Postgres中有效的时间序列查询

I have a table in my PG db that looks somewhat like this:

我的PG数据库中有一个表,看起来有点像这样:

id | widget_id | for_date | score |

Each referenced widget has a lot of these items. It's always 1 per day per widget, but there are gaps.

每个引用的小部件都有很多这些项目。每个小部件每天总是1个,但是存在差距。

What I want to get is a result that contains all the widgets for each date since X. The dates are brought in via generate series:

我想得到的结果是包含自X以来每个日期的所有小部件。日期通过生成系列引入:

 SELECT date.date::date
   FROM generate_series('2012-01-01'::timestamp with time zone,'now'::text::date::timestamp with time zone, '1 day') date(date)
 ORDER BY date.date DESC;

If there is no entry for a date for a given widget_id, I want to use the previous one. So say widget 1337 doesn't have an entry on 2012-05-10, but on 2012-05-08, then I want the resultset to show the 2012-05-08 entry on 2012-05-10 as well:

如果没有给定widget_id的日期条目,我想使用前一个。所以说小工具1337在2012-05-10没有条目,但在2012-05-08,那么我希望结果集在2012-05-10也显示2012-05-08条目:

Actual data:
widget_id | for_date   | score
1312      | 2012-05-07 | 20
1337      | 2012-05-07 | 12
1337      | 2012-05-08 | 41
1337      | 2012-05-11 | 500

Desired output based on generate series:
widget_id | for_date   | score
1336      | 2012-05-07 | 20
1337      | 2012-05-07 | 12
1336      | 2012-05-08 | 20
1337      | 2012-05-08 | 41
1336      | 2012-05-09 | 20
1337      | 2012-05-09 | 41
1336      | 2012-05-10 | 20
1337      | 2012-05-10 | 41
1336      | 2012-05-11 | 20
1337      | 2012-05-11 | 500

Eventually I want to boil this down into a view so I have consistent data sets per day that I can query easily.

最终我想把它归结为一个视图,所以我每天都有一致的数据集,我可以轻松查询。

Edit: Made the sample data and expected resultset clearer

编辑:使样本数据和预期结果集更清晰

4 个解决方案

#1

SQL Fiddle

select
    widget_id,
    for_date,
    case
        when score is not null then score
        else first_value(score) over (partition by widget_id, c order by for_date)
        end score
from (
    select
        a.widget_id,
        a.for_date,
        s.score,
        count(score) over(partition by a.widget_id order by a.for_date) c
    from (
        select widget_id, g.d::date for_date
        from (
            select distinct widget_id
            from score
            ) s
            cross join
            generate_series(
                (select min(for_date) from score),
                (select max(for_date) from score),
                '1 day'
            ) g(d)
        ) a
        left join
        score s on a.widget_id = s.widget_id and a.for_date = s.for_date
) s
order by widget_id, for_date

#2

First of all, you can have a much simpler generate_series() table expression. Equivalent to yours (except for descending order, that contradicts the rest of your question anyways):

首先,您可以使用更简单的generate_series()表表达式。相当于你的(除了降序,这与你的其余问题相矛盾):

SELECT generate_series('2012-01-01'::date, now()::date, '1d')::date

The type date is coerced to timestamptz automatically on input. The return type is timestamptz either way. I use a subquery below, so I can cast to the output to date right away.

类型日期在输入时自动强制为timestamptz。返回类型是时间戳两种方式。我在下面使用子查询,所以我可以立即转换为输出到日期。

Next, max() as window function returns exactly what you need: the highest value since frame start ignoring NULL values. Building on that, you get a radically simple query.

接下来,max()作为窗口函数准确返回所需内容:自帧起始忽略NULL值的最高值。在此基础上,您将获得一个极其简单的查询。

For a given widget_id

Most likely faster than involving CROSS JOIN or WITH RECURSIVE:

最有可能比涉及CROSS JOIN或WITH RECURSIVE更快:

SELECT a.day, s.*
FROM  (
   SELECT d.day
         ,max(s.for_date) OVER (ORDER BY d.day) AS effective_date
   FROM  (
      SELECT generate_series('2012-01-01'::date, now()::date, '1d')::date
      ) d(day)
   LEFT   JOIN score s ON s.for_date = d.day
                      AND s.widget_id = 1337 -- "for a given widget_id"
   ) a
LEFT   JOIN score s ON s.for_date = a.effective_date
                   AND s.widget_id = 1337
ORDER  BY a.day;

->sqlfiddle

With this query you can put any column from score you like into the final SELECT list. I put s.* for simplicity. Pick your columns.

使用此查询,您可以将您喜欢的任何列放入最终的SELECT列表中。我把s。*简单化了。选择你的专栏。

If you want to start your output with the first day that actually has a score, simply replace the last LEFT JOIN with JOIN.

如果您想在实际有分数的第一天开始输出,只需用JOIN替换最后一个LEFT JOIN。

Generic form for all widget_id's

Here I use a CROSS JOIN to produce a row for every widget on every date ..

