在 psql 中查询类似时间序列的计数器

Question

我在 psql 中有以下仅附加表：

CREATE TABLE IF NOT EXISTS data (
    id UUID DEFAULT gen_random_uuid () PRIMARY KEY,

    test_id UUID NOT NULL,
    user_id UUID NOT NULL,
    completed BOOL NOT NULL DEFAULT False,

    inserted_at TIMESTAMPTZ NOT NULL DEFAULT (NOW() AT TIME ZONE 'UTC'),
);

CREATE INDEX some_idx ON data (user_id, test_id, inserted_at DESC);
CREATE INDEX some_idx2 ON data (test_id, inserted_at DESC);

对于给定的test_id ，单个user_id可能有多个条目，但只能completed一个（ completed的条目也是最后一个）。

我正在查询给定的test_id 。 我需要的是过去一周中每一天的类似时间序列的数据。 对于每一天，我应该有以下内容：

• total - 唯一用户的总条目 WHERE inserted_at < "day"
• 已完成- 唯一用户的总完成条目，其中inserted_at < "day"

最终， total和completed就像计数器一样，我只是想计算过去一周每一天的值。 例如：

| date       | total | completed |
|------------|-------|-----------|
| 2022.01.19 | 100   | 50        |
| 2022.01.18 | 90    | 45        |
| ...        |       |           |

什么是具有高效查询计划的查询？ 我可以考虑添加新索引或修改现有索引。

PS：我这里有一个工作版本：

SELECT date, entered, completed
FROM (
     SELECT d::date AS date
     FROM generate_series('2023-01-12', now(),INTERVAL '1 day') AS d
) AS dates
cross join lateral (
    SELECT COUNT(DISTINCT user_id) AS entered,
           COUNT(1) FILTER (WHERE completed) AS completed // no need for distinct as completed is guaranteed to be once per user
    FROM data
    WHERE
        test_id = 'someId' AND
        inserted_at < dates.date
) AS vals

我不认为这是一个好的/高性能的解决方案，因为它会在每次横向连接迭代时重新扫描表。 这是查询计划：

+---------------------------------------------------------------------------------------------------------------------------->
| QUERY PLAN                                                                                                                 >
|---------------------------------------------------------------------------------------------------------------------------->
| Nested Loop  (cost=185.18..185218.25 rows=1000 width=28) (actual time=0.928..7.687 rows=8 loops=1)                         >
|   ->  Function Scan on generate_series d  (cost=0.01..10.01 rows=1000 width=8) (actual time=0.009..0.012 rows=8 loops=1)   >
|   ->  Aggregate  (cost=185.17..185.18 rows=1 width=16) (actual time=0.957..0.957 rows=1 loops=8)                           >
|         ->  Bitmap Heap Scan on data  (cost=12.01..183.36 rows=363 width=38) (actual time=0.074..0.197 rows=779 loops>
|               Recheck Cond: ((test_id = 'someId'::uuid) AND (inserted_at < (d.d)::date))  >
|               Heap Blocks: exact=629                                                                                       >
|               ->  Bitmap Index Scan on some_idx2  (cost=0.00..11.92 rows=363 width=0) (actual time=>
|                     Index Cond: ((test_id = 'someId'::uuid) AND (inserted_at < (d.d)::date>
| Planning Time: 0.261 ms                                                                                                    >
| Execution Time: 7.733 ms                                                                                                   >
+---------------------------------------------------------------------------------------------------------------------------->

我确定我在这里缺少一些方便的功能，这些功能会有所帮助。 感谢所有帮助：祈祷：

Answer 1

聊天后我们找到了解决方案

   SELECT
     date_trunc('day',inserted_at) AS adate,
     COUNT(user_id) OVER (
        ORDER BY inserted_at ASC 
        ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
     ) -
     SUM(CASE WHEN completed THEN 1 ELSE 0 END) OVER (
        ORDER BY date_trunc('day', inserted_at) ASC 
        ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
     ) as user_cnt,
     SUM(CASE WHEN completed THEN 1 ELSE 0 END) OVER (
        ORDER BY date_trunc('day', inserted_at) ASC 
        ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
     ) AS completed
    FROM executions
    WHERE journey_id = 'd2be0e01-19b1-403e-8659-ce6222f074fd'
    ORDER BY date_trunc('day', inserted_at) ASC

您可以看到我们使用了相同的 SUM window function 两次。 一旦给出第二列的预期结果，所有 SQL 引擎都会对此进行优化以进行计算。 （完成的项目被计为一个额外的用户）。

---

下面的先前答案

好的，当我查看它时，您毕竟不需要 window function——只是 SUM() 和 GROUP BY 中 CASE 语句的技巧

SELECT COUNT(DISTINCT user_id) AS entered,
       SUM(CASE WHEN completed THEN 1 ELSE 0 END) AS completed
FROM data
WHERE test_id = 'someId' 
GROUP BY inserted_at

---

要获得给定日期的所有先验信息，如下所示：

SELECT date_trunc(day,inserted_at) AS date,
   DENSE_RANK()
     OVER (PARTITION BY user_id ORDER BY inserted_at ASC 
             BETWEEN ROWS UNBOUNDED PRECEDING AND CURRENT ROW) as user_cnt,
   SUM(CASE WHEN completed THEN 1 ELSE 0 END) 
     OVER (ORDER BY inserted_at ASC 
             BETWEEN ROWS UNBOUNDED PRECEDING AND CURRENT ROW) AS completed
FROM data
WHERE test_id = 'someId'
ORDER BY inserted_at ASC