那么使用条件聚合和过滤器可以更简单地完成：

select sum(case when date = trunc(sysdate) then numbers else null end) as current_sum, 
       sum(case when date = trunc(sysdate-7) then numbers else null end) as previous_sum, 
       trunc(sysdate)         as curr_date, 
       trunc(sysdate-7)       as prev_date,
       (current_sum - previous_sum)*100/current_sum as percentage
  from t1 where date = trunc(sysdate) or date = trunc(sysdate-7)

Answer 2

您可以使用 window（解析）函数来执行此操作，这应该是最快的方法。您的实际聚合查询有点不清楚，但我认为是：

select date as current_date, sum(numbers) as current_sum
from t1 
group by date;

如果您有所有日期的值，则使用：

select date as current_date, sum(numbers) as current_sum, 
       lag(sum(numbers), 7) over (order by date) as prev_7_sum
from t1 
group by date;

如果您没有所有天的数据，则使用 window 框架：

select date as current_date, sum(numbers) as current_sum, 
       max(sum(numbers), 7) over (order by date range between '7' day preceding and '7' day preceding) as prev_7_sum
from t1 
group by date;

使用 LAG 比较今天和 7 天前（不是两者之间）的数据

using LAG to compare the data from today and 7 days ago (not between)

sql

oracle

lag

window-functions