Presto SQL 查询，用于在算术渐进的时间间隔内计算满足一组条件的行数答案

【问题标题】：Presto SQL query that counts the number of rows, that satisfy a set of conditions, within arithmetically progressing time intervalsPresto SQL 查询，用于在算术渐进的时间间隔内计算满足一组条件的行数
【发布时间】：2021-11-19 19:29:23
【问题描述】：

我想弄清楚满足一组条件的行数如何随着时间的推移而改变。为此，我想计算在开始日期满足条件，然后执行相同的计算直到今天的每一天。我想要的输出表会类似于下表（不包括未命名的列）：

|------------+------------+----------------------------|
|            |       date | rows_staisfying_conditions |
|------------+------------+----------------------------|
| start date | 2021-09-01 |                       2367 |
|            | 2021-09-02 |                       2784 |
|            | 2021-09-03 |                       3011 |
|            | 2021-09-04 |                       3601 |
| today      | 2021-09-05 |                       4155 |
|------------+------------+----------------------------|

生成上表的一种简单方法是每天有一个 CTE，然后加入 CTE（见下面的代码）。问题是这很冗长而且无法扩展。

WITH day0 AS (
    SELECT count(*) AS day0
    FROM (
        SELECT DISTINCT account_id
        FROM default_table
        WHERE
            secret_column = 'secret value'
            AND lower(device_os) LIKE '%android%'
            AND from_iso8601_timestamp(timestamp) < from_iso8601_timestamp('2021-09-01T00:00:00.0000000Z') + interval '0' day
    )
),
day1 AS (
    SELECT count(*) AS day1
    FROM (
        SELECT DISTINCT account_id
        FROM default_table
        WHERE
            secret_column = 'secret value'
            AND lower(device_os) LIKE '%android%'
            AND from_iso8601_timestamp(timestamp) < from_iso8601_timestamp('2021-09-01T00:00:00.0000000Z') + interval '1' day
    )
),
⋮
day4 AS (
    SELECT count(*) AS day1
    FROM (
        SELECT DISTINCT account_id
        FROM default_table
        WHERE
            secret_column = 'secret value'
            AND lower(device_os) LIKE '%android%'
            AND from_iso8601_timestamp(timestamp) < from_iso8601_timestamp('2021-09-01T00:00:00.0000000Z') + interval '4' day
    )
),
SELECT *
FROM
    day0
    FULL JOIN day1 ON TRUE
    FULL JOIN day2 ON TRUE
    FULL JOIN day3 ON TRUE
    FULL JOIN day4 ON TRUE

有人对我如何计算上表有什么建议吗？可扩展的方式？

【问题讨论】：

标签： sql amazon-athena presto

【解决方案1】：

您可以使用sum 窗口函数按日期分组并使用下一个frame 排序总和：

WITH dataset (date, condition) AS
(
  VALUES
  (date '2021-09-01', true),
  (date '2021-09-01', true),
  (date '2021-09-01', false),
  (date '2021-09-02', true),
  (date '2021-09-03', true)
)

SELECT date, sum(cnt) over (order by date range between unbounded preceding and current row) rows_staisfying_conditions
FROM (
         SELECT date, sum(case when condition then 1 else 0 end) cnt
         FROM dataset
         GROUP BY date
     )

输出：

date	rows_staisfying_conditions
2021-09-01	2
2021-09-02	3
2021-09-03	4

【讨论】：