Question

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  |
+-------------+-----------+-----------------------+

故障日志每小时收集一次。我对以下信息感兴趣：

每个客户每天发生的故障总数
每位客户每天连续出现故障的次数
每位客户每天不连续发生故障的次数

设备可能有多个小时的故障，这可能表示硬件故障。另一方面，如果某个设备的故障不跨越多个小时，则可能是该设备的错误使用。

结果应如下所示：

+-------------+-----------+---------------------+-----------------+------------+-----------------------+
| 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，被认为是“非连续的”，此后不久，又在2019-02-12T03：00：00发生了另一个故障，被认为是“连续”。

我想创建一个查询，生成该结果。我尝试过的

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

通过这种方式，我可以按客户每天获得的总故障数量。我想我必须使用LEAD运算符来获取（非）连续计数，但是我不确定如何。有任何想法吗？结果应按天“滚动”。

Answer 1

以下是用于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

您可以使用下面的示例中的虚拟数据来测试，玩游戏

#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

有结果

如何在bigquery中按天计算（非）连续记录？

1 个答案: