按月求和，然后将月作为列

本文介绍了按月求和，然后将月作为列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我每个月都有时间序列数据，我想对每个ID的值求和，按月分组，然后将月名作为列而不是行.

I have time series data on a monthly basis and I would like to sum values for each ID, grouped by month and then have the month names as columns rather than as rows.

+----+------------+-------+-------+
| 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    |

所需结果

+----+------------+-----+-----+-----+
| 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

我可以轻松地按月分组，对值求和.这使我能够:

Current Approach

I can easily group by month, summing the values. Which gets me to:

-----------------------------------
| 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    |

但是我现在需要那些月份作为列名.不知道从这里去哪里.

But I now need those months as column names. Not sure where to go from here.

就语言而言，此查询将在postgres中运行.
以上月份只是示例，显然真实数据集要大得多，涵盖了数千个ID的所有12个月

任何来自SQL专家的想法都非常感谢！

Any ideas from an SQL guru very much appreciated!

tablefunc 模块

我会为此使用crosstab().如果还没有，请安装附加模块tablefunc:

tablefunc module

I would use crosstab() for this. Install the additional module tablefunc if you don't have already:

CREATE EXTENSION tablefunc

此处的基本知识:
PostgreSQL交叉表查询

Basics here:
PostgreSQL Crosstab Query

如何处理多余的列:
使用Tablefunc在多个列上进行透视

How to deal with extra columns:
Pivot on Multiple Columns using Tablefunc

高级用法:
使用CASE和GROUP BY进行旋转的动态替代方案

Advanced usage:
Dynamic alternative to pivot with CASE and GROUP BY

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);

我在基表中使用了实际的date，因为我假设只是为了简化您的问题而将其隐藏.但是只有月份名称，ORDER BY就没什么了.

I am using an actual date in the base table, since I am assuming are just hiding that in a effort to simplify your question. But with just month names, there would be nothing to ORDER BY.

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，但是生成的查询要快得多，短得多并且用途更多.

Installing the tablefunc module requires some overhead and some learning, but the resulting queries are much faster and shorter and more versatile.

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tablefunc