Sql 基于另一个表中的组比例在表中创建组
背景: 我有两张桌子。表A的数据格式为示例格式(12组A-L,A=最高,L=最低): 表B中的数据格式示例如下:Sql 基于另一个表中的组比例在表中创建组,sql,oracle,Sql,Oracle,背景: 我有两张桌子。表A的数据格式为示例格式(12组A-L,A=最高,L=最低): 表B中的数据格式示例如下: ID | SCORE ---- | ---- 1 | 0.12 2 | 0.37 3 | 0.21 4 | 0.55 5 | 0.01 6 | 0.90 7 | 0.10 8 | 0.71 ... 我使用以下公式计算了表A中各组的比例大小: CREATE TABLE table_a_gro
ID | SCORE
---- | ----
1 | 0.12
2 | 0.37
3 | 0.21
4 | 0.55
5 | 0.01
6 | 0.90
7 | 0.10
8 | 0.71
...
我使用以下公式计算了表A中各组的比例大小:
CREATE TABLE table_a_group_pct AS
SELECT band
, count(*) * 100.0 / sum(count(*)) over() AS pct
FROM table_a
GROUP BY band;
输出:
BAND | PCT
---- | ----
A | 12
B | 15
C | 11
D | 9
E | 10
F | 8
G | 11
H | 10
I | 6
J | 4
K | 3
L | 1
我希望为表B创建12个有序(按分数)组,其比例大小与表A中的组相同
例如,表A中12%的行具有group=A,那么前12%的行(基于分数)将被赋予group=A,依此类推
我想我可以通过使用NTILE(100)
函数找到每个分数的百分比位置,然后在时使用案例,根据表a中每个组的累积百分比创建手动组来解决这个问题。(即,如果a组的ID占前12%,那么我在表B中找到第88个百分位,并执行以下操作:
CASE WHEN score_pct > 88 then 'A'
WHEN score_pct BETWEEN 88 and 73 then 'B' ...
END AS group`
然而,我试图了解是否有更聪明的方法来解决这个问题
其他资料:
表A和表B的大小不同,并且没有完全相同的ID,我只是尝试创建相似比例的组
我的预期输出如下:
ID | SCORE | BAND
---- | ---- | ----
1 | 0.12 | K/11
2 | 0.37 | G/7
3 | 0.21 | H/8
4 | 0.55 | E/5
5 | 0.01 | L/12
6 | 0.90 | A/1
7 | 0.10 | K/11
8 | 0.71 | B/2
[编辑我的问题以增加清晰度]这可以通过使用cume_dist分析函数以及一些时髦的连接(在12c之前)来实现,如:
(注意:我已经修改了表a中的数据,使其包含前8个等级;这与您的示例数据不匹配,因此当我的输出与您的输出不匹配时,请不要感到惊讶。)
这是一个在Oracle上运行的示例。你能包括预期的输出吗?不……编辑你的问题,我们看不懂你的想法。不知何故,这个问题对我来说毫无意义。你先说有12个有序的条带,然后显示8行的表格。也许你应该问另一个简化版的问题你的问题和更清楚的解释。什么是“表a的比例大小”例如?表格是示例数据,有数十万行。如果我理解你的意思,你想根据分数百分比按比例分配分数?即,如果10%的分数是A,那么前10%的分数应该是A,如果12%的分数是B,那么接下来的12%的分数应该是B等
ID | SCORE | BAND
---- | ---- | ----
1 | 0.12 | K/11
2 | 0.37 | G/7
3 | 0.21 | H/8
4 | 0.55 | E/5
5 | 0.01 | L/12
6 | 0.90 | A/1
7 | 0.10 | K/11
8 | 0.71 | B/2
WITH table_a AS (SELECT 1 ID, 'A' band FROM dual UNION ALL
SELECT 2 ID, 'B' band FROM dual UNION ALL
SELECT 3 ID, 'A' band FROM dual UNION ALL
SELECT 4 ID, 'C' band FROM dual UNION ALL
SELECT 5 ID, 'D' band FROM dual UNION ALL
SELECT 6 ID, 'E' band FROM dual UNION ALL
SELECT 7 ID, 'D' band FROM dual UNION ALL
SELECT 8 ID, 'F' band FROM dual),
table_b AS (SELECT 1 ID, 0.12 score FROM dual UNION ALL
SELECT 2 ID, 0.37 score FROM dual UNION ALL
SELECT 3 ID, 0.21 score FROM dual UNION ALL
SELECT 4 ID, 0.55 score FROM dual UNION ALL
SELECT 5 ID, 0.01 score FROM dual UNION ALL
SELECT 6 ID, 0.90 score FROM dual UNION ALL
