如何获得平均穿越的城市数

Question

我是数据库的初学者。我需要写一些 SQL 查询。

table 是：

远征(id, number, id_captain, id_ship, id_heros)
城市(id, name)
英雄联盟(id, family_name, first_name)
步骤(id, index, id_expedition, id_city)

示例数据：

'Table expedition'

id	个数	id_captain	id_ship	id_hero
1	1	1	10	8
2	2	2	1	5
3	3	1	8	3
4	4	10	9	6
5	5	5	7	4
6	6	6	5	4
7	7	7	3	7
8	8	8	2	8
9	9	9	1	3
10	10	1	4	2
11	11	6	3	1
12	12	8	6	1
13	13	5	8	6
14	14	4	9	9
15	15	3	10	4
16	16	10	2	2
17	17	9	3	3
18	18	8	7	7
19	19	9	8	10
20	20	7	2	2

table 'heros'

id	family_name	first_name
1	familyname1	名字 1
2	familyname2	名字 2
3	familyname3	名字 3
4	familyname4	名字 4
5	familyname5	名字 5
6	familyname6	firstname6
7	familyname7	名字 7
8	familyname8	名字 8
9	familyname9	名字 9
10	familyname10	名字 10

query1:旅行最少的家庭（以姓氏为准）（最少的城市不同路口）。

我已经为第一个查询做了这个：

select expedition.id, id_hero, heros.family_name as Famille_expedition, count(distinct id_city) as city_count
from expedition, step, heros 
where expedition.id=step.id and expedition.id_hero=heros.id
group by id_hero
having city_count = 
(select count(distinct id_city) as min_city_count
from expedition, step
where expedition.id=step.id
group by id_hero
order by min_city_count asc
limit 1);

query2:一次探险穿越城市的平均值

第二个不知道怎么回答

Answer 1

那么，首先问问自己，您需要哪些信息来回答您的问题？

从你的问题来看，我想说平均交叉次数只是 steps table 中所有条目的总和除以 expeditions 的数量，因为在每一步中，访问一个城市，所有访问的平均值就是您要查找的内容：

SELECT (
       (SELECT COUNT(s.id_city)
          FROM step AS s) /
       (SELECT COUNT(e.id)
          FROM expedition AS e) ) AS total_average__cities

也就是说，这取决于您如何准确定义城市数量和交汇。想象以下 table step 的示例数据：

id	idx	id_expedition	id_city
1	1	1	1
2	2	1	5
3	3	1	3
4	1	2	5
5	2	2	9
6	1	3	8
7	2	3	5
8	3	3	9
9	4	3	5
10	5	3	8

table列出了三个探险的步骤。探险 1 从一个城市经过另一个城市到达第三个城市。 Expedition 2 直接从一个城市前往另一个城市。探险 3 穿过几个城市，沿途访问一个城市两次，还 returns 到达它开始的城市。

所有这些步骤的平均城市数是 (3 + 2 + 5 [cities in all steps]) / 3 [expeditions] = 3.3333。也就是上面查询的结果。

现在，如果您将城市数量定义为唯一个城市每个探险 , 远征 3 只访问了 3 个城市而不是 5 个。那么你的平均值计算为 (3 + 2 + 3 [unique cities/expedition in all steps]) / 3 [expeditions] = 2.6666。根据查询需要计算每个探险中的不同城市，然后再计算平均值：

SELECT (
       (SELECT SUM(cnt) FROM (SELECT COUNT(DISTINCT s.id_city) AS cnt
                                FROM step AS s
                            GROUP BY s.id_expedition) t) /
       (SELECT COUNT(e.id)
          FROM expedition AS e) ) AS total_average__cities

现在，如果您将穿越定义为只覆盖沿途个城市，探险1只穿越1个城市，探险2穿越none 完全没有。

那么您的查询也需要看起来不同。您需要过滤所有城市以排除每次探险的第一个和最后一个。子查询可能如下所示：

SELECT s.* FROM step s
JOIN (   SELECT id_expedition, 
                MAX(idx) AS max_idx, 
                MIN(idx) AS min_idx
           FROM step s
       GROUP BY id_expedition) minmax
  ON s.id_expedition = minmax.id_expedition
 AND s.idx > minmax.min_idx
 AND s.idx < minmax.max_idx

因此，如果您希望穿越的城市数不包括起点和终点，则平均值将计算为 (1 + 0 + 3 [intermediate cities in all steps]) / 3 [expeditions] = 1.3333。相应的查询将是

SELECT (
       (SELECT COUNT(s.id_city) 
          FROM step s
          JOIN (   SELECT id_expedition, 
                          MAX(idx) as max_idx, 
                          MIN(idx) as min_idx
                     FROM step s
                 GROUP BY id_expedition) minmax 
            ON s.id_expedition = minmax.id_expedition
           AND s.idx > minmax.min_idx
           AND s.idx < minmax.max_idx) /
       (SELECT COUNT(e.id)
          FROM expedition AS e) ) AS total_average__cities

最后，如果您想排除开始和停止并且只想计算独特的城市，您的平均值将被计算作为 (1 + 0 + 2 [unique intermediate cities in all steps]) / 3 [expeditions] = 1。以下查询结合了上述两种方法：

SELECT (
       (SELECT SUM(cnt) FROM (SELECT COUNT(DISTINCT id_city) AS cnt
                                FROM step s
                                JOIN (   SELECT id_expedition, 
                                                MAX(idx) AS max_idx, 
                                                MIN(idx) AS min_idx
                                           FROM step s
                                       GROUP BY id_expedition) minmax
                                  ON s.id_expedition = minmax.id_expedition
                                 AND s.idx > minmax.min_idx 
                                 AND s.idx < minmax.max_idx
                            GROUP BY s.id_expedition) t) /
       (SELECT COUNT(e.id)
          FROM expedition AS e) ) AS total_average_cities

您可以在 this db<>fiddle 中测试所有这些查询。

如何获得平均穿越的城市数

How to get the average number of cities crossed

mysql

sql

database

count

average

id	个数	id_captain	id_ship	id_hero
1	1	1	10	8
2	2	2	1	5
3	3	1	8	3
4	4	10	9	6
5	5	5	7	4
6	6	6	5	4
7	7	7	3	7
8	8	8	2	8
9	9	9	1	3
10	10	1	4	2
11	11	6	3	1
12	12	8	6	1
13	13	5	8	6
14	14	4	9	9
15	15	3	10	4
16	16	10	2	2
17	17	9	3	3
18	18	8	7	7
19	19	9	8	10
20	20	7	2	2

id	个数	id_captain	id_ship	id_hero
1	1	1	10	8
2	2	2	1	5
3	3	1	8	3
4	4	10	9	6
5	5	5	7	4
6	6	6	5	4
7	7	7	3	7
8	8	8	2	8
9	9	9	1	3
10	10	1	4	2
11	11	6	3	1
12	12	8	6	1
13	13	5	8	6
14	14	4	9	9
15	15	3	10	4
16	16	10	2	2
17	17	9	3	3
18	18	8	7	7
19	19	9	8	10
20	20	7	2	2

id	个数	id_captain	id_ship	id_hero
1	1	1	10	8
2	2	2	1	5
3	3	1	8	3
4	4	10	9	6
5	5	5	7	4
6	6	6	5	4
7	7	7	3	7
8	8	8	2	8
9	9	9	1	3
10	10	1	4	2
11	11	6	3	1
12	12	8	6	1
13	13	5	8	6
14	14	4	9	9
15	15	3	10	4
16	16	10	2	2
17	17	9	3	3
18	18	8	7	7
19	19	9	8	10
20	20	7	2	2