python怎样用自定义函数对列表排序?

Posted 2023-05-06

比如举一个列表a=[7,4,5,1,8,3,6],对它进行升序排列，不能用自带的sort函数，也不能用有关的取最大、最小值的函数（比如min,max），一句话就是不能投机取巧，老师说要用循环，然后一定得是自定义函数，其实本质就是对数字大小进行比较。初学者求帮助啊。

您好！我写了一个升序的方法供您参考：

请注意，要在 idle 中运行这个方法，并输入你希望的值

下面讲解一下：

方法体的第一句用于把你输入的数字储存在列表中，这句可以用于其他许多场合

后续的代码就是不断地递归，从而达到升序的效果

有不懂的可以问我哦～望采纳！

参考技术A

MySQL巧用自定义函数进行查询优化

用户自定义变量是一个很容易被遗忘的MySQL特性,但是用的好,发挥其潜力,在很多场景都可以写出非常高效的查询语句。

一. 实现一个按照actorid排序的列

 1 mysql> set @rownum :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> select actor_id ,@rownum :=@rownum + 1 as rownum
 5 -> from sakila.actor limit 3;
 6 +----------+--------+
 7 | actor_id | rownum |
 8 +----------+--------+
 9 | 58 | 1 |
10 | 92 | 2 |
11 | 182 | 3 |
12 +----------+--------+
13 3 rows in set (0.00 sec)

 

二. 扩展一下,现在需要获取演过最多电影的前十位,针对数量作一个排名,如果数量一样,则排名相同

 1 mysql> set @curr_cnt :=0 ,@pre_cnt :=0 ,@rank :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> select actor_id, 
 5 -> @prev_cnt :=@curr_cnt as dummy,
 6 -> @curr_cnt := cnt as cnt,
 7 -> @rank := IF(@prev_cnt <> @curr_cnt,@rank+1,@rank) as rank
 8 -> FROM(
 9 -> SELECT actor_id ,count(*) as cnt
10 -> FROM sakila.film_actor
11 -> GROUP BY actor_id
12 -> ORDER BY cnt DESC
13 -> LIMIT 10
14 -> )as der;
15 +----------+-------+-----+------+
16 | actor_id | dummy | cnt | rank |
17 +----------+-------+-----+------+
18 | 107 | 0 | 42 | 1 |
19 | 102 | 42 | 41 | 2 |
20 | 198 | 41 | 40 | 3 |
21 | 181 | 40 | 39 | 4 |
22 | 23 | 39 | 37 | 5 |
23 | 81 | 37 | 36 | 6 |
24 | 158 | 36 | 35 | 7 |
25 | 144 | 35 | 35 | 7 |
26 | 37 | 35 | 35 | 7 |
27 | 106 | 35 | 35 | 7 |
28 +----------+-------+-----+------+
29 10 rows in set (0.00 sec)

 

三. 避免重复查询刚更新的数据

如果想要高效的更新一条记录的时间戳 ,又想返回更新的数据

 1 mysql> create table t2 (id int,lastUpdated datetime);
 2 Query OK, 0 rows affected (0.03 sec)
 3 
 4 mysql> insert into t2 (id ,lastupdated)values(1,sysdate());
 5 Query OK, 1 row affected (0.02 sec)
 6 
 7 mysql> select * from t2;
 8 +------+---------------------+
 9 | id | lastUpdated |
10 +------+---------------------+
11 | 1 | 2017-07-24 16:03:34 |
12 +------+---------------------+
13 1 row in set (0.01 sec)
14 
15 mysql> update t2 set lastUpdated=NOW() WHERE id =1 and @now :=Now();
16 Query OK, 1 row affected (0.02 sec)
17 Rows matched: 1 Changed: 1 Warnings: 0
18 
19 
20 mysql> select @now, sysdate();
21 +---------------------+---------------------+
22 | @now | sysdate() |
23 +---------------------+---------------------+
24 | 2017-07-24 16:05:42 | 2017-07-24 16:06:06 |
25 +---------------------+---------------------+
26 1 row in set (0.00 sec)

 

四. 统计更新和插入的数量
使用 INSERT ON DUPLICATE KEY UPDATE 时,查询插入成功的条数,冲突的条数

 

 1 mysql> set @x :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> INSERT INTO t3(c1,c2) values(1,2),(1,3),(2,2)
 5 -> ON DUPLICATE KEY UPDATE
 6 -> c2=VALUES(c2)+(0*(@x:=@x+1));
 7 Query OK, 4 rows affected (0.01 sec)
 8 Records: 3 Duplicates: 1 Warnings: 0
 9 
10 mysql> select @x;
11 +------+
12 | @x |
13 +------+
14 | 1 |
15 +------+
16 1 row in set (0.00 sec)
17 
18 mysql> select * from t3;
19 +----+------+
20 | c1 | c2 |
21 +----+------+
22 | 1 | 3 |
23 | 2 | 2 |
24 +----+------+
25 2 rows in set (0.00 sec)

 

五. 确定取值的顺序
想要获取sakila.actor中的一个结果

错误的查询一:
下面的查询看起来好像只返回一个结果,实际呢:

