PostgreSQL中的条件累积平均值

时间:2014-11-16 20:18:13

标签: sql postgresql

我有一个简单的表,即:

    DROP TABLE IF EXISTS running_averages;

    CREATE TABLE running_averages
    (
            avg_id          SERIAL NOT NULL PRIMARY KEY,
            num1             integer,
            num2             integer   DEFAULT 0     

    );

    INSERT INTO running_averages(num1, num2)
    SELECT 100, 100 UNION ALL
    SELECT 200, 175 UNION ALL
    SELECT -400, NULL UNION ALL
    SELECT 300, 200 UNION ALL
    SELECT -100, NULL;

在上表中," num2"如果列" num1"列应该使用累计平均值上一行进行更新。是一个负值。我目前的疑问是:

    SELECT *,
            num1 * num2 AS current_total,
            SUM(num1 * num2) OVER(order by avg_id) AS cumulative_sum,
            SUM(num1)  OVER(order by avg_id) AS culmulative_num1,

            CASE WHEN num1 > 0 THEN
            SUM(num1 * num2) OVER(order by avg_id) 
            / 
            SUM(num1)  OVER(order by avg_id) 
            ELSE
            0
            END AS cumulative_average
    FROM running_averages;

结果:

avg_id  num1  num2    current_total cumulative_sum   cumulative_num1 cumulative_average
1       100   100     10,000        10,000           100             100
2       200   175     35,000        45,000           300             150
3       -400          NULL          45,00            -100            0
4       300   200     60,000        105,000          200             525
5       -100          NULL          105,000          100               0

如果当前行的num1列为负数,我无法找出带来前一行的累积平均值的方法。而不是上述内容,预期输出应为:

avg_id  num1  num2    current_total cumulative_sum   cumulative_num1 cumulative_average
1       100   100     10,000        10,000           100             100
2       200   175     35,000        45,000           300             150
3       -400  150     -60,000       -15,00           -100            150
4       300   200     60,000        45,000           200             225
5       -100  225     -22,500       22,500           100             225

在这种情况下,如何获取最后一行的列值?

编辑:

我编辑了上面的SQL脚本。我非常喜欢Gordon Linoff的答案。但遗憾的是,根据脚本更改产生了错误的结果:

avg_id  num1  num2    new_num2
1       100   100     100
2       200   175     175
3       -400  150     150 (Correct)
4       300   200     200
5       -100  225     50  (Incorrect)

编辑2

我也测试过Multisync的答案,它也会产生错误的结果:

avg_id  num1  num2              current_total cumulative_sum   cumulative_num1 cumulative_average
1       100   100               10,000        10,000           100             100
2       200   175               35,000        45,000           300             150
3       -400  150 (Correct)     -60,000       -15,00           -100            150
4       300   200               60,000        45,000           200             225
5       -100  175 (Incorrect)   -17,500       27,500           100             275

编辑3

我接受了Multisync的更新答案,因为它产生了正确的结果。我还想知道如何改进这样的查询,我们有很多聚合和窗口函数。有关此主题的任何参考都会有所帮助。

2 个答案:

答案 0 :(得分:2)

我只能想到一个递归查询:

with recursive tmp (avg_id, num1, num2, sum_m, sum_num1, last_id) as (
  select avg_id, num1, num2, num1 * num2, num1, avg_id
  from running_averages where avg_id = 1  
  union all
  select r.avg_id, r.num1, 
         case when r.num1 < 0 then t.sum_m / t.sum_num1 else r.num2 end, 
         t.sum_m + case when r.num1 < 0 then t.sum_m / t.sum_num1 else r.num2 end * r.num1,
         t.sum_num1 + r.num1,
         r.avg_id 
  from running_averages r join tmp t on r.avg_id = t.last_id + 1
)
select avg_id, num1, num2, 
       num1 * num2 AS current_total,
       SUM(num1 * num2) OVER(order by avg_id) AS cumulative_sum,
       SUM(num1) OVER(order by avg_id) AS culmulative_num1,
       SUM(num1 * num2) OVER(order by avg_id) 
       / SUM(num1) OVER(order by avg_id) AS cumulative_average
from tmp;

avg_id必须包含相应的数字(您可以使用row_number()代替,我没有使用它进行简化)

num2在计算期间正在发生变化,这就是为什么我无法想到除递归查询之外的任何事情(上一步的输出是下一步的输入)

答案 1 :(得分:2)

让我们专注于此:

  

在上表中,“num2”列应该用。更新   如果列“num1”是a,则前一行的累积平均值   负值

这应该不会太困难:

select ra.*,
       (case when num1 >= 0 then num2
             else avg(num1) over (order by avg_id rows between unbounded preceding and 1 preceding)
        end) as new_num2
from running_averages ra;

我认为您可以使用new_num2进行剩余的计算。