8

假设我有一个给定的时间范围。为了解释，让我们考虑一些简单的事情，比如 2018 年全年。我想从 ClickHouse 查询数据作为每个季度的总和聚合，因此结果应该是 4 行。

问题是我只有两个季度的数据，所以在使用时GROUP BY quarter，只返回两行。

SELECT
     toStartOfQuarter(created_at) AS time,
     sum(metric) metric
 FROM mytable
 WHERE
     created_at >= toDate(1514761200) AND created_at >= toDateTime(1514761200)
    AND
     created_at <= toDate(1546210800) AND created_at <= toDateTime(1546210800)
 GROUP BY time
 ORDER BY time

1514761200– 2018-01-01
1546210800–2018-12-31

这将返回：

time       metric
2018-01-01 345
2018-04-01 123

我需要：

time       metric
2018-01-01 345
2018-04-01 123
2018-07-01 0
2018-10-01 0

这是简化的示例，但在实际用例中，聚合将是例如。5 分钟而不是宿舍，并且 GROUP BY 至少还有一个属性，就像GROUP BY attribute1, time这样期望的结果是

time        metric  attribute1
2018-01-01  345     1
2018-01-01  345     2
2018-04-01  123     1
2018-04-01  123     2
2018-07-01  0       1
2018-07-01  0       2
2018-10-01  0       1
2018-10-01  0       2

有没有办法以某种方式填充整个给定的间隔？就像 InfluxDBfill对 group 或 TimescaleDb 的time_bucket()函数有参数一样，generate_series() 我尝试搜索 ClickHouse 文档和 github 问题，似乎这还没有实现，所以问题可能是是否有任何解决方法。

4

4 回答 4

5

您可以使用“数字”函数生成零值。然后使用 UNION ALL 加入您的查询和零值，并且已经根据获得的数据进行了 GROUP BY。

因此，您的查询将如下所示：

SELECT SUM(metric),
       time
  FROM (
        SELECT toStartOfQuarter(toDate(1514761200+number*30*24*3600))  time,
               toUInt16(0) AS metric
          FROM numbers(30)

     UNION ALL 

          SELECT toStartOfQuarter(created_at) AS time,
               metric
          FROM mytable
         WHERE created_at >= toDate(1514761200)
           AND created_at >= toDateTime(1514761200)
           AND created_at <= toDate(1546210800)
           AND created_at <= toDateTime(1546210800)
       )
 GROUP BY time
 ORDER BY time

注意 UInt16(0) - 零值必须与metrics

于 2018-05-14T08:04:51.223 回答
5

从 ClickHouse 19.14 开始，您可以使用该WITH FILL子句。它可以通过这种方式填充宿舍：

WITH
    (
        SELECT toRelativeQuarterNum(toDate('1970-01-01'))
    ) AS init
SELECT
    -- build the date from the relative quarter number
    toDate('1970-01-01') + toIntervalQuarter(q - init) AS time,
    metric
FROM
(
    SELECT
        toRelativeQuarterNum(created_at) AS q,
        sum(rand()) AS metric
    FROM
    (
        -- generate some dates and metrics values with gaps
        SELECT toDate(arrayJoin(range(1514761200, 1546210800, ((60 * 60) * 24) * 180))) AS created_at
    )
    GROUP BY q
    ORDER BY q ASC WITH FILL FROM toRelativeQuarterNum(toDate(1514761200)) TO toRelativeQuarterNum(toDate(1546210800)) STEP 1
)

┌───────time─┬─────metric─┐
│ 2018-01-01 │ 2950782089 │
│ 2018-04-01 │ 2972073797 │
│ 2018-07-01 │          0 │
│ 2018-10-01 │  179581958 │
└────────────┴────────────┘
于 2020-12-17T16:21:55.273 回答
2

numbers()在某些情况下，作为函数的替代方案range，数组函数可能很有用。

示例：对于每对 (id1,id2) 应生成前 7 天的日期。

SELECT
  id1,
  id2,
  arrayJoin(
    arrayMap( x -> today() - 7 + x, range(7) )
  ) as date2
FROM table
WHERE date >= now() - 7
GROUP BY id1, id2

该选择的结果可以在 UNION ALL 中用于填充数据中的“漏洞”。

SELECT id1, id2, date, sum(column1)
FROM (
  SELECT
    id1,
    id2,
    date,
    column1 
  FROM table
  WHERE date >= now() - 7

  UNION ALL 

  SELECT
    id1,
    id2,
    arrayJoin(
      arrayMap( x -> today() - 7 + x, range(7) )
    ) as date2,
    0 as column1
  FROM table
  WHERE date >= now() - 7
  GROUP BY id1, id2
)
GROUP BY id1, id2, date
ORDER BY date, id1, id2
于 2018-09-03T11:23:43.807 回答
2

感谢@filimonov 和@mikhail，这是我在小时桶中的做法（需要在 Grafana 中可视化）

SELECT t, SUM(metric) as metric FROM (
    SELECT 
        arrayJoin(
          arrayMap( x -> toStartOfHour(addHours(toDateTime($from),x)),
              range(toUInt64(
                  dateDiff('hour', 
                      toDateTime($from), 
                      toDateTime($to)) + 1)))
        ) as t,
        0 as metric

    UNION ALL

    SELECT
        toStartOfHour(my_date) as t,
        COUNT(metric)
        FROM my_table
        WHERE t BETWEEN toDateTime($from) AND toDateTime($to)
        GROUP BY t
)
GROUP BY t ORDER BY t

因此，例如，对于从 2019-01-01 到 2019-01-02 的范围，它将为您提供：

SELECT t, SUM(metric) as metric FROM (
    SELECT 
        arrayJoin(
          arrayMap( x -> toStartOfHour(addHours(toDateTime('2019-01-01 00:00:00'),x)),
              range(toUInt64(
                  dateDiff('hour', 
                      toDateTime('2019-01-01 00:00:00'), 
                      toDateTime('2019-01-02 00:00:00')) + 1)))
        ) as t,
        0 as metric

    UNION ALL

    SELECT
        toStartOfHour(my_date) as t,
        COUNT(1) as metric
        FROM my_table
        WHERE t BETWEEN toDateTime('2019-01-01 00:00:00') AND toDateTime('2019-01-02 00:00:00')
        GROUP BY t
)
GROUP BY t ORDER BY t;

t                  |metric|
-------------------|------|
2019-01-01 00:00:00|     0|
2019-01-01 01:00:00|     0|
2019-01-01 02:00:00|     0|
2019-01-01 03:00:00|     0|
2019-01-01 04:00:00|     0|
2019-01-01 05:00:00|     0|
2019-01-01 06:00:00|     0|
2019-01-01 07:00:00|105702|
2019-01-01 08:00:00|113315|
2019-01-01 09:00:00|149837|
2019-01-01 10:00:00|185314|
2019-01-01 11:00:00|246106|
2019-01-01 12:00:00|323036|
2019-01-01 13:00:00|     0|
2019-01-01 14:00:00|409160|
2019-01-01 15:00:00|379113|
2019-01-01 16:00:00|256634|
2019-01-01 17:00:00|286601|
2019-01-01 18:00:00|280039|
2019-01-01 19:00:00|248504|
2019-01-01 20:00:00|218642|
2019-01-01 21:00:00|186152|
2019-01-01 22:00:00|148478|
2019-01-01 23:00:00|109721|
2019-01-02 00:00:00|     0|
于 2019-07-06T09:44:39.330 回答