Question

我有一个Postgres 9.1数据库。我试图生成每周的记录数（对于给定的日期范围）并将其与上一年进行比较。

我有以下代码用于生成系列：

select generate_series('2013-01-01', '2013-01-31', '7 day'::interval) as series

但是，我不知道如何将计算的记录加入到生成的日期。

因此，使用以下记录作为示例：

Pt_ID      exam_date
======     =========
1          2012-01-02
2          2012-01-02
3          2012-01-08
4          2012-01-08
1          2013-01-02
2          2013-01-02
3          2013-01-03
4          2013-01-04
1          2013-01-08
2          2013-01-10
3          2013-01-15
4          2013-01-24

我希望将记录返回为：

  series        thisyr      lastyr
===========     =====       =====
2013-01-01        4           2
2013-01-08        3           2
2013-01-15        1           0
2013-01-22        1           0
2013-01-29        0           0

不确定如何在子搜索中引用日期范围。感谢您的帮助。

Answer 1

简单的方法是通过@jpw演示的CROSS JOIN来解决这个问题。但是，有一些隐藏的问题：

无条件CROSS JOIN的效果随着行数的增加而迅速恶化。在聚合中处理这个巨大的派生表之前，总行数乘以您要测试的周数。索引无法提供帮助。
1月1日开始的几周会导致不一致。 ISO周可能是另一种选择。见下文。

以下所有问题都会大量使用exam_date 上的索引。一定要有一个。

仅加入相关行

应该更快：

SELECT d.day, d.thisyr , count(t.exam_date) AS lastyr FROM ( SELECT d.day::date, (d.day - '1 year'::interval)::date AS day0 -- for 2nd join , count(t.exam_date) AS thisyr FROM generate_series('2013-01-01'::date , '2013-01-31'::date -- last week overlaps with Feb. , '7 days'::interval) d(day) -- returns timestamp LEFT JOIN tbl t ON t.exam_date >= d.day::date AND t.exam_date < d.day::date + 7 GROUP BY d.day ) d LEFT JOIN tbl t ON t.exam_date >= d.day0 -- repeat with last year AND t.exam_date < d.day0 + 7 GROUP BY d.day, d.thisyr ORDER BY d.day;

这是从1月1日开始的几个星期，就像你原来的一样。正如评论的那样，这产生了一些不一致的地方：每周从不同的一天开始，自从我们在年底切断，一年的最后一周只有1或2天（闰年）。

与ISO周期相同

根据要求，请考虑 ISO周，从周一开始，始终为7天。但他们跨越了多年的边界。 Per documentation on EXTRACT()：


周

当天的一周中的星期数。根据定义（ISO 8601），星期一和星期的第一周开始   年份包含当年的1月4日。换句话说，第一个   一年的星期四是在那一年的第1周。

在ISO定义中，1月初的日期可能是上一年的第52周或第53周的一部分，并且   12月下旬可以成为明年第一周的一部分。对于   例如，2005-01-01是2004年第53周的一部分，并且   2006-01-01是2005年第52周的一部分，而2012-12-31是。isoyear   2013年第一周的一部分。建议使用week   字段与SELECT w AS isoweek , day::text AS thisyr_monday, thisyr_ct , day0::text AS lastyr_monday, count(t.exam_date) AS lastyr_ct FROM ( SELECT w, day , date_trunc('week', '2012-01-04'::date)::date + 7 * w AS day0 , count(t.exam_date) AS thisyr_ct FROM ( SELECT w , date_trunc('week', '2013-01-04'::date)::date + 7 * w AS day FROM generate_series(0, 4) w ) d LEFT JOIN tbl t ON t.exam_date >= d.day AND t.exam_date < d.day + 7 GROUP BY d.w, d.day ) d LEFT JOIN tbl t ON t.exam_date >= d.day0 -- repeat with last year AND t.exam_date < d.day0 + 7 GROUP BY d.w, d.day, d.day0, d.thisyr_ct ORDER BY d.w, d.day;一起获得一致的结果。

以上查询用ISO周重写：

date_trunc('week', '2012-01-04'::date)::date

1月4日始终是今年的第一个ISO周。因此，此表达式获取给定年份的第一个ISO周的星期一日期：

EXTRACT()

