Question

我有一个表格，其中包含以下形式的每月数据：

Date | Value
2018-06 | 100
2018-07 | 105
2017-06 | 90
2017-07 | 92

获取此数据并计算每个月的同比回报的最佳方法是什么？我希望它看起来像：

Date | YoY growth
2018-06 | 0.11111
2018-07 | 0.1413

Answer 1

使用DISTINCT ON month仅选择月份。然后使用反向列表中的LEAD窗口函数（通过ORDER BY d DESC）获得当前年份之前的年份

实时测试：http://sqlfiddle.com/#!17/5be39/2

select
    distinct on ( date_part('month', (d || '-01')::date) )  -- get the months only

    -- ORDER BY d DESC sort the list from most recent to oldest.
    -- so LEAD here actually refers to previous year

    d,
    v,               
    lead(v) over(partition by date_part('month', (d || '-01')::date) order by d desc), 

    ( v / lead(v) over(partition by date_part('month', (d || '-01')::date) order by d desc) )
    - 1 as YoyGrowth


from progress

输出：

|       d |   v | lead |          yoygrowth |
|---------|-----|------|--------------------|
| 2018-06 | 100 |   90 | 0.1111111111111111 |
| 2018-07 | 105 |   92 |  0.141304347826087 |

Answer 2

我只是解析日期并使用lag()：

select date, value, prev_value,
       (value - prev_value) / prev_value as YOY_growth
from (select t.*,
             lag(value) over (partition by right(date, 2)
                              order by left(date, 4)
                             ) as prev_value
      from t
     ) t
where prev_value is null

SQL-计算月度回报的逐年增长

2 个答案: