两行之间的匹配数 mysql
The number of matches between two rows mysql
所以,这就是挑战:
我有两个 table:
标准具:
+-----+-----+-----+-----+----+
| e1 | e2 | e3 | e4 | e5 |
+-----+-----+-----+-----+----+
| 01 | 02 | 03 | 04 | 05 |
+-----+-----+-----+-----+----+
和候选人:
+-----+----+-----+-----+-----+----+----+
| ID | c1 | c2 | c3 | c4 | c5 | nn |
+-----+----+-----+-----+-----+----+----+
| 00 | 03 | 08 | 02 | 01 | 06 | ** |
+-----+----+-----+-----+-----+----+----+
| 01 | 05 | 04 | 03 | 02 | 01 | ** |
+-----+----+-----+-----+-----+----+----+
| 02 | 06 | 07 | 08 | 09 | 10 | ** |
+-----+----+-----+-----+-----+----+----+
| 03 | 08 | 06 | 09 | 02 | 07 | ** |
+-----+----+-----+-----+-----+----+----+
我应该使用什么请求来查找并保存(在 nn 列中)两行(e1、e2、e3、e4、e5 和 c1、c2、c3、c4、c5)之间的匹配数在 table 个候选人中排行?
下一个结果应该是:
候选人:
|-----|----|-----|-----|-----|-----|----|
| ID | c1 | c2 | c3 | c4 | c5 | nn |
|-----|----|-----|-----|-----|-----|----|
| 00 | 03 | 08 | 02 | 01 | 06 | 03 |
|-----|----|-----|-----|-----|-----|----|
| 01 | 05 | 04 | 03 | 02 | 01 | 05 |
|-----|----|-----|-----|-----|-----|----|
| 02 | 06 | 07 | 08 | 09 | 10 | 00 |
|-----|----|-----|-----|-----|-----|----|
| 03 | 08 | 06 | 09 | 02 | 07 | 01 |
|-----|----|-----|-----|-----|-----|----|
nn 的结果是:
0 - no matches
1,2,3,4,5 - numbers of matches
我怎样才能做到这一点?
objective是在master行和client的每一行之间建立一个最大部分匹配table而不考虑各自的列标识。
想法是通过以另一种方式表示列内容来从列 ID 中抽象出来。正如您指出的值域是 {1, ..., 10}
,可以选择前 10 个素数 {p_1, ...,p_10} = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 }
,映射 i
到 p_i
。比较将基于映射列值的乘积。这种方法利用素数分解的独特性,即。每个正整数分解成唯一的多组质数。
一次通过独立 sql 更新语句写起来相当麻烦,因此我们创建一个临时 table 包含映射值的乘积:
CREATE TEMPORARY TABLE t_pp (
id NUMBER
, mp_candidates NUMBER
, mp_etalon NUMBER
, nn NUMBER
);
INSERT INTO t_pp ( id, mp_candidates, mp_etalon )
SELECT id
, CASE c1
WHEN 1 THEN 2
WHEN 2 THEN 3
WHEN 3 THEN 5
WHEN 4 THEN 7
WHEN 5 THEN 11
WHEN 6 THEN 13
WHEN 7 THEN 17
WHEN 8 THEN 19
WHEN 9 THEN 23
WHEN 10 THEN 29
ELSE 31
END
* CASE c2 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE c3 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE c4 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE c5 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
mp_candidates
, CASE e1
WHEN 1 THEN 2
WHEN 2 THEN 3
WHEN 3 THEN 5
WHEN 4 THEN 7
WHEN 5 THEN 11
WHEN 6 THEN 13
WHEN 7 THEN 17
WHEN 8 THEN 19
WHEN 9 THEN 23
WHEN 10 THEN 29
ELSE 31
END
* CASE e2 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE e3 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE e4 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE e5 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
mp_etalon
, 0 nn
FROM candidates
CROSS JOIN etalon
;
现在进行 #2 - 计算匹配项:
UPDATE t_pp
SET nn =
CASE WHEN mp_candidates MOD 2 = 0 AND mp_etalon MOD 2 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 3 = 0 AND mp_etalon MOD 3 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 5 = 0 AND mp_etalon MOD 5 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 7 = 0 AND mp_etalon MOD 7 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 11 = 0 AND mp_etalon MOD 11 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 13 = 0 AND mp_etalon MOD 13 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 17 = 0 AND mp_etalon MOD 17 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 19 = 0 AND mp_etalon MOD 19 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 23 = 0 AND mp_etalon MOD 23 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 29 = 0 AND mp_etalon MOD 29 = 0 THEN 1 ELSE 0 END
;
最后将结果转成原图table并清理:
UPDATE candidates c
set nn = ( SELECT p.nn FROM t_pp p WHERE p.id = c.id )
;
DELETE TEMPORARY TABLE t_pp;
更多注意事项:
- 所示方案假定单元格值在每一行中是唯一的。但是,它可以轻松扩展以允许多次出现值。
- 原则上,这可以包含在单个 sql 语句中 - 出于显而易见的原因,不建议这样做。
- 除 mysql 以外的 Rdbms 遵循 sql 标准并提供
WITH
子句,避免需要临时数组 table.
