在 BigQuery 中使用 LEAD

Question

假设我的 table 结构是这样的

我计划按（USER 和 SEQUENCE）对其进行分组，并获取下一个序列的 LEAD 时间戳。这是我正在寻找的输出

如果可能的话，我可以不用 JOIN 使用 LEAD 函数来解决这个问题吗？

Answer 1

不确定bigquery，但一般SQL会写成：

select user, sequence, LEAD (max_timestamp,1) OVER (PARTITION BY user ORDER BY sequence) as timestamp
from (
    select user, sequence, max(timestamp) as max_timestamp
    from table
    group by user, sequence) q1;

请注意保留字，例如 table、用户、时间戳等

编辑：是的，忘了这个答案吧，对所需的输出不够关注。米哈伊尔答对了！

Answer 2

以下适用于 BigQuery 标准 SQL

我将提供两个选项 - 使用 JOIN（只是为了证明我 understood/reversed-engineered 正确地预期了逻辑）然后 JOIN-less 版本（注意我使用 ts 作为字段名称而不是timestamp)

Using JOIN

#standardSQL
SELECT a.user, a.sequence, MIN(b.ts) ts 
FROM (
  SELECT user, sequence, MAX(ts) AS max_ts
  FROM `project.dataset.table`
  GROUP BY user, sequence
) a
LEFT JOIN `project.dataset.table` b
ON a.user = b.user AND b.sequence = a.sequence + 1
WHERE a.max_ts <= IFNULL(b.ts, a.max_ts)
GROUP BY user, sequence
-- ORDER BY user, sequence

JOIN-less version

#standardSQL
SELECT
  user, sequence, 
  (
    SELECT ts FROM UNNEST(arr_ts) ts 
    WHERE max_ts < ts ORDER BY ts LIMIT 1
  ) ts
FROM (
  SELECT
    user, sequence, max_ts,
    LEAD(arr_ts) OVER (PARTITION BY user ORDER BY sequence) arr_ts
  FROM (
  SELECT 
      user, sequence, MAX(ts) max_ts, 
      ARRAY_AGG(ts ORDER BY ts) arr_ts
    FROM `project.dataset.table`
    GROUP BY user, sequence
  )
)
-- ORDER BY user, sequence

以上两个版本都可以使用以下虚拟数据进行测试/播放

WITH `project.dataset.table` AS (
  SELECT 'user1' user, 2 sequence, 'T1' ts UNION ALL
  SELECT 'user1', 2, 'T2' UNION ALL
  SELECT 'user1', 1, 'T3' UNION ALL
  SELECT 'user1', 1, 'T4' UNION ALL
  SELECT 'user1', 3, 'T5' UNION ALL
  SELECT 'user1', 2, 'T6' UNION ALL
  SELECT 'user1', 3, 'T7' UNION ALL
  SELECT 'user1', 3, 'T8' 
)

并且 returns 都低于结果

user    sequence    ts   
user1   1           T6   
user1   2           T7   
user1   3           null

在 BigQuery 中使用 LEAD

Using LEAD in BigQuery

sql

lead

google-bigquery