计算 SQL 中值转换之间的行数

Question

我有包含 user_id、时间戳和是/否答案的行。我想计算每个 ID 有多少条“NO”的条纹（连续行）。

示例：

user_id	timestamp	response	no_streak
1	2021-01-20 13:59:26	YES	0
1	2021-01-20 14:01:27	NO	1
1	2021-01-20 14:03:21	NO	2
1	2021-01-20 14:07:29	NO	3
1	2021-01-20 14:09:22	YES	0
1	2021-01-20 14:11:26	YES	0
1	2021-01-20 14:13:30	NO	1
1	2021-01-20 14:17:26	NO	2
1	2021-01-20 14:19:29	YES	0
1	2021-01-20 14:25:30	NO	1
1	2021-01-20 14:27:23	NO	2
1	2021-01-20 14:31:23	NO	3
1	2021-01-20 14:35:27	NO	4
1	2021-01-20 14:39:24	YES	0
2	2021-01-20 14:39:24	NO	1
2	2021-01-20 14:47:28	NO	2
2	2021-01-20 14:49:22	NO	3
2	2021-01-20 14:51:25	NO	4
2	2021-01-20 14:53:29	NO	5
2	2021-01-20 14:55:22	NO	6
2	2021-01-20 14:57:22	YES	0

最终我想知道每个用户的连续多长时间：

user_id	streak length
1	0
1	3
1	2
1	4
2	0
2	6

我可以使用 LAG() 找到“否”到“是”的过渡位置，反之亦然，但我很难计算每个过渡之间的行数.

Answer 1

计算每行“是”的数量，使相邻的“否”具有相同的分组值。然后过滤聚合：

select t.user_id, count(*), min(timestamp), max(timestamp)
from (select t.*,
             sum(case when response = 'YES' then 1 else 0 end) over (partition by user_id order by timestamp) as grp
      from t
     ) t
where response = 'NO'
group by user_id, grp;

注意：这不是 return 长度 0 的条纹。我不确定“连胜”这个词是否合适。但是要获取它们，请删除 where 过滤器并使用条件聚合：

select t.user_id, sum(case when response = 'NO' then 1 else 0 end),
       min(timestamp), max(timestamp)
from (select t.*,
             sum(case when response = 'YES' then 1 else 0 end) over (partition by user_id order by timestamp) as grp
      from t
     ) t
group by user_id, grp;

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