填充基于当前行和以前行的值的MySQL

Question

我有一个表tb1这样一列：

ID  DATE_  STATs 
1  2007-01  0.2 
1  2007-02  0.12 
1  2007-03  0.42 
1  2007-04  0.23 
1  2007-05  0.26 
1  2007-06  0.17 
2  2007-01  0.33 
2  2007-02  0.14 
2  2007-03  0.21 
2  2007-04  0.35 
2  2007-05  0.67 
2  2007-06  0.07 

如何添加一个附加列的计算：

(1+current.STATs)/(1+priorMonth.STATs) - 1 

每个ID ？

Answer 1

初步：使事情更容易，使date_和实际date。在第一天表示一个月的情况很常见。这就是我所做的。

选项1：使用子查询

SELECT 
    id, date_, 
    (1.0 + stats)/(1.0 + (SELECT stats FROM t t_prev WHERE t_prev.id = t.id AND t_prev.date_ = t.date_ - interval 1 month)) - 1.0 AS r 
FROM 
    t 
ORDER BY 
    id, date_ ; 

选项2：前（左）与同桌加入，一个月

SELECT 
    curr.id, curr.date_, (1.0 + curr.stats)/(1 + prev.stats) - 1.0 AS r 
FROM 
    t AS curr 
    LEFT JOIN t AS prev 
     ON prev.id = curr.id AND prev.date_ = curr.date_ - interval 1 month 
ORDER BY 
    curr.id, curr.date_ ; 

在这两种情况下，你会获得：

 
id | date_  |     r 
-: | :--------- | -------------------: 
1 | 2007-01-01 |     null 
1 | 2007-02-01 | -0.06666667121979919 
1 | 2007-03-01 | 0.26785713418538926 
1 | 2007-04-01 | -0.13380280596423388 
1 | 2007-05-01 | 0.024390232674120105 
1 | 2007-06-01 | -0.07142856298120104 
2 | 2007-01-01 |     null 
2 | 2007-02-01 | -0.14285715085991468 
2 | 2007-03-01 | 0.06140350246565207 
2 | 2007-04-01 | 0.11570248045838927 
2 | 2007-05-01 | 0.23703705486119708 
2 | 2007-06-01 | -0.35928144335037204 

你c检查一切在dbfiddle here