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?
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)
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
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。
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。