Question

我有一个代表消息的BigQuery表，每条消息都属于一个对话并有一个日期：

conversation    date    
1               2018-06-22 23:16:46.456 UTC
2               2018-06-05 00:07:12.178 UTC
1               2018-06-22 23:16:46.456 UTC
4               2018-06-05 00:07:12.178 UTC
3               2018-06-22 23:51:28.540 UTC
3               2018-06-23 00:02:59.285 UTC
4               2018-06-04 23:21:59.500 UTC

我需要获取在对话中花费的平均时间

我使用此查询来获取它：

SELECT conversation, timestamp_diff(MAX(date), MIN(date), MINUTE) minutes
FROM `Message`
GROUP BY conversation

但是由于某些对话需要几天的时间，因此当消息之间的间隔大于1小时时，必须将它们分成较小的块：

conversation    date    
2               2018-06-22 00:01:46.456 UTC   # group 1
2               2018-06-22 00:07:12.178 UTC   # group 1
2               2018-06-22 00:16:46.456 UTC   # group 1
2               2018-06-22 01:07:42.178 UTC   # group 1
      there is a gap here
2               2018-06-22 12:51:28.540 UTC   # group 2
2               2018-06-22 13:00:40.486 UTC   # group 2
      there is another gap here
2               2018-06-22 19:54:30.031 UTC   # group 3

我认为使用解析函数可以实现： https://cloud.google.com/bigquery/docs/reference/standard-sql/analytic-function-concepts

但是我不知道该怎么做，任何帮助都会得到真正的帮助。

Answer 1

下面是BigQuery标准SQL

当消息之间的间隔大于1小时时，必须将它们切成小块：

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 2 conversation, TIMESTAMP '2018-06-22 00:01:46.456 UTC' dt UNION ALL   # group 1
  SELECT 2, '2018-06-22 00:07:12.178 UTC' UNION ALL   # group 1
  SELECT 2, '2018-06-22 00:16:46.456 UTC' UNION ALL   # group 1
  SELECT 2, '2018-06-22 01:07:42.178 UTC' UNION ALL   # group 1
  SELECT 2, '2018-06-22 12:51:28.540 UTC' UNION ALL   # group 2
  SELECT 2, '2018-06-22 13:00:40.486 UTC' UNION ALL   # group 2
  SELECT 2, '2018-06-22 19:54:30.031 UTC'             # group 3
), conversation_groups AS (
  SELECT 
    conversation, dt, 
    SUM(flag) OVER(PARTITION BY conversation ORDER BY dt) conversation_group
  FROM (
    SELECT 
      conversation, dt, 
      SIGN(IFNULL(TIMESTAMP_DIFF(dt, LAG(dt) OVER(PARTITION BY conversation ORDER BY dt), HOUR), 0)) flag
    FROM `project.dataset.table`
  )
)
SELECT *
FROM conversation_groups
ORDER BY conversation, dt

结果为

Row conversation    dt                              conversation_group   
1   2               2018-06-22 00:01:46.456 UTC     0    
2   2               2018-06-22 00:07:12.178 UTC     0    
3   2               2018-06-22 00:16:46.456 UTC     0    
4   2               2018-06-22 01:07:42.178 UTC     0    
5   2               2018-06-22 12:51:28.540 UTC     1    
6   2               2018-06-22 13:00:40.486 UTC     1    
7   2               2018-06-22 19:54:30.031 UTC     2

我需要获取在对话中花费的平均时间

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 2 conversation, TIMESTAMP '2018-06-22 00:01:46.456 UTC' dt UNION ALL   # group 1
  SELECT 2, '2018-06-22 00:07:12.178 UTC' UNION ALL   # group 1
  SELECT 2, '2018-06-22 00:16:46.456 UTC' UNION ALL   # group 1
  SELECT 2, '2018-06-22 01:07:42.178 UTC' UNION ALL   # group 1
  SELECT 2, '2018-06-22 12:51:28.540 UTC' UNION ALL   # group 2
  SELECT 2, '2018-06-22 13:00:40.486 UTC' UNION ALL   # group 2
  SELECT 2, '2018-06-22 19:54:30.031 UTC'             # group 3
), conversation_groups AS (
  SELECT 
    conversation, dt, 
    SUM(flag) OVER(PARTITION BY conversation ORDER BY dt) conversation_group
  FROM (
    SELECT 
      conversation, dt, 
      SIGN(IFNULL(TIMESTAMP_DIFF(dt, LAG(dt) OVER(PARTITION BY conversation ORDER BY dt), HOUR), 0)) flag
    FROM `project.dataset.table`
  )
)
SELECT conversation, AVG(IF(duration = 0, NULL, duration)) avg_duration
FROM (
  SELECT
    conversation, conversation_group,
    TIMESTAMP_DIFF(MAX(dt), MIN(dt), MINUTE) duration
  FROM conversation_groups
  GROUP BY conversation, conversation_group
)
GROUP BY conversation
ORDER BY conversation

结果为

Row conversation    avg_duration     
1   2               37.0

注意：您可以根据自己的特殊需求/愿景来调整计算平均值的逻辑-但上面的操作方式是-首先计算每组的持续时间，然后计算这些组的持续时间的平均值注意-如果持续时间为零，则将其替换为NULL，因此不会影响平均计算。持续时间以MINUTE计算，但您可以选择SECOND，也可以选择

有间隔约束的BigQuery平均时间

1 个答案: