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sql - 滚动总和直到达到某个值，加上计算的持续时间

问题描述

我有一个要求，我需要知道何时sum(value)达到某个点并计算持续时间。下面是示例表。

create table sample (dt timestamp, value real);

insert into sample values
     ('2019-01-20 00:29:43 ',0.29)
    ,('2019-01-20 00:35:06 ',0.31)
    ,('2019-01-20 00:35:50 ',0.41)
    ,('2019-01-20 00:36:32 ',0.26)
    ,('2019-01-20 00:37:20 ',0.33)
    ,('2019-01-20 00:41:30 ',0.42)
    ,('2019-01-20 00:42:28 ',0.35)
    ,('2019-01-20 00:43:14 ',0.52)
    ,('2019-01-20 00:44:18 ',0.25);

现在我的要求是计算以下行的累积总和，以查看何时sum(value)达到 1.0 以上。这可能只需要 1 行或 n 行。到达该行后，我需要计算当前行与sum(value)达到 1.0 以上的行之间的时间差。

基本上我想要的输出格式如下。
对于第 1 行，sum(value)在第 3 行达到累积。
对于第 2 行，sum(value)在第 4 行达到累积值，依此类推。

         dt         | value | sum(value)| time_at_sum(value)_1| Duration
---------------------+--------+------------------------------------------
 2019-01-20 00:29:43| 0.29  |   1.01    | 2019-01-20 00:35:50 | 00:06:07
 2019-01-20 00:35:06| 0.31  |   1.31    | 2019-01-20 00:37:20 | 00:02:14 
 2019-01-20 00:35:50| 0.41  |   1.00    | 2019-01-20 00:37:20 | 00:01:30 
 2019-01-20 00:36:32| 0.26  |   1.01    | 2019-01-20 00:41:30 | 00:04:58 
 2019-01-20 00:37:20| 0.33  |   1.10    | 2019-01-20 00:42:28 | 00:05:08 
 2019-01-20 00:41:30| 0.42  |   1.29    | 2019-01-20 00:43:14 | 00:01:44 
 2019-01-20 00:42:28| 0.35  |   1.12    | 2019-01-20 00:44:18 | 00:01:50 
 2019-01-20 00:43:14| 0.52  |   NULL    |  -                  | -
 2019-01-20 00:44:18| 0.25  |   NULL    |  -                  | -

任何人都有关于如何处理上述要求的想法或指示？

标签： sqlpostgresqlsqlcommandcumulative-sumrolling-computation

解决方案

WITH tmp AS (
    SELECT *
        , sum(value) OVER (ORDER BY dt rows between current row and unbounded following) as forward_sum
    FROM sample
    ORDER BY dt)
SELECT t1.dt, t1.value
    , (t2.value + t1.forward_sum - t2.forward_sum) as "sum(value)"
    , t2.dt as "time_at_sum(value)_1" 
    , t2.dt - t1.dt as "Duration"
FROM tmp t1
LEFT JOIN LATERAL (
    SELECT * 
    FROM tmp t
    WHERE t1.forward_sum - t.forward_sum < 1
        AND (t.value + t1.forward_sum - t.forward_sum) >= 0.999
    ORDER BY dt DESC 
    LIMIT 1
    ) t2
ON TRUE

产量

| dt                  | value | sum(value) | time_at_sum(value)_1 | Duration |
|---------------------+-------+------------+----------------------+----------|
| 2019-01-20 00:29:43 |  0.29 |       1.01 | 2019-01-20 00:35:50  | 00:06:07 |
| 2019-01-20 00:35:06 |  0.31 |       1.31 | 2019-01-20 00:37:20  | 00:02:14 |
| 2019-01-20 00:35:50 |  0.41 |          1 | 2019-01-20 00:37:20  | 00:01:30 |
| 2019-01-20 00:36:32 |  0.26 |       1.01 | 2019-01-20 00:41:30  | 00:04:58 |
| 2019-01-20 00:37:20 |  0.33 |        1.1 | 2019-01-20 00:42:28  | 00:05:08 |
| 2019-01-20 00:41:30 |  0.42 |       1.29 | 2019-01-20 00:43:14  | 00:01:44 |
| 2019-01-20 00:42:28 |  0.35 |       1.12 | 2019-01-20 00:44:18  | 00:01:50 |
| 2019-01-20 00:43:14 |  0.52 |            |                      |          |
| 2019-01-20 00:44:18 |  0.25 |            |                      |          |

