查找前两条记录和最后一条记录的特殊 SQL 条件查询？

Question

我有一个像这样的数据集：

 student_id     date     project_id
 1              1/1/18   15
 1              1/1/18   17
 1              2/2/18   16 
 1              3/3/18   15 
 1              3/3/18   12 
 2              2/3/18   3
 2              4/3/18   4
 2              5/3/18   6 
 2              5/3/18   4 

我想用他们所做的前两个项目和他们所做的最后一个项目的日期来查找学生 ID，

student_id    project_id1st    date1st     project_id2nd      date2nd projectlast    datelast
1             15               1/1/18      17                 1/1/18  12               3/3/18
2             3                2/3/18      4                  4/3/18  4                5/3/18

首先我想在 pandas 中解决它，但得到了一些不好的结果。然后我尝试用 SQL 解决它。

WITH abc AS (
  SELECT student_id, project_id, date, 
         ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY date) rn
  FROM table 
)
SELECT student_id, 
       SUM(CASE WHEN rn = 1 THEN abc.project_id END) as firstid,
       SUM(CASE WHEN rn = 2 THEN abc.project_id END) as secondsid,
       MIN(CASE WHEN rn = 1 THEN date END) as first,
       MIN(CASE WHEN rn = 2 THEN date END) as second
FROM abc
GROUP BY 1;

我得到了很好的结果，但不知何故它使用ROW_NUMBER() 弄乱了数据集的顺序。例如，对于学生 1，rownumber 中值为 1 的 project_id 17，project_id 15 将成为第二个开始日期。

Answer 1

您的问题是您没有指示正确顺序的列。在row_number 窗口框架中，您提供order by date。但是当框架中有很多行时，结果是完全随机的。

您必须告诉数据库如何处理同一帧中的行。你想接project_id吗？没有像“原始订单”这样的指标。

例如

SELECT * FROM table

从不给出专门的顺序（例如插入数据集的顺序）。结果集可以完全随机排序。您定义的任何框架窗口也是如此。

因此，您将需要一种方法（列或算法）来确保您期望的顺序。

---

如果您选择ORDER BY date, project_id，那么（例如）3/3/18 的框架将使用project_id 12, 15 进行排序，这不是您的“原始订单”。如果您要订购它DESC 您的1/1/18 框架订购错误，因为第一个ID 将是17。所以project_id 不是一个好的订单标准。但是没有其他订购的可能性。那是因为您需要另一列。

要获得您的“原始订单”，一个自动增量 ID 列（键入 serial - 或者如果是 Postgres 10 和更高版本的 GENERATED AS IDENTITY - 会有所帮助）。

---

当您有特定顺序（例如插入 ID 列）时，这可能是您的查询：

WITH abc AS (
    SELECT insert_id, student_id, project_id, date, 
        ROW_NUMBER() OVER (PARTITION BY student_id ORDER BY insert_id) rn_asc,        -- A
        ROW_NUMBER() OVER (PARTITION BY student_id ORDER BY insert_id DESC) rn_desc   -- B
    FROM projects
)
SELECT 
    student_id,
    MAX(project_id) FILTER (WHERE rn_asc = 1),
    MAX(date) FILTER (WHERE rn_asc = 1),
    MAX(project_id) FILTER (WHERE rn_asc = 2),
    MAX(date) FILTER (WHERE rn_asc = 2),
    MAX(project_id) FILTER (WHERE rn_desc = 1),
    MAX(date) FILTER (WHERE rn_desc = 1)
FROM abc
GROUP BY student_id

A：按 ID 升序对学生帧进行排序，并给出行号 1 和 2，这有助于过滤前两行。

B：以相同的降序排列以获得最后一行（在本例中为 row number == 1）

demo: db<>fiddle

Answer 2

我会尝试通过简单的连接来完成，例如：

select min(date), max(date), t.bez, min(x.date) from table t
left join (
    SELECT date, student_id,
    ROW_NUMBER() OVER (PARTITION BY student_id ORDER BY date) rn
    FROM table
    group by date, student_id) x on x.student_id = t.student_id and x.rn = 2
group by t.student_id

您可以在 ROW_Number() 中简单地按日期和 student_id 分组，这样您就不会两次获得相同的日期

Answer 3

使用带有 2 列 project_id 和 date 的 order by

[Demo]

WITH abc AS (
  SELECT student_id, project_id, date, 
         ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY date,project_id) rn
  FROM table 
)
SELECT student_id, 
       SUM(CASE WHEN rn = 1 THEN abc.project_id END) as firstid,
       SUM(CASE WHEN rn = 2 THEN abc.project_id END) as secondsid,
       MIN(CASE WHEN rn = 1 THEN date END) as first,
       MIN(CASE WHEN rn = 2 THEN date END) as second
FROM abc
GROUP BY 1;

Answer 4

请尝试转换为整数。

与 abc AS (SELECT student_id, project_id, date,

CAST(ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY date) AS INT) rn FROM table )