Oracle 获取当前组和上一个组（分区）平均值的差异

Question

我使用的是 Oracle 12.1.0.2.0 我想要当前组（分区）的平均值差异 - 前一组（分区）的平均值 我获取当前组平均值的代码是

    with rws as (
  select rownum x, mod(rownum, 2) y from dual connect by level <= 10
), avgs as (
  select x, y, avg(x) over (partition by y) mean from rws
)
  select x, y, mean
  from avgs;

现在我想要类似的东西：

X Y MEAN PREV_MEAN MEAN_DIFF 4 0 6 8 0 6 2 0 6 6 0 6 10 0 6 9 1 5 6 -1 7 1 5 3 1 5 1 1 5 5 1 5 2 2 3 5 -3 3 2 3 5 2 3 1 2 3 4 2 3

1. AVG（此分区组）- Avg（前一个分区组） 在这种情况下，我需要 ( 5 - 6 ) 在 GROUP_MEAN_DIFFERENCE 列中进行计算。

2. 另外，我如何才能始终获得第一组的平均差异。 在上面的例子中，我需要 (5 - 6) 和 (3 - 6)

你能帮忙吗？

Answer 1

将函数lag() 与ignore nulls 子句一起使用：

select id, val, av, av - lag(av ignore nulls) over (order by id) diff
  from (select id, val, 
               case when row_number() over (partition by id order by null) = 1 
                    then avg(val) over (partition by id) end av 

          from t)
  order by id

测试：

with t (id, val) as (select 1, 44.520 from dual union all
                     select 1, 47.760 from dual union all
                     select 1, 50.107 from dual union all
                     select 1, 48.353 from dual union all
                     select 1, 47.640 from dual union all
                     select 2, 48.353 from dual union all
                     select 2, 50.447 from dual union all
                     select 2, 51.967 from dual union all
                     select 2, 45.800 from dual union all
                     select 2, 46.913 from dual )
select id, val, av, av - lag(av ignore nulls) over (order by id) diff
  from (select id, val, 
               case when row_number() over (partition by id order by null) = 1 
                    then avg(val) over (partition by id) end av 

          from t)
  order by id

输出：

ID   VAL      AV       DIFF
---  -------  -------  -------
1     44.520   47.676   
1     47.760
1     50.107        
1     48.353        
1     47.640    
2     48.353   48.696     1.02
2     50.447        
2     51.967        
2     45.800    
2     46.913