MySQL选择最频繁的组
https://stackoverflow.com/questions/1407723
-
05-07-2019 - |
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如何获取MySQL中每个标记最常出现的类别？理想情况下，我想模拟一个计算模式的聚合函数。列。
SELECT
t.tag
, s.category
FROM tags t
LEFT JOIN stuff s
USING (id)
ORDER BY tag;
+------------------+----------+
| tag | category |
+------------------+----------+
| automotive | 8 |
| ba | 8 |
| bamboo | 8 |
| bamboo | 8 |
| bamboo | 8 |
| bamboo | 8 |
| bamboo | 8 |
| bamboo | 10 |
| bamboo | 8 |
| bamboo | 9 |
| bamboo | 8 |
| bamboo | 10 |
| bamboo | 8 |
| bamboo | 9 |
| bamboo | 8 |
| banana tree | 8 |
| banana tree | 8 |
| banana tree | 8 |
| banana tree | 8 |
| bath | 9 |
+-----------------------------+
解决方案
SELECT t1.*
FROM (SELECT tag, category, COUNT(*) AS count
FROM tags INNER JOIN stuff USING (id)
GROUP BY tag, category) t1
LEFT OUTER JOIN
(SELECT tag, category, COUNT(*) AS count
FROM tags INNER JOIN stuff USING (id)
GROUP BY tag, category) t2
ON (t1.tag = t2.tag AND (t1.count < t2.count
OR t1.count = t2.count AND t1.category < t2.category))
WHERE t2.tag IS NULL
ORDER BY t1.count DESC;
我同意这对单个SQL查询来说太过分了。在子查询中使用 GROUP BY 会让我感到畏缩。您可以使用视图使外观更简单：
CREATE VIEW count_per_category AS
SELECT tag, category, COUNT(*) AS count
FROM tags INNER JOIN stuff USING (id)
GROUP BY tag, category;
SELECT t1.*
FROM count_per_category t1
LEFT OUTER JOIN count_per_category t2
ON (t1.tag = t2.tag AND (t1.count < t2.count
OR t1.count = t2.count AND t1.category < t2.category))
WHERE t2.tag IS NULL
ORDER BY t1.count DESC;
但它基本上是在幕后做同样的工作。
您评论说您可以在应用程序代码中轻松执行类似的操作。那你为什么不这样做呢？做更简单的查询以获得每个类别的计数：
SELECT tag, category, COUNT(*) AS count
FROM tags INNER JOIN stuff USING (id)
GROUP BY tag, category;
并在应用程序代码中对结果进行排序。
其他提示
SELECT tag, category
FROM (
SELECT @tag <> tag AS _new,
@tag := tag AS tag,
category, COUNT(*) AS cnt
FROM (
SELECT @tag := ''
) vars,
stuff
GROUP BY
tag, category
ORDER BY
tag, cnt DESC
) q
WHERE _new
在您的数据上，返回以下内容：
'automotive', 8
'ba', 8
'bamboo', 8
'bananatree', 8
'bath', 9
这是测试脚本：
CREATE TABLE stuff (tag VARCHAR(20) NOT NULL, category INT NOT NULL);
INSERT
INTO stuff
VALUES
('automotive',8),
('ba',8),
('bamboo',8),
('bamboo',8),
('bamboo',8),
('bamboo',8),
('bamboo',8),
('bamboo',10),
('bamboo',8),
('bamboo',9),
('bamboo',8),
('bamboo',10),
('bamboo',8),
('bamboo',9),
('bamboo',8),
('bananatree',8),
('bananatree',8),
('bananatree',8),
('bananatree',8),
('bath',9);
（编辑：在ORDER BY中忘记了DESC）
使用子查询中的LIMIT很容易。 MySQL仍然没有LIMIT-in-subqueries限制吗？下面的例子是使用PostgreSQL。
=> select tag, (select category from stuff z where z.tag = s.tag group by tag, category order by count(*) DESC limit 1) AS category, (select count(*) from stuff z where z.tag = s.tag group by tag, category order by count(*) DESC limit 1) AS num_items from stuff s group by tag;
tag | category | num_items
------------+----------+-----------
