如何在bigquery中按天计算（非）连续记录？答案

【问题标题】：How to count (non) consecutive records by day in bigquery?如何在bigquery中按天计算（非）连续记录？
【发布时间】：2019-03-26 14:26:44
【问题描述】：

记录设备的每个故障。每个条目都包含一个 customer_id、device_id 和时间戳：

+-------------+-----------+-----------------------+
| customer_id | device_id |  timestamp            |
+-------------+-----------+-----------------------+
| 1           | 1         |  2019-02-12T01:00:00  |
| 2           | 2         |  2019-02-12T01:00:00  |
| 1           | 1         |  2019-02-12T02:00:00  |
| 1           | 1         |  2019-02-12T03:00:00  |
+-------------+-----------+-----------------------+

每小时收集一次故障日志。我对以下信息感兴趣：

每位客户每天的故障总数
每位客户每天的连续故障次数
每位客户每天的非连续故障数

设备可能出现故障数小时，这可能表示硬件故障。另一方面，如果设备出现不超过数小时的故障，则可能只是设备使用不当。

结果应该是这样的：

+-------------+-----------+---------------------+-----------------+------------+-----------------------+
| customer_id | device_id | total | consecutive | non consecutive |  day       | last_recording        |
+-----+-------------------+-------+-------------+-----------------+------------------------------------+
| 1           | 1         | 3     |  1          | 2               | 2019-02-12 |  2019-02-12T03:00:00  |
| 2           | 2         | 1     |  0          | 1               | 2019-02-12 |  2019-02-12T01:00:00  |
+-------------+-----------+-------+-------------+-----------------+------------+-----------------------+

在上面的示例中，设备 1 在 2019-02-12T02:00:00 报告了一个故障，这被认为是“非连续的”，然后在 2019-02-12T03:00:00 报告了另一个故障，这被认为是“连续”。

我想创建一个查询，生成这样的结果。我试过的

SELECT customer_id, device_id, COUNT(customer_id) AS count, FORMAT_TIMESTAMP("%Y-%m-%d", TIMESTAMP(timestamp)) as day
FROM `malfunctions`
GROUP BY day, customer_id, device_id

这样我可以得到客户每天的故障总数。我想我必须使用 LEAD 运算符来获得（非）连续计数，但我不确定如何。有任何想法吗？结果应该是按天“滚动”的。

【问题讨论】：

标签： google-bigquery

【解决方案1】：

以下是 BigQuery 标准 SQL

#standardSQL
SELECT customer_id, device_id, day, SUM(batch_count) total, 
  SUM(batch_count) - COUNTIF(batch_count = 1) consecutive,
  COUNTIF(batch_count = 1) non_consecutive, 
  ARRAY_AGG(STRUCT(batch AS batch, batch_count AS batch_count, first_recording AS first_recording, last_recording AS last_recording)) details
FROM (
  SELECT customer_id, device_id, day, batch, 
    COUNT(1) batch_count,
    MIN(ts) first_recording,
    MAX(ts) last_recording
  FROM (
    SELECT customer_id, device_id, ts, day,
      COUNTIF(gap) OVER(PARTITION BY customer_id, device_id, day ORDER BY  ts) batch
    FROM (
      SELECT customer_id, device_id, ts, DATE(ts) day,
        IFNULL(TIMESTAMP_DIFF(ts, LAG(ts) OVER(PARTITION BY customer_id, device_id, DATE(ts) ORDER BY  ts), HOUR), 777) > 1 gap
      FROM `project.dataset.malfunctions`
    )
  )
  GROUP BY customer_id, device_id, day, batch
)
GROUP BY customer_id, device_id, day

你可以像下面的例子一样使用虚拟数据测试，玩上面的例子

#standardSQL
WITH `project.dataset.malfunctions` AS (
  SELECT 1 customer_id, 1 device_id, TIMESTAMP '2019-02-12T01:00:00' ts UNION ALL
  SELECT 1, 1, '2019-02-12T02:00:00' UNION ALL
  SELECT 1, 1, '2019-02-12T03:00:00' UNION ALL
  SELECT 1, 1, '2019-02-12T04:00:00' UNION ALL
  SELECT 1, 1, '2019-02-12T09:00:00' UNION ALL
  SELECT 1, 1, '2019-02-12T10:00:00' UNION ALL
  SELECT 1, 1, '2019-02-13T03:00:00' UNION ALL
  SELECT 2, 2, '2019-02-12T01:00:00' 
)
SELECT customer_id, device_id, day, SUM(batch_count) total, 
  SUM(batch_count) - COUNTIF(batch_count = 1) consecutive,
  COUNTIF(batch_count = 1) non_consecutive, 
  ARRAY_AGG(STRUCT(batch AS batch, batch_count AS batch_count, first_recording AS first_recording, last_recording AS last_recording)) details
FROM (
  SELECT customer_id, device_id, day, batch, 
    COUNT(1) batch_count,
    MIN(ts) first_recording,
    MAX(ts) last_recording
  FROM (
    SELECT customer_id, device_id, ts, day,
      COUNTIF(gap) OVER(PARTITION BY customer_id, device_id, day ORDER BY  ts) batch
    FROM (
      SELECT customer_id, device_id, ts, DATE(ts) day,
        IFNULL(TIMESTAMP_DIFF(ts, LAG(ts) OVER(PARTITION BY customer_id, device_id, DATE(ts) ORDER BY  ts), HOUR), 777) > 1 gap
      FROM `project.dataset.malfunctions`
    )
  )
  GROUP BY customer_id, device_id, day, batch
)
GROUP BY customer_id, device_id, day
-- ORDER BY customer_id, device_id, day

结果

【讨论】：