如何获得 PostgreSQL 中的两个平均值之间的差异，平均值在列上，最终表按两列分组？

Posted 2023-04-12

【中文标题】如何获得 PostgreSQL 中的两个平均值之间的差异，平均值在列上，最终表按两列分组？

【英文标题】：How to get difference between two average values in PostgreSQL, where the averages are on a column, and the final table grouped by two columns?

【发布时间】：2020-10-10 14:59:22

【问题描述】：

我想知道value 的两个平均值之间的差异，其中每个平均值都通过条件isCool 过滤为True 或False，最终结果由town 和@ 分组987654326@，例如

table

| id | value | isCool | town   | season |
|----|-------|--------|--------|--------|
| 0  | 1     | True   | TownA  | spring |
| 1  | 2     | False  | TownA  | winter |
| 2  | 3     | True   | TownB  | spring |
| 3  | 4     | False  | TownA  | winter |
| 4  | 5     | False  | TownB  | spring |
| 5  | 6     | True   | TownB  | winter |

我想以表格结束：

| category | difference_of_is_cool_averages |
|----------|--------------------------------|
| TownA    | 2                              | <-- ABS(1 - (2 + 4)/2)
| TownB    | 0.5                            | <-- ABS(5 - (3 + 6)/2)
| spring   | 3                              | <-- ABS(5 - (3 + 1)/2)
| winter   | 3                              | <-- ABS(6 - (4 + 2)/2)

我已经尝试过了，但是我的 PostgreSQL 技能有限，而且我没有走多远，很遗憾。我试过了

SELECT
   AVG(value), town
   (SELECT id, value, town, season
   FROM table
   WHERE isCool = 'True') AS TableSummary1
GROUP BY town;

但这远不是我想要的。请问有人可以帮忙吗？使用 PostgreSQL 甚至可以做到这一点吗？

【参考方案1】：

这里Union All 会帮助你。只需将town 分组，然后将season 分组并合并它们，即可计算平均值的差异。您可以编写如下查询：

select
town "Category",
round(abs(avg(value) filter (where iscool='t') - avg(value) filter (where iscool='f')),2) "difference_of_is_cool_averages"
from town
group by town

union all

select
season,
round(abs(avg(value) filter (where iscool='t') - avg(value) filter (where iscool='f')),2)
from town
group by season

DEMO

【参考方案2】：

您可以反透视，然后计算每组的两个条件平均值之间的差异：

select x.category, 
    abs(avg(t.value) filter(where not t.iscool) - avg(t.value) filter(where t.iscool)) diff
from mytable t
cross join lateral (values (town), (season)) as x(category)
group by x.category

如果您希望能够按照所需结果对结果集进行排序，那么我们需要跟踪原始列：

select x.category, 
    abs(avg(t.value) filter(where not t.iscool) - avg(t.value) filter(where t.iscool)) diff
from mytable t
cross join lateral (values (town, 1), (season, 2)) as x(category, grp)
group by x.category, x.grp
order by x.grp

Demo on DB Fiddle：

类别 |差异 :------- | ---------------------: 镇B | 0.5000000000000000 镇A | 2.00000000000000000000 冬天| 高分辨率照片| CLIPARTO 3.0000000000000000 弹簧 | 3.0000000000000000