Question

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，比如整个2018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。

问题是我的数据只有两个季度，因此在使用GROUP BY quarter时，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

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，但在实际使用情况下，聚合将是例如。 5分钟而不是季度，GROUP BY至少会有一个属性，如GROUP BY attribute1, time，所以期望的结果是

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补整个给定的间隔？就像InfluxDB对组有fill参数或者time_bucket() generate_series()函数{{3}}我尝试搜索ClickHouse文档和github问题时，似乎还没有实现是否有任何解决方法。

Answer 1

您可以使用＆＃34;数字＆＃34;生成零值。功能。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创建了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toUInt16（0） - 零值必须与metrics

的类型相同

Answer 2

在某些情况下，DECLARE @parchesQueryAdd decimal(18,2), @parchesQueryRemove decimal(18,2), @topupQuery decimal(18,2), @Balance decimal(18,2), @totalamount decimal(18,2) SELECT @Balance = SUM(CASE WHEN AmountType = 10 THEN Amount * CurrentBalanceCurrency ELSE 0 END) - SUM(CASE WHEN AmountType = 20 THEN Amount * CurrentBalanceCurrency ELSE 0 END) FROM UserBalance WHERE BalanceForId = @userId GROUP BY BalanceForId SET @topupQuery = (SELECT SUM(Amount * Quentity) from TopUpRecords where TopupById = @userId) SET @totalamount= @Balance - @topupQuery PRINT @totalamount作为numbers()函数的替代方法，数组函数可能会有用。

示例：对于每对（id1，id2），应生成前7天的日期。

range

该选择的结果可以在UNION ALL中使用，以填充数据中的“空洞”。

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

Answer 3

在ClickHouse 19.14中，您可以使用WITH FILL子句。它可以这样填充宿舍：

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 │
└────────────┴────────────┘

Answer 4

这是我如何进行数小时的工作（需要在Grafana中可视化）感谢@filimonov和@mikhail

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

例如，从2019-01-01到2019-01-02，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|

如何按ClickHouse中的时间段进行分组，并使用nulls / 0s填充缺失的数据

4 个答案: