前「n」結果爲每個關鍵字

Question

我有一個查詢來獲取前「n」誰在特定關鍵字的評論的用戶，

SELECT `user` , COUNT(*) AS magnitude 
FROM `results` 
WHERE `keyword` = "economy" 
GROUP BY `user` 
ORDER BY magnitude DESC 
LIMIT 5 

我有大約6000關鍵字，並想運行此查詢讓我成爲我們擁有數據的每個關鍵字的頂級'n'用戶。援助感謝。

Answer 1

因爲你沒有給該架構results，我會認爲它的這個或非常相似（可能額外列）：

create table results (
    id int primary key, 
    user int, 
    foreign key (user) references <some_other_table>(id), 
    keyword varchar(<30>) 
); 

第1步：的總量，除以keyword/user在你的榜樣查詢，但所有關鍵字：

create view user_keyword as (
    select 
    keyword, 
    user, 
    count(*) as magnitude 
    from results 
    group by keyword, user 
); 

步驟2：秩每個關鍵字組內的每個用戶（請注意使用子查詢的排名的行）：

create view keyword_user_ranked as (
    select 
    keyword, 
    user, 
    magnitude, 
    (select count(*) 
    from user_keyword 
    where l.keyword = keyword and magnitude >= l.magnitude 
    ) as rank 
    from 
    user_keyword l 
); 

步驟3：在秩是只選擇行小於某一數目：

select * 
from keyword_user_ranked 
where rank <= 3; 

---

實施例：

個

基礎數據使用：

mysql> select * from results; 
+----+------+---------+ 
| id | user | keyword | 
+----+------+---------+ 
| 1 | 1 | mysql | 
| 2 | 1 | mysql | 
| 3 | 2 | mysql | 
| 4 | 1 | query | 
| 5 | 2 | query | 
| 6 | 2 | query | 
| 7 | 2 | query | 
| 8 | 1 | table | 
| 9 | 2 | table | 
| 10 | 1 | table | 
| 11 | 3 | table | 
| 12 | 3 | mysql | 
| 13 | 3 | query | 
| 14 | 2 | mysql | 
| 15 | 1 | mysql | 
| 16 | 1 | mysql | 
| 17 | 3 | query | 
| 18 | 4 | mysql | 
| 19 | 4 | mysql | 
| 20 | 5 | mysql | 
+----+------+---------+ 

分組由關鍵字和用戶：

mysql> select * from user_keyword order by keyword, magnitude desc; 
+---------+------+-----------+ 
| keyword | user | magnitude | 
+---------+------+-----------+ 
| mysql | 1 |   4 | 
| mysql | 2 |   2 | 
| mysql | 4 |   2 | 
| mysql | 3 |   1 | 
| mysql | 5 |   1 | 
| query | 2 |   3 | 
| query | 3 |   2 | 
| query | 1 |   1 | 
| table | 1 |   2 | 
| table | 2 |   1 | 
| table | 3 |   1 | 
+---------+------+-----------+ 

用戶關鍵字內排名：

mysql> select * from keyword_user_ranked where rank <= 2 order by keyword, rank asc; 
+---------+------+-----------+------+ 
| keyword | user | magnitude | rank | 
+---------+------+-----------+------+ 
| mysql | 1 |   4 | 1 | 
| query | 2 |   3 | 1 | 
| query | 3 |   2 | 2 | 
| table | 1 |   2 | 1 | 
+---------+------+-----------+------+ 

：

mysql> select * from keyword_user_ranked order by keyword, rank asc; 
+---------+------+-----------+------+ 
| keyword | user | magnitude | rank | 
+---------+------+-----------+------+ 
| mysql | 1 |   4 | 1 | 
| mysql | 2 |   2 | 3 | 
| mysql | 4 |   2 | 3 | 
| mysql | 3 |   1 | 5 | 
| mysql | 5 |   1 | 5 | 
| query | 2 |   3 | 1 | 
| query | 3 |   2 | 2 | 
| query | 1 |   1 | 3 | 
| table | 1 |   2 | 1 | 
| table | 3 |   1 | 3 | 
| table | 2 |   1 | 3 | 
+---------+------+-----------+------+ 

僅從每個關鍵字頂部2

---

注意，當有關係 - 看到用戶2和4的關鍵字「的MySQL」的例子 - 各方的配合獲得了「最後一個」級別，即如果第二和第三綁，兩者都是分配等級3.

---

性能：向關鍵字和用戶列添加索引將有所幫助。我用類似的方法查詢了一個表，兩列（在600000行表中）有4000和1300個不同的值。您可以添加像這樣的索引：

alter table results add index keyword_user (keyword, user); 

在我的情況下，查詢時間從大約6秒下降到大約2秒。

Answer 2

您可以使用這樣的模式（從Within-group quotas (Top N per group)):

SELECT tmp.ID, tmp.entrydate 
FROM ( 
    SELECT 
    ID, entrydate, 
    IF(@prev <> ID, @rownum := 1, @rownum := @rownum+1) AS rank, 
    @prev := ID 
    FROM test t 
    JOIN (SELECT @rownum := NULL, @prev := 0) AS r 
    ORDER BY t.ID 
) AS tmp 
WHERE tmp.rank <= 2 
ORDER BY ID, entrydate; 
+------+------------+ 
| ID | entrydate | 
+------+------------+ 
| 1 | 2007-05-01 | 
| 1 | 2007-05-02 | 
| 2 | 2007-06-03 | 
| 2 | 2007-06-04 | 
| 3 | 2007-07-01 | 
| 3 | 2007-07-02 | 
+------+------------+