在这里,我使用CROSS JOIN为每个日期的每个小部件生成一行..

SELECT a.day, a.widget_id, s.score
FROM  (
   SELECT d.day, w.widget_id
         ,max(s.for_date) OVER (PARTITION BY w.widget_id
                                ORDER BY d.day) AS effective_date
   FROM  (SELECT generate_series('2012-05-05'::date
                                ,'2012-05-15'::date, '1d')::date AS day) d
   CROSS  JOIN (SELECT DISTINCT widget_id FROM score) AS w
   LEFT   JOIN score s ON s.for_date = d.day AND s.widget_id = w.widget_id
   ) a
JOIN  score s ON s.for_date = a.effective_date
             AND s.widget_id = a.widget_id  -- instead of LEFT JOIN
ORDER BY a.day, a.widget_id;

->sqlfiddle

#3

Using your table structure, I created the following Recursive CTE which starts with your MIN(For_Date) and increments until it reaches the MAX(For_Date). Not sure if there is a more efficient way, but this appears to work well:

使用您的表结构,我创建了以下递归CTE,它以MIN(For_Date)开始并递增,直到达到MAX(For_Date)。不确定是否有更有效的方法,但这似乎运作良好:

WITH RECURSIVE nodes_cte(widgetid, for_date, score) AS (
-- First Widget Using Min Date
 SELECT 
    w.widgetId, 
    w.for_date, 
    w.score
 FROM widgets w 
  INNER JOIN ( 
      SELECT widgetId, Min(for_date) min_for_date
      FROM widgets
      GROUP BY widgetId
   ) minW ON w.widgetId = minW.widgetid 
        AND w.for_date = minW.min_for_date
UNION ALL
 SELECT 
    n.widgetId,
    n.for_date + 1 for_date,
    coalesce(w.score,n.score) score
 FROM nodes_cte n
  INNER JOIN (
      SELECT widgetId, Max(for_date) max_for_date
      FROM widgets 
      GROUP BY widgetId
   ) maxW ON n.widgetId = maxW.widgetId
  LEFT JOIN widgets w ON n.widgetid = w.widgetid 
    AND n.for_date + 1 = w.for_date
  WHERE n.for_date + 1 <= maxW.max_for_date
)
SELECT * 
FROM nodes_cte 
ORDER BY for_date

Here is the SQL Fiddle.

这是SQL小提琴。

And the returned results (format the date however you'd like):

并返回结果(格式化日期,但你喜欢):

WIDGETID   FOR_DATE                     SCORE
1337       May, 07 2012 00:00:00+0000   12
1337       May, 08 2012 00:00:00+0000   41
1337       May, 09 2012 00:00:00+0000   41
1337       May, 10 2012 00:00:00+0000   41
1337       May, 11 2012 00:00:00+0000   500

Please note, this assumes your For_Date field is a Date -- if it includes a Time -- then you may need to use Interval '1 day' in the query above instead.

请注意,这假设您的For_Date字段是日期 - 如果它包含时间 - 那么您可能需要在上面的查询中使用Interval“1天”。

Hope this helps.

希望这可以帮助。

#4

The data:

DROP SCHEMA tmp CASCADE;
CREATE SCHEMA tmp ;
SET search_path=tmp;

CREATE TABLE widget
        ( widget_id INTEGER NOT NULL
        , for_date DATE NOT NULL
        , score INTEGER
         , PRIMARY KEY (widget_id,for_date)
        );
INSERT INTO widget(widget_id , for_date , score) VALUES
 (1312, '2012-05-07', 20)
, (1337, '2012-05-07', 12)
, (1337, '2012-05-08', 41)
, (1337, '2012-05-11', 500)
        ;

The query:

SELECT w.widget_id AS widget_id
        , cal::date AS for_date
        -- , w.for_date AS org_date
        , w.score AS score
FROM generate_series( '2012-05-07'::timestamp , '2012-05-11'::timestamp
                 , '1day'::interval) AS cal
        -- "half cartesian" Join;
        -- will be restricted by the NOT EXISTS() below
LEFT JOIN widget w ON w.for_date <= cal
WHERE NOT EXISTS (
        SELECT * FROM widget nx
        WHERE nx.widget_id = w.widget_id
        AND nx.for_date <= cal
        AND nx.for_date > w.for_date
        )
ORDER BY cal, w.widget_id
        ;

The result:

 widget_id |  for_date  | score 
-----------+------------+-------
      1312 | 2012-05-07 |    20
      1337 | 2012-05-07 |    12
      1312 | 2012-05-08 |    20
      1337 | 2012-05-08 |    41
      1312 | 2012-05-09 |    20
      1337 | 2012-05-09 |    41
      1312 | 2012-05-10 |    20
      1337 | 2012-05-10 |    41
      1312 | 2012-05-11 |    20
      1337 | 2012-05-11 |   500
(10 rows)