每日HiveSQL_求解运动员最大连胜的次数_15

1.现需要从运动员比赛结果表中统计每个运动员最大连胜的次数

需求结果：

2.所用到的表和数据

--表创建
CREATE TABLE athlete_results
(
    athlete_id INT,
    match_time TIMESTAMP,
    result     VARCHAR(10) -- 'win', 'lose', 'draw'
);

--数据装载
INSERT INTO athlete_results
VALUES (1, '2023-11-01', 'win'),
       (1, '2023-11-02', 'win'),
       (1, '2023-11-03', 'lose'),
       (1, '2023-11-04', 'win'),
       (1, '2023-11-05', 'draw'),
       (1, '2023-11-06', 'win'),
       (1, '2023-11-07', 'win'),
       (1, '2023-11-08', 'win'),
       (2, '2023-11-01', 'win'),
       (2, '2023-11-02', 'lose'),
       (2, '2023-11-03', 'draw'),
       (2, '2023-11-04', 'win'),
       (2, '2023-11-05', 'win'),
       (2, '2023-11-06', 'win');

3.答案

解析：由结果可知，此结果是对每个运动员进行分组得到的每个运动员的最大连胜数，此题关键在于如何根据特定的分组条件进行分组，得到每个可以代表运动员和比赛结果的分组字段。

思路一：

首先使用row_number()开窗根据运动员id进行分区，日期进行排序得到分组字段rn1，row_number()开窗根据运动员id和比赛结果进行分区，日期进行排序得到分组字段rn2,如下图rn1 - rn2可以唯一表述每个运动员在一次连续胜利、失败、平局的分组字段。

3.1?使用row_number()开两个窗，一个根据运动员id分区，日期升序排序，得到group1，另一个是根据运动员id和比赛结果进行分区，日期升序排序，得到group2

select athlete_id,
       to_date(match_time)                                                              match_time,
       result,
       row_number() over (partition by athlete_id order by to_date(match_time))         group1,
       row_number() over (partition by athlete_id, result order by to_date(match_time)) group2
from athlete_results;

运行结果：?

3.2筛选出比赛结果为”win“，然后再根据运动员id和win_group(group1 - group2)分组，count(*)得到运动员每个分组阶段的连胜次数

select athlete_id,
       group1 - group2 win_group,
       count(*)        win_count
from (
         select athlete_id,
                to_date(match_time)                                                              match_time,
                result,
                row_number() over (partition by athlete_id order by to_date(match_time))         group1,
                row_number() over (partition by athlete_id, result order by to_date(match_time)) group2
         from athlete_results
     ) t
where result = 'win'
group by athlete_id, group1 - group2;

运行结果：

3.3最后根据以上运行结果根据运动员id进行分组得到最终结果

select athlete_id,
       max(win_count) max_win_count
from (
         select athlete_id,
                group1 - group2 win_group,
                count(*)        win_count
         from (
                  select athlete_id,
                         to_date(match_time)                                                              match_time,
                         result,
                         row_number() over (partition by athlete_id order by to_date(match_time))         group1,
                         row_number() over (partition by athlete_id, result order by to_date(match_time)) group2
                  from athlete_results
              ) t
         where result = 'win'
         group by athlete_id, group1 - group2
     ) t
group by athlete_id;

运行结果：?

思路二：?

思考打破连续胜利的条件，仔细思考可知，一个运动员每次失败或者平局都会开启一个新的分组，怎么描述这个分组呢？很容易可以想到sum(if())结构，如下图?

3.1使用sum(if())开窗得到分组字段win_group

select athlete_id,
       to_date(match_time) match_day,
       result,
       sum(if(result = 'lose' or result = 'draw', 1, 0))
           over (partition by athlete_id order by to_date(match_time)) win_group
from athlete_results;

运行结果：

3.2根据以上结果筛选出result = ‘win’的结果，之后进行根据运动员id和win_group分组，count(*)得到连胜的次数。

select athlete_id,
       count(*) win_count
from (
         select athlete_id,
                result,
                sum(if(result = 'lose' or result = 'draw', 1, 0))
                    over (partition by athlete_id order by match_time) win_group
         from athlete_results) t
where result = 'win'
group by athlete_id, win_group;

运行结果：

3.3根据运动员id进行分组，得到最终结果

select athlete_id,
       max(win_count) max_win_count
from (
         select athlete_id,
                count(*) win_count
         from (
                  select athlete_id,
                         result,
                         sum(if(result = 'lose' or result = 'draw', 1, 0))
                             over (partition by athlete_id order by match_time) win_group
                  from athlete_results) t
         where result = 'win'
         group by athlete_id, win_group
     ) t
group by athlete_id;

运行结果：