Postgres - 如何返回丢失数据的0计数行？

Question

我有不均匀分布的数据(wrt日期)几年(2003-2008).我想查询一组给定的开始和结束日期的数据,按PostgreSQL 8.3中任何支持的时间间隔(日,周,月,季,年)对数据进行分组(http://www.postgresql.org/docs /8.3/static/functions-datetime.html#FUNCTIONS-DATETIME-TRUNC).

问题是某些查询会在所需的时间段内提供连续的结果,如下所示:

select to_char(date_trunc('month',date), 'YYYY-MM-DD'),count(distinct post_id) 
from some_table where category_id=1 and entity_id = 77  and entity2_id = 115 
and date <= '2008-12-06' and date >= '2007-12-01' group by 
date_trunc('month',date) order by date_trunc('month',date);
          to_char   | count 
        ------------+-------
         2007-12-01 |    64
         2008-01-01 |    31
         2008-02-01 |    14
         2008-03-01 |    21
         2008-04-01 |    28
         2008-05-01 |    44
         2008-06-01 |   100
         2008-07-01 |    72
         2008-08-01 |    91
         2008-09-01 |    92
         2008-10-01 |    79
         2008-11-01 |    65
        (12 rows)

但有些人因为没有数据而错过了一些间隔,因为这个:

select to_char(date_trunc('month',date), 'YYYY-MM-DD'),count(distinct post_id) 
from some_table where category_id=1 and entity_id = 75  and entity2_id = 115 
and date <= '2008-12-06' and date >= '2007-12-01' group by 
date_trunc('month',date) order by date_trunc('month',date);

        to_char   | count 
    ------------+-------

     2007-12-01 |     2
     2008-01-01 |     2
     2008-03-01 |     1
     2008-04-01 |     2
     2008-06-01 |     1
     2008-08-01 |     3
     2008-10-01 |     2
    (7 rows)

所需结果集的位置是:

  to_char   | count 
------------+-------
 2007-12-01 |     2
 2008-01-01 |     2
 2008-02-01 |     0
 2008-03-01 |     1
 2008-04-01 |     2
 2008-05-01 |     0
 2008-06-01 |     1
 2008-07-01 |     0
 2008-08-01 |     3
 2008-09-01 |     0
 2008-10-01 |     2
 2008-11-01 |     0
(12 rows)

缺少条目的计数为0.

我之前已经看过Stack Overflow的讨论,但它们似乎并没有解决我的问题,因为我的分组时间是(日,周,月,季,年)之一,并由应用程序决定运行时.因此,左边连接与日历表或序列表的方法对我来说无济于事.

我目前的解决方案是使用日历模块在Python(在Turbogears应用程序中)填补这些空白.

有一个更好的方法吗.

Answer 1

^{这个问题很老了.但是,由于其他用户选择它作为新副本的主人,我正在添加一个正确的答案.}

适当的解决方案

SELECT *
FROM  (
   SELECT day::date
   FROM   generate_series(timestamp '2007-12-01'
                        , timestamp '2008-12-01'
                        , interval  '1 month') day
   ) d
LEFT   JOIN (
   SELECT date_trunc('month', date_col)::date AS day
        , count(*) AS some_count
   FROM   tbl
   WHERE  date_col >= date '2007-12-01'
   AND    date_col <= date '2008-12-06'
-- AND    ... more conditions
   GROUP  BY 1
   ) t USING (day)
ORDER  BY day;

• LEFT JOIN当然可以使用.

• generate_series() 可以快速生成一张时间戳表,速度非常快.

• 在加入之前聚合通常会更快.我最近在sqlfiddle.com上提供了一个相关答案的测试用例:

  • PostgreSQL - 按数组排序

• 将timestampto date(::date)转换为基本格式.更多用途to_char().

• GROUP BY 1是语法简写,用于引用第一个输出列.也可以GROUP BY day,但这可能与同名的现有列冲突.或者GROUP BY date_trunc('month', date_col)::date这对我来说太长了.

• 使用可用的interval参数date_trunc().

• count()永远不会产生NULL(0没有行),但LEFT JOIN确实如此.
  要返回0而不是NULL在外部SELECT,请使用COALESCE(some_count, 0) AS some_count.手册.

• 对于更通用的解决方案或任意时间间隔,请考虑以下相关的答案:

  • 在Rails + Postgres中按任意时间间隔计算记录的最佳方法

Answer 2

您可以创建去年(例如)的所有前几天的列表

select distinct date_trunc('month', (current_date - offs)) as date 
from generate_series(0,365,28) as offs;
          date
------------------------
 2007-12-01 00:00:00+01
 2008-01-01 00:00:00+01
 2008-02-01 00:00:00+01
 2008-03-01 00:00:00+01
 2008-04-01 00:00:00+02
 2008-05-01 00:00:00+02
 2008-06-01 00:00:00+02
 2008-07-01 00:00:00+02
 2008-08-01 00:00:00+02
 2008-09-01 00:00:00+02
 2008-10-01 00:00:00+02
 2008-11-01 00:00:00+01
 2008-12-01 00:00:00+01

然后你可以加入那个系列.