如何獲得 PostgreSQL 中的兩個平均值之間的差異，其中平均值在一列上，而最終表格按兩列分組？

Question

我想知道兩個平均值之間的差異value ，其中每個平均值都按條件isCool過濾為True或False ，最終結果按town和season分組，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 |

我想以表格結束：

| 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)

我已經嘗試過了，但不幸的是，我的 PostgreSQL 技能有限，我根本沒有走多遠。 我試過

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

但這遠非我想要的。 有人可以幫忙嗎？ PostgreSQL 甚至可能嗎？

Answer 1

您可以取消透視，然后計算每組的兩個條件平均值之間的差異：

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

如果您希望能夠按照所需結果中所示對結果集進行排序，那么我們需要跟蹤原始列：

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

DB Fiddle 演示：

category |                   diff
:------- | ---------------------:
TownB    |     0.5000000000000000
TownA    | 2.00000000000000000000
winter   |     3.0000000000000000
spring   |     3.0000000000000000

Answer 2

在這里， Union All將為您提供幫助。 簡單地通過對town進行分組然后通過對season進行分組並合並它們來計算一次平均值的差異。 您可以像下面這樣編寫查詢：

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

演示