- 上述
CASE
表达式的 ELSE
分支中的值 31
是一个虚拟值。
所以,这就是挑战:
我有两个 table:
标准具:
+-----+-----+-----+-----+----+
| e1 | e2 | e3 | e4 | e5 |
+-----+-----+-----+-----+----+
| 01 | 02 | 03 | 04 | 05 |
+-----+-----+-----+-----+----+
和候选人:
+-----+----+-----+-----+-----+----+----+
| ID | c1 | c2 | c3 | c4 | c5 | nn |
+-----+----+-----+-----+-----+----+----+
| 00 | 03 | 08 | 02 | 01 | 06 | ** |
+-----+----+-----+-----+-----+----+----+
| 01 | 05 | 04 | 03 | 02 | 01 | ** |
+-----+----+-----+-----+-----+----+----+
| 02 | 06 | 07 | 08 | 09 | 10 | ** |
+-----+----+-----+-----+-----+----+----+
| 03 | 08 | 06 | 09 | 02 | 07 | ** |
+-----+----+-----+-----+-----+----+----+
我应该使用什么请求来查找并保存(在 nn 列中)两行(e1、e2、e3、e4、e5 和 c1、c2、c3、c4、c5)之间的匹配数在 table 个候选人中排行?
下一个结果应该是:
候选人:
|-----|----|-----|-----|-----|-----|----|
| ID | c1 | c2 | c3 | c4 | c5 | nn |
|-----|----|-----|-----|-----|-----|----|
| 00 | 03 | 08 | 02 | 01 | 06 | 03 |
|-----|----|-----|-----|-----|-----|----|
| 01 | 05 | 04 | 03 | 02 | 01 | 05 |
|-----|----|-----|-----|-----|-----|----|
| 02 | 06 | 07 | 08 | 09 | 10 | 00 |
|-----|----|-----|-----|-----|-----|----|
| 03 | 08 | 06 | 09 | 02 | 07 | 01 |
|-----|----|-----|-----|-----|-----|----|
nn 的结果是:
0 - no matches
1,2,3,4,5 - numbers of matches
我怎样才能做到这一点?
objective是在master行和client的每一行之间建立一个最大部分匹配table而不考虑各自的列标识。
想法是通过以另一种方式表示列内容来从列 ID 中抽象出来。正如您指出的值域是 {1, ..., 10}
,可以选择前 10 个素数 {p_1, ...,p_10} = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 }
,映射 i
到 p_i
。比较将基于映射列值的乘积。这种方法利用素数分解的独特性,即。每个正整数分解成唯一的多组质数。
一次通过独立 sql 更新语句写起来相当麻烦,因此我们创建一个临时 table 包含映射值的乘积:
CREATE TEMPORARY TABLE t_pp (
id NUMBER
, mp_candidates NUMBER
, mp_etalon NUMBER
, nn NUMBER
);
INSERT INTO t_pp ( id, mp_candidates, mp_etalon )
SELECT id
, CASE c1
WHEN 1 THEN 2
WHEN 2 THEN 3
WHEN 3 THEN 5
WHEN 4 THEN 7
WHEN 5 THEN 11
WHEN 6 THEN 13
WHEN 7 THEN 17
WHEN 8 THEN 19
WHEN 9 THEN 23
WHEN 10 THEN 29
ELSE 31
END
* CASE c2 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE c3 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE c4 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE c5 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
mp_candidates
, CASE e1
WHEN 1 THEN 2
WHEN 2 THEN 3
WHEN 3 THEN 5
WHEN 4 THEN 7
WHEN 5 THEN 11
WHEN 6 THEN 13
WHEN 7 THEN 17
WHEN 8 THEN 19
WHEN 9 THEN 23
WHEN 10 THEN 29
ELSE 31
END
* CASE e2 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE e3 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE e4 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
* CASE e5 WHEN 2 THEN 3 WHEN 3 THEN 5 WHEN 4 THEN 7 WHEN 5 THEN 11 WHEN 6 THEN 13 WHEN 7 THEN 17 WHEN 8 THEN 19 WHEN 9 THEN 23 WHEN 10 THEN 29 ELSE 31 END
mp_etalon
, 0 nn
FROM candidates
CROSS JOIN etalon
;
现在进行 #2 - 计算匹配项:
UPDATE t_pp
SET nn =
CASE WHEN mp_candidates MOD 2 = 0 AND mp_etalon MOD 2 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 3 = 0 AND mp_etalon MOD 3 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 5 = 0 AND mp_etalon MOD 5 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 7 = 0 AND mp_etalon MOD 7 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 11 = 0 AND mp_etalon MOD 11 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 13 = 0 AND mp_etalon MOD 13 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 17 = 0 AND mp_etalon MOD 17 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 19 = 0 AND mp_etalon MOD 19 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 23 = 0 AND mp_etalon MOD 23 = 0 THEN 1 ELSE 0 END
+ CASE WHEN mp_candidates MOD 29 = 0 AND mp_etalon MOD 29 = 0 THEN 1 ELSE 0 END
;
最后将结果转成原图table并清理:
UPDATE candidates c
set nn = ( SELECT p.nn FROM t_pp p WHERE p.id = c.id )
;
DELETE TEMPORARY TABLE t_pp;
更多注意事项:
- 所示方案假定单元格值在每一行中是唯一的。但是,它可以轻松扩展以允许多次出现值。
- 原则上,这可以包含在单个 sql 语句中 - 出于显而易见的原因,不建议这样做。
- 除 mysql 以外的 Rdbms 遵循 sql 标准并提供
WITH
子句,避免需要临时数组 table. - 上述
CASE
表达式的ELSE
分支中的值31
是一个虚拟值。