按月计算并将月份作为列

Question

背景

我每月都有时间序列数据，我想按月分组每个ID的值，然后将月份名称作为列而不是行。

实施例

+----+------------+-------+-------+
| id | extra_info | month | value |
+----+------------+-------+-------+
| 1  | abc        | jan   | 10    |
| 1  | abc        | feb   | 20    |
| 2  | def        | jan   | 10    |
| 2  | def        | feb   | 5     |
| 1  | abc        | jan   | 15    |
| 3  | ghi        | mar   | 15    |

期望结果

+----+------------+-----+-----+-----+
| id | extra_info | jan | feb | mar |
+----+------------+-----+-----+-----+
| 1  | abc        | 25  | 20  | 0   |
| 2  | def        | 10  | 5   | 0   |
| 3  | ghi        | 0   | 0   | 15  |

当前方法

我可以轻松按月分组，总结价值观。这让我：

-----------------------------------
| id | extra_info | month | value |
+----+------------+-------+-------+
| 1  | abc        | jan   | 25    |
| 1  | abc        | feb   | 20    |
| 2  | def        | jan   | 10    |
| 2  | def        | feb   | 5     |
| 3  | ghi        | mar   | 15    |

但我现在需要那些月作为列名。不知道从哪里开始。

其他信息

• 就语言而言，此查询将在postgres中运行。
• 上述几个月只是示例，显然真实数据集要大得多，涵盖数千个ID的所有12个月

非常感谢SQL大师的任何想法！

Answer 1

您可以使用带有CASE表达式的聚合函数将行转换为列：

select id,
  extra_info,
  sum(case when month = 'jan' then value else 0 end) jan,
  sum(case when month = 'feb' then value else 0 end) feb,
  sum(case when month = 'mar' then value else 0 end) mar,
  sum(case when month = 'apr' then value else 0 end) apr,
  sum(case when month = 'may' then value else 0 end) may,
  sum(case when month = 'jun' then value else 0 end) jun,
  sum(case when month = 'jul' then value else 0 end) jul,
  sum(case when month = 'aug' then value else 0 end) aug,
  sum(case when month = 'sep' then value else 0 end) sep,
  sum(case when month = 'oct' then value else 0 end) oct,
  sum(case when month = 'nov' then value else 0 end) nov,
  sum(case when month = 'dec' then value else 0 end) "dec"
from yt
group by id, extra_info

请参阅SQL Fiddle with Demo

Answer 2

tablefunc模块

我会使用crosstab()。如果您还没有安装附加模块tablefunc：

CREATE EXTENSION tablefunc

基础知识：
PostgreSQL Crosstab Query

如何处理额外的列：
Pivot on Multiple Columns using Tablefunc

高级用法：
Dynamic alternative to pivot with CASE and GROUP BY

设置

CREATE TEMP TABLE tbl
   (id int, extra_info varchar(3), month date, value int);

INSERT INTO tbl (id, extra_info, month, value)
VALUES
   (1, 'abc', '2012-01-01', 10),
   (1, 'abc', '2012-02-01', 20),
   (2, 'def', '2012-01-01', 10),
   (2, 'def', '2012-02-01', 5),
   (1, 'abc', '2012-01-01', 15),
   (3, 'ghi', '2012-03-01', 15);

我在基表中使用实际的date，因为我假设只是为了简化你的问题而隐藏它。但只有月份名称，ORDER BY就没有了。

查询

SELECT * FROM crosstab(
     $$SELECT id, extra_info, to_char(month, 'mon'), sum(value) AS value
       FROM   tbl
       GROUP  BY 1,2,month
       ORDER  BY 1,2,month$$

    ,$$VALUES
      ('jan'::text), ('feb'), ('mar'), ('apr'), ('may'), ('jun')
    , ('jul'),       ('aug'), ('sep'), ('oct'), ('nov'), ('dec')$$
   )
AS ct (id  int, extra text
   , jan int, feb int, mar int, apr int, may int, jun int
   , jul int, aug int, sep int, oct int, nov int, dec int);

结果：

 id | extra | jan | feb | mar | apr | may | jun | jul | aug | sep | oct | nov | dec
----+-------+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----
  1 | abc   |  25 |  20 |     |     |     |     |     |     |     |     |     |
  2 | def   |  10 |   5 |     |     |     |     |     |     |     |     |     |
  3 | ghi   |     |     |  15 |     |     |     |     |     |     |     |     |

安装tablefunc模块需要一些开销和一些学习，但结果查询更快，更短，更通用。