使用EXTRACT()
简化
由于ISO周与SELECT w AS isoweek , COALESCE(thisyr_ct, 0) AS thisyr_ct , COALESCE(lastyr_ct, 0) AS lastyr_ct FROM generate_series(1, 5) w LEFT JOIN ( SELECT EXTRACT(week FROM exam_date)::int AS w, count(*) AS thisyr_ct FROM tbl WHERE EXTRACT(isoyear FROM exam_date)::int = 2013 GROUP BY 1 ) t13 USING (w) LEFT JOIN ( SELECT EXTRACT(week FROM exam_date)::int AS w, count(*) AS lastyr_ct FROM tbl WHERE EXTRACT(isoyear FROM exam_date)::int = 2012 GROUP BY 1 ) t12 USING (w);返回的周数一致，我们可以简化查询。首先，简短而简单的形式：

WITH params AS ( -- enter parameters here, once SELECT date_trunc('week', '2012-01-04'::date)::date AS last_start , date_trunc('week', '2013-01-04'::date)::date AS this_start , date_trunc('week', '2014-01-04'::date)::date AS next_start , 1 AS week_1 , 5 AS week_n -- show weeks 1 - 5 ) SELECT w.w AS isoweek , p.this_start + 7 * (w - 1) AS thisyr_monday , COALESCE(t13.ct, 0) AS thisyr_ct , p.last_start + 7 * (w - 1) AS lastyr_monday , COALESCE(t12.ct, 0) AS lastyr_ct FROM params p , generate_series(p.week_1, p.week_n) w(w) LEFT JOIN ( SELECT EXTRACT(week FROM t.exam_date)::int AS w, count(*) AS ct FROM tbl t, params p WHERE t.exam_date >= p.this_start -- only relevant dates AND t.exam_date < p.this_start + 7 * (p.week_n - p.week_1 + 1)::int -- AND t.exam_date < p.next_start -- don't cross over into next year GROUP BY 1 ) t13 USING (w) LEFT JOIN ( -- same for last year SELECT EXTRACT(week FROM t.exam_date)::int AS w, count(*) AS ct FROM tbl t, params p WHERE t.exam_date >= p.last_start AND t.exam_date < p.last_start + 7 * (p.week_n - p.week_1 + 1)::int -- AND t.exam_date < p.this_start GROUP BY 1 ) t12 USING (w);

优化查询

同样有更多细节并针对性能进行了优化

JOIN LATERAL

在索引支持下，这应该非常快，并且可以很容易地适应选择的间隔。上次查询中generate_series()的隐式{{1}}需要 Postgres 9.3 。

SQL Fiddle.

Answer 2

使用cross join应该可行，我只是要粘贴下面的SQL Fiddle的markdown输出。对于2013-01-08系列来说，你的样本输出似乎不正确：thisyr应该是2，而不是3.这可能不是最好的方法，但是我的Postgresql知识还有很多不足之处。

SQL Fiddle

PostgreSQL 9.2.4架构设置：

CREATE TABLE Table1
    ("Pt_ID" varchar(6), "exam_date" date);

INSERT INTO Table1
    ("Pt_ID", "exam_date")
VALUES
    ('1', '2012-01-02'),('2', '2012-01-02'),
    ('3', '2012-01-08'),('4', '2012-01-08'),
    ('1', '2013-01-02'),('2', '2013-01-02'),
    ('3', '2013-01-03'),('4', '2013-01-04'),
    ('1', '2013-01-08'),('2', '2013-01-10'),
    ('3', '2013-01-15'),('4', '2013-01-24');

查询1 ：

select 
  series, 
  sum (
    case 
      when exam_date 
        between series and series + '6 day'::interval
      then 1 
      else 0 
    end
  ) as thisyr,
  sum (
    case 
      when exam_date + '1 year'::interval 
        between series and series + '6 day'::interval
      then 1 else 0 
    end
  ) as lastyr

from table1
cross join generate_series('2013-01-01', '2013-01-31', '7 day'::interval) as series
group by series
order by series

<强> Results ：

|                         SERIES | THISYR | LASTYR |
|--------------------------------|--------|--------|
| January, 01 2013 00:00:00+0000 |      4 |      2 |
| January, 08 2013 00:00:00+0000 |      2 |      2 |
| January, 15 2013 00:00:00+0000 |      1 |      0 |
| January, 22 2013 00:00:00+0000 |      1 |      0 |
| January, 29 2013 00:00:00+0000 |      0 |      0 |

每周的总记录数

2 个答案:

仅加入相关行

与ISO周期相同

使用`EXTRACT()`

优化查询