SELECT 7 ID, 0.10 score FROM dual UNION ALL
SELECT 8 ID, 0.71 score FROM dual),
-- end of data set-up, see the rest of the query below:
a_pc AS (SELECT DISTINCT band,
cume_dist() OVER (ORDER BY band) pc_cume_dist
FROM table_a),
b_pc AS (SELECT id,
score,
cume_dist() OVER (ORDER BY score DESC) pc_cume_dist
FROM table_b)
SELECT b_pc.id,
b_pc.score,
b_pc.pc_cume_dist,
min(a_pc.band) band
FROM b_pc
INNER JOIN a_pc ON (a_pc.band = CASE WHEN b_pc.pc_cume_dist <= a_pc.pc_cume_dist AND a_pc.band = 'A' THEN 'A'
WHEN b_pc.pc_cume_dist <= a_pc.pc_cume_dist AND a_pc.band = 'B' THEN 'B'
WHEN b_pc.pc_cume_dist <= a_pc.pc_cume_dist AND a_pc.band = 'C' THEN 'C'
WHEN b_pc.pc_cume_dist <= a_pc.pc_cume_dist AND a_pc.band = 'D' THEN 'D'
WHEN b_pc.pc_cume_dist <= a_pc.pc_cume_dist AND a_pc.band = 'E' THEN 'E'
WHEN b_pc.pc_cume_dist <= a_pc.pc_cume_dist AND a_pc.band = 'F' THEN 'F'
END)
GROUP BY b_pc.id, b_pc.score, b_pc.pc_cume_dist
ORDER BY b_pc.score DESC;
ID SCORE PC_CUME_DIST BAND
---------- ---------- ------------ ----
6 0.9 0.125 A
8 0.71 0.25 A
4 0.55 0.375 B
2 0.37 0.5 C
3 0.21 0.625 D
1 0.12 0.75 D
7 0.1 0.875 E
5 0.01 1 F
WITH table_a AS (SELECT 1 ID, 'A' band FROM dual UNION ALL
SELECT 2 ID, 'B' band FROM dual UNION ALL
SELECT 3 ID, 'A' band FROM dual UNION ALL
SELECT 4 ID, 'C' band FROM dual UNION ALL
SELECT 5 ID, 'D' band FROM dual UNION ALL
SELECT 6 ID, 'E' band FROM dual UNION ALL
SELECT 7 ID, 'D' band FROM dual UNION ALL
SELECT 8 ID, 'F' band FROM dual),
table_b AS (SELECT 1 ID, 0.12 score FROM dual UNION ALL
SELECT 2 ID, 0.37 score FROM dual UNION ALL
SELECT 3 ID, 0.21 score FROM dual UNION ALL
SELECT 4 ID, 0.55 score FROM dual UNION ALL
SELECT 5 ID, 0.01 score FROM dual UNION ALL
SELECT 6 ID, 0.90 score FROM dual UNION ALL
SELECT 7 ID, 0.10 score FROM dual UNION ALL
SELECT 8 ID, 0.71 score FROM dual),
a_pc AS (SELECT DISTINCT band,
cume_dist() OVER (ORDER BY band) pc_cume_dist
FROM table_a),
b_pc AS (SELECT id,
score,
cume_dist() OVER (ORDER BY score DESC) pc_cume_dist
FROM table_b)
SELECT b_pc.id,
b_pc.score,
b_pc.pc_cume_dist,
a_pc2.band
FROM b_pc,
lateral (SELECT MIN(band) band
FROM a_pc
WHERE a_pc.pc_cume_dist >= b_pc.pc_cume_dist) a_pc2
order by b_pc.score desc
ID SCORE PC_CUME_DIST BAND
---------- ---------- ------------ ----
6 0.9 0.125 A
8 0.71 0.25 A
4 0.55 0.375 B
2 0.37 0.5 C
3 0.21 0.625 D
1 0.12 0.75 D
7 0.1 0.875 E
5 0.01 1 F