首先计算value列上的累积和：

SELECT *
    , sum(value) OVER (ORDER BY dt rows between current row and unbounded following) as forward_sum
FROM sample
ORDER BY dt

产生

| dt                  | value | forward_sum |
|---------------------+-------+-------------|
| 2019-01-20 00:29:43 |  0.29 |        3.14 |
| 2019-01-20 00:35:06 |  0.31 |        2.85 |
| 2019-01-20 00:35:50 |  0.41 |        2.54 |
| 2019-01-20 00:36:32 |  0.26 |        2.13 |
| 2019-01-20 00:37:20 |  0.33 |        1.87 |
| 2019-01-20 00:41:30 |  0.42 |        1.54 |
| 2019-01-20 00:42:28 |  0.35 |        1.12 |
| 2019-01-20 00:43:14 |  0.52 |        0.77 |
| 2019-01-20 00:44:18 |  0.25 |        0.25 |

请注意，从 中减去两个值forward_sum对应于values 的部分和。例如，

0.29 + 0.31 + 0.41 = 3.14 - 2.13

所以差异forward_sums将发挥重要作用，我们希望将这些差异与 1 进行比较。我们将希望使用以下连接条件将这个表与其自身连接起来：

t1.forward_sum - t.forward_sum < 1

让我们看看如果我们使用 LEFT JOIN LATERAL 会发生什么。关于 LEFT JOIN LATERAL 的关键是，必须对左侧表中的每一行评估 LATERAL 连接右侧的子查询一次：

WITH tmp AS (
    SELECT *
        , sum(value) OVER (ORDER BY dt rows between current row and unbounded following) as forward_sum
    FROM sample
    ORDER BY dt)
SELECT t1.*, t2.*
FROM tmp t1
LEFT JOIN LATERAL (
    SELECT * 
    FROM tmp t
    WHERE t1.forward_sum - t.forward_sum < 1
    ORDER BY dt DESC 
    LIMIT 1
    ) t2
ON TRUE

产量

| dt                  | value | forward_sum | dt                  | value | forward_sum |
|---------------------+-------+-------------+---------------------+-------+-------------|
| 2019-01-20 00:29:43 |  0.29 |        3.14 | 2019-01-20 00:35:50 |  0.41 |        2.54 |
| 2019-01-20 00:35:06 |  0.31 |        2.85 | 2019-01-20 00:37:20 |  0.33 |        1.87 |
| 2019-01-20 00:35:50 |  0.41 |        2.54 | 2019-01-20 00:37:20 |  0.33 |        1.87 |
| 2019-01-20 00:36:32 |  0.26 |        2.13 | 2019-01-20 00:41:30 |  0.42 |        1.54 |
| 2019-01-20 00:37:20 |  0.33 |        1.87 | 2019-01-20 00:42:28 |  0.35 |        1.12 |
| 2019-01-20 00:41:30 |  0.42 |        1.54 | 2019-01-20 00:43:14 |  0.52 |        0.77 |
| 2019-01-20 00:42:28 |  0.35 |        1.12 | 2019-01-20 00:44:18 |  0.25 |        0.25 |
| 2019-01-20 00:43:14 |  0.52 |        0.77 | 2019-01-20 00:44:18 |  0.25 |        0.25 |
| 2019-01-20 00:44:18 |  0.25 |        0.25 | 2019-01-20 00:44:18 |  0.25 |        0.25 |

请注意，我们已经猜到了匹配所需日期的连接条件的方式。现在只需编写正确的值表达式即可获得所需的列sum(value), time_at_sum(value)_1。

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