ba | 8 | 1
automotive | 8 | 1
bananatree | 8 | 4
bath | 9 | 1
bamboo | 8 | 9
(5 rows)
只有在需要计数时才需要第三栏。
这适用于更简单的情况：
SELECT action，COUNT（action）AS ActionCount
从日志
GROUP BY行动
ORDER BY ActionCount DESC;
这是一个hacky方法，利用 max 聚合函数，因为在MySQL（或窗口函数等）中没有模式聚合函数可以允许：
SELECT
tag,
convert(substring(max(concat(lpad(c, 20, '0'), category)), 21), int)
AS most_frequent_category
FROM (
SELECT tag, category, count(*) AS c
FROM tags INNER JOIN stuff using (id)
GROUP BY tag, category
) as grouped_cats
GROUP BY tag;
基本上它利用了这样一个事实，即我们可以找到每个类别的计数的词汇最大值。
使用命名类别更容易看到：
create temporary table tags (id int auto_increment primary key, tag character varying(20));
create temporary table stuff (id int, category character varying(20));
insert into tags (tag) values ('automotive'), ('ba'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('bamboo'), ('banana tree'), ('banana tree'), ('banana tree'), ('banana tree'), ('bath');
insert into stuff (id, category) values (1, 'cat-8'), (2, 'cat-8'), (3, 'cat-8'), (4, 'cat-8'), (5, 'cat-8'), (6, 'cat-8'), (7, 'cat-8'), (8, 'cat-10'), (9, 'cat-8'), (10, 'cat-9'), (11, 'cat-8'), (12, 'cat-10'), (13, 'cat-8'), (14, 'cat-9'), (15, 'cat-8'), (16, 'cat-8'), (17, 'cat-8'), (18, 'cat-8'), (19, 'cat-8'), (20, 'cat-9');
在这种情况下，我们不应该在 most_frequent_category 列上进行整数转换：
SELECT
tag,
substring(max(concat(lpad(c, 20, '0'), category)), 21) AS most_frequent_category
FROM (
SELECT tag, category, count(*) AS c
FROM tags INNER JOIN stuff using (id)
GROUP BY tag, category
) as grouped_cats
GROUP BY tag;
+-------------+------------------------+
| tag | most_frequent_category |
+-------------+------------------------+
| automotive | cat-8 |
| ba | cat-8 |
| bamboo | cat-8 |
| banana tree | cat-8 |
| bath | cat-9 |
+-------------+------------------------+
为了更深入地了解正在发生的事情，这里的 grouped_cats 内部选择看起来像（我已添加 order by tag，c desc ） ：
+-------------+----------+---+
| tag | category | c |
+-------------+----------+---+
| automotive | cat-8 | 1 |
| ba | cat-8 | 1 |
| bamboo | cat-8 | 9 |
| bamboo | cat-10 | 2 |
| bamboo | cat-9 | 2 |
| banana tree | cat-8 | 4 |
| bath | cat-9 | 1 |
+-------------+----------+---+
如果我们省略 substring 位，我们可以看到 count（*）列的最大值如何沿着它的相关类别拖动：
SELECT
tag,
max(concat(lpad(c, 20, '0'), category)) AS xmost_frequent_category
FROM (
SELECT tag, category, count(*) AS c
FROM tags INNER JOIN stuff using (id)
GROUP BY tag, category
) as grouped_cats
GROUP BY tag;
+-------------+---------------------------+
| tag | xmost_frequent_category |
+-------------+---------------------------+
| automotive | 00000000000000000001cat-8 |
| ba | 00000000000000000001cat-8 |
| bamboo | 00000000000000000009cat-8 |
| banana tree | 00000000000000000004cat-8 |
| bath | 00000000000000000001cat-9 |
+-------------+---------------------------+
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