查找分区中最大的连续数字组

Question

我有以下数据由player_id和match_date排序。 我想找出连续运行次数最多的记录组（连续3次从2014-04-03到2014-04-12 4次运行）

 player_id  match_date  runs
    1       2014-04-01    5
    1       2014-04-02    55       
    1       2014-04-03    4       
    1       2014-04-10    4       
    1       2014-04-12    4       
    1       2014-04-14    3       
    1       2014-04-19    4       
    1       2014-04-20    44               
    2       2014-04-01    23
    2       2014-04-02    23       
    2       2014-04-03    23       
    2       2014-04-10    23       
    2       2014-04-12    4       
    2       2014-04-14    3       
    2       2014-04-19    23       
    2       2014-04-20    1   

我想出了以下SQL：

select *,row_number() over (partition by ranked.player_id,ranked.runs
order by ranked.match_date) as R from (
select player_id ,match_date,runs from players order by 1,2 desc )
ranked order by ranked.player_id, match_date asc

但这继续从之前的连续运行中排名（2014-04-19的4次运行，预计玩家1将获得排名1，但由于已经发生了3次同一分区，因此获得了排名4）。 类似地，2014-04-19的玩家2的23次运行预计将获得等级1，但是因为此玩家已经有4次出现23次运行，所以获得等级5。

当运行值从上一行更改时，如何将等级重置为1？

SQLFiddle上提供了架构，数据，SQL和输出。

Answer 1

select p1.player_id, p1.match_date, p1.runs, count(p2.match_date) from players p1
join players p2 on p1.player_id = p2.player_id
    and p1.match_date >= p2.match_date
    and p1.runs = p2.runs
    and not exists (
        select 1 from players p3
        where p3.runs <> p2.runs
        and p3.player_id = p2.player_id
        and p3.match_date < p1.match_date
        and p3.match_date > p2.match_date
    )
group by p1.player_id, p1.match_date, p1.runs
order by p1.player_id, p1.match_date

http://sqlfiddle.com/#!15/78a77/1

Answer 2

您可以使用窗口功能执行此操作。

select player_id, runs, count(*) as numruns
from (select p.*,
             (row_number() over (partition by player_id order by match_date) -
              row_number() over (partition by player_id, runs order by match_date)
             ) as grp
      from players p
     ) pg
group by grp, player_id, runs
order by numruns desc
limit 1;

关键的观察是“按序列运行”具有以下属性：如果按日期枚举行（对于每个玩家）并且您枚举每个玩家的行和按日期运行，则运行时差异是不变的是完全相同的。 这形成了一个可用于聚合的组，以识别您想要的玩家。

这是SQL小提琴。