根据不同的列组合获取行集

Question

鉴于此数据集，每只股票都有年度价值快照。

+----+------+------+-------+-------+
| ID | Name | Year | Stock | Value |
+----+------+------+-------+-------+
|  1 | John | 2019 | ABC   |   123 |
|  1 | John | 2020 | ABC   |   123 |
|  1 | John | 2021 | ABC   |   123 |
|  1 | John | 2021 | XYZ   |   200 |
| 1  | John | 2022 | ABC   |   123 |
|  1 | John | 2022 | XYZ   |   200 |
|  1 | John | 2023 | ABC   |   630 |
|  1 | John | 2023 | XYZ   |   200 |
+----+------+------+-------+-------+

2019年，约翰只持有价值123的ABC

2020年，约翰也只持有ABC，价值123（未变）

2021年，John持有ABC，但也收购了XYZ，价值200

2022 年，John 持有 ABC 和 XYZ，这两个值都没有变化。

2023年，约翰持有ABC和XYZ，ABC的价值增加到630，XYZ的价值保持在200。

我想返回行以便

• 每年，如果 John 的投资组合自去年以来没有任何变化，则不会返回任何行
• 如果 John 的投资组合自去年以来发生任何变化，则会列出他目前持有的所有资产

例如，

+----+------+------+-------+-------+
| ID | Name | Year | Stock | Value |
+----+------+------+-------+-------+
|  1 | John | 2019 | ABC   |   123 |
|  1 | John | 2021 | ABC   |   123 |
|  1 | John | 2021 | XYZ   |   200 |
|  1 | John | 2023 | ABC   |   630 |
|  1 | John | 2023 | XYZ   |   200 |
+----+------+------+-------+-------+

我将如何做到这一点，无论是通过 PL/SQL 中的函数还是纯 SQL 中的函数？

Answer 1

如果每个用户的行数不多，那么listagg() 提供了一个方便的解决方案：

select ny.*
from (select name, year,
             listagg(stock || ':' || value, ',') within group (order by stock) as stocks,
             lag(listagg(stock || ':' || value, ',') within group (order by stock)) as prev_stocks,
             lag(year) over (partition by name order by year) as prev_year
      from t
      group by name, year
     ) ny
where prev_year is null or prev_year <> year - 1 or prev_stocks <> stocks;

或者，您可以单独检查每一行并使用分析函数将信息投影到名称/年份中的所有行：

select t.*
from (select t.*,
             sum(case when prev_nsv_year = year then 0 else 1 end) over (partition by name, year) as num_diff,
             lag(cnt) over (partition by name order by year) as prev_cnt
      from (select t.*,
                   lag(year) over (partition by name, stock, value over order by year) as prev_nsv_year,
                   count(*) over (partition by name, year) as cnt
            from t
           ) t
     ) t
where cnt <> prev_cnt or prev_cnt is null or
      num_diff > 0;