 1 mysql> set @row_num :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> SELECT actor_id,@row_num :=@row_num+1 AS cnt
 5 -> FROM sakila.actor 
 6 -> WHERE @row_num <=1 
 7 -> ;
 8 +----------+------+
 9 | actor_id | cnt |
10 +----------+------+
11 | 58 | 1 |
12 | 92 | 2 |
13 +----------+------+
14 2 rows in set (0.00 sec)
15 
16 看一下执行计划:
17 +----+-------------+-------+-------+---------------+---------------------+---------+------+------+--------------------------+
18 | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
19 +----+-------------+-------+-------+---------------+---------------------+---------+------+------+--------------------------+
20 | 1 | SIMPLE | actor | index | NULL | idx_actor_last_name | 137 | NULL | 200 | Using where; Using index |
21 +----+-------------+-------+-------+---------------+---------------------+---------+------+------+--------------------------+
22 1 row in set (0.00 sec)

这是因为where 和 select 是在 查询的不同阶段执行的造成的。

错误的查询二：
如果加上按照 first_name 排序呢 :

 1 mysql> set @row_num :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> SELECT actor_id,@row_num :=@row_num+1 AS cnt 
 5 -> FROM sakila.actor 
 6 -> WHERE @row_num <=1 
 7 -> order by first_name;
 8 +----------+------+
 9 | actor_id | cnt |
10 +----------+------+
11 | 71 | 1 |
12 | 132 | 2 |
13 | 165 | 3 |
14 | 173 | 4 |
15 | 125 | 5 |
16 | 146 | 6 |
17 | 29 | 7 |
18 | 65 | 8 |
19 | 144 | 9 |
20 | 76 | 10 |
21 | 49 | 11 |
22 | 34 | 12 |
23 | 190 | 13 |
24 | 196 | 14 |
25 | 83 | 15 |
26 .. ...
27 返回了所有行,再看下查询计划:
28 
29 +----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------+
30 | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
31 +----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------+
32 | 1 | SIMPLE | actor | ALL | NULL | NULL | NULL | NULL | 200 | Using where; Using filesort |
33 +----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------+
34 1 row in set (0.00 sec)

 

可以看出原因是 Using where 是在排序操作之前取值的,所以输出了全部的行。

解决这个问题的方法是:让变量的赋值和取值发生在执行查询的统一阶段:

正确的查询:

 1 mysql> set @row_num :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> SELECT actor_id,@row_num AS cnt
 5 -> FROM sakila.actor 
 6 -> WHERE (@row_num :=@row_num+1) <=1 
 7 -> ;
 8 +----------+------+
 9 | actor_id | cnt |
10 +----------+------+
11 | 58 | 1 |
12 +----------+------+
13 1 row in set (0.00 sec)
14 
15 看一下执行计划
16 
17 +----+-------------+-------+-------+---------------+---------------------+---------+------+------+--------------------------+
18 | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
19 +----+-------------+-------+-------+---------------+---------------------+---------+------+------+--------------------------+
20 | 1 | SIMPLE | actor | index | NULL | idx_actor_last_name | 137 | NULL | 200 | Using where; Using index |
21 +----+-------------+-------+-------+---------------+---------------------+---------+------+------+--------------------------+
22 1 row in set (0.00 sec)

 

想一想 如果加上ORDER BY 该怎么写？

 1 mysql> set @row_num :=0;
 2 Query OK, 0 rows affected (0.00 sec)
 3 
 4 mysql> SELECT actor_id,first_name ,@row_num AS row_num
 5 -> FROM sakila.actor 
 6 -> WHERE @row_num<=1
 7 -> ORDER BY first_name , least(0, @row_num :=@row_num+1) 
 8 -> ;
 9 
10 +----------+------------+---------+
11 | actor_id | first_name | row_num |
12 +----------+------------+---------+
13 | 2 | NICK | 2 |
14 | 1 | PENELOPE | 1 |
15 +----------+------------+---------+
16 2 rows in set (0.00 sec)
17 
18 
19 mysql> select @row_num;
20 +----------+
21 | @row_num |
22 +----------+
23 | 2 |
24 +----------+
25 1 row in set (0.00 sec)
26 
27 看一下执行计划:
28 
29 +----+-------------+-------+------+---------------+------+---------+------+------+----------------------------------------------+
30 | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
31 +----+-------------+-------+------+---------------+------+---------+------+------+----------------------------------------------+
32 | 1 | SIMPLE | actor | ALL | NULL | NULL | NULL | NULL | 200 | Using where; Using temporary; Using filesort |
33 +----+-------------+-------+------+---------------+------+---------+------+------+----------------------------------------------+
34 1 row in set (0.00 sec)
35 
36 SELECT actor_id,first_name ,@row_num:=@row_num+1 AS row_num 
37 FROM sakila.actor 
38 WHERE @row_num<=1 
39 ORDER BY first_name , least(0, @row_num :=@row_num+1)

 

六. UNION的巧妙改写

假设有两张用户表,一张主用户表,存放着活跃用户；一些归档用户表,存放着长期不活跃的用户。现在需要查找id 为123的客户。
先看下这个语句

1 select id from users where id= 123
2 union all
3 select id from users_archived where id =123

上面的语句是可以执行的，但是效率不好，因为两张表都必须查询一次

引入自定义变量的改写:

1 SELECT GREATEST(@found:=-1,id) AS id ,‘users‘ AS which_tbl 
2 FROM users WHERE id =123
3 UNION ALL
4 SEELCT id,‘users_archived‘ FROM users_archived WHERE id = 123 AND @found IS NULL
5 UNION ALL
6 SELECT 1,‘reset‘ FROM DUAL WHERE (@found:=NULL) IS NOT NULL

 

上面的改写非常巧妙：
第一段，如果在users查询到记录，则为@found赋值,也不会查询第二段;如果没有查询到记录,@found 为 null ,执行第二段。
第三段没有输出 ,只是简单的重置@found 为null。另外 GREATEST(@found:=-1,id) 也不会影响输出！