1

我在 SQL 中的表是这样的： -

RN   Name   value1  value2  Timestamp
1    Mark   110     210     20160119
1    Mark   106     205     20160115
1    Mark   103     201     20160112
2    Steve  120     220     20151218
2    Steve  111     210     20151210
2    Steve  104     206     20151203

期望的输出：-

RN  Name    value1Lag1 value1lag2   value2lag1  value2lag2
1   Mark       4             3            5        4
2   Steve      9             7            10       4

RN 1 的差值是从最近到第二次计算，然后从第二次到第三次计算

值1lag1 = 110-106 =4

值1lag2 = 106-103 = 3

value2lag1 = 210-205 = 5

value2lag2 = 205-201 = 4

其他RN也类似。

注意：对于每个 RN，有 3 行且只有 3 行。

我通过类似帖子的帮助尝试了几种方法，但没有运气。

4

4 回答 4

1

这里还有其他答案，但我认为您的问题是需要分析函数，特别是LAG()：

select
    rn,
    name,
    -- calculate the differences
    value1 - v1l1 value1lag1,
    v1l1 - v1l2 value1lag2,
    value2 - v2l1 value2lag1,
    v2l1 - v2l2 value2lag2
 from (
     select 
       rn, 
       name, 
       value1, 
       value2, 
       timestamp, 
       -- these two are the values from the row before this one ordered by timestamp (ascending)
       lag(value1) over(partition by rn, name order by timestamp asc) v1l1,
       lag(value2) over(partition by rn, name order by timestamp asc) v2l1
       -- these two are the values from two rows before this one ordered by timestamp (ascending)
       lag(value1, 2) over(partition by rn, name order by timestamp asc) v1l2,
       lag(value2, 2) over(partition by rn, name order by timestamp asc) v2l2

    from (
      select
      1 rn, 'Mark' name, 110 value1, 210 value2, '20160119' timestamp
      from dual
      union all
      select
      1 rn, 'Mark' name, 106 value1, 205 value2, '20160115' timestamp
      from dual
      union all
      select
      1 rn, 'Mark' name, 103 value1, 201 value2, '20160112' timestamp
      from dual
      union all
      select
      2 rn, 'Steve' name, 120 value1, 220 value2, '20151218' timestamp
      from dual
      union all
      select
      2 rn, 'Steve' name, 111 value1, 210 value2, '20151210' timestamp
      from dual
      union all
      select
      2 rn, 'Steve' name, 104 value1, 206 value2, '20151203' timestamp
      from dual
    ) data
)
where 
-- return only the rows that have defined values
v1l1 is not null and 
v1l2 is not null and
v2l1 is not null and 
v2l1 is not null

这种方法的好处是 Oracle 在内部进行了所有必要的缓冲，避免了自连接等。对于大数据集，从性能的角度来看，这可能很重要。

例如，该查询的解释计划类似于

-------------------------------------------------------------------------
| Id  | Operation        | Name | Rows  | Bytes | Cost (%CPU)| Time     |
-------------------------------------------------------------------------
|   0 | SELECT STATEMENT |      |     6 |   150 |    13   (8)| 00:00:01 |
|*  1 |  VIEW            |      |     6 |   150 |    13   (8)| 00:00:01 |
|   2 |   WINDOW SORT    |      |     6 |   138 |    13   (8)| 00:00:01 |
|   3 |    VIEW          |      |     6 |   138 |    12   (0)| 00:00:01 |
|   4 |     UNION-ALL    |      |       |       |            |          |
|   5 |      FAST DUAL   |      |     1 |       |     2   (0)| 00:00:01 |
|   6 |      FAST DUAL   |      |     1 |       |     2   (0)| 00:00:01 |
|   7 |      FAST DUAL   |      |     1 |       |     2   (0)| 00:00:01 |
|   8 |      FAST DUAL   |      |     1 |       |     2   (0)| 00:00:01 |
|   9 |      FAST DUAL   |      |     1 |       |     2   (0)| 00:00:01 |
|  10 |      FAST DUAL   |      |     1 |       |     2   (0)| 00:00:01 |
-------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - filter("V1L1" IS NOT NULL AND "V1L2" IS NOT NULL AND "V2L1" IS

请注意，没有连接，只有一个 WINDOW SORT 缓冲来自“数据源”（在我们的例子中，VIEW 3 是我们的 SELECT ... FROM DUAL 的 UNION ALL）的必要数据来分区和计算不同的滞后。

于 2016-02-02T08:48:38.457 回答
1

我假设 RN 和 Name 在这里链接。这有点乱，但如果每个 RN 总是有 3 个值并且你总是想按这个顺序检查它们，那么这样的东西应该可以工作。

SELECT
    t1.Name
    , AVG(CASE WHEN table_ranked.Rank = 1 THEN table_ranked.value1 ELSE NULL END) - AVG(CASE WHEN table_ranked.Rank = 2 THEN table_ranked.value1 ELSE NULL END)   value1Lag1
    , AVG(CASE WHEN table_ranked.Rank = 2 THEN table_ranked.value1 ELSE NULL END) - AVG(CASE WHEN table_ranked.Rank = 3 THEN table_ranked.value1 ELSE NULL END) value1Lag2
    , AVG(CASE WHEN table_ranked.Rank = 1 THEN table_ranked.value2 ELSE NULL END) - AVG(CASE WHEN table_ranked.Rank = 2 THEN table_ranked.value2 ELSE NULL END) value2Lag1
    , AVG(CASE WHEN table_ranked.Rank = 2 THEN table_ranked.value2 ELSE NULL END) - AVG(CASE WHEN table_ranked.Rank = 3 THEN table_ranked.value2 ELSE NULL END) value2Lag2
FROM table t1
INNER JOIN
(
    SELECT
        t1.Name
        , t1.value1
        , t1.value2
        , COUNT(t2.TimeStamp) Rank
    FROM table t1
    INNER JOIN table t2
        ON t2.name = t1.name
        AND t1.TimeStamp <= t2.TimeStamp
    GROUP BY t1.Name, t1.value1, t1.value2
) table_ranked
    ON table_ranked.Name = t1.Name
GROUP BY t1.Name
于 2016-02-02T07:54:39.280 回答
0 
-- test data
with data(rn,
name,
value1,
value2,
timestamp) as
 (select 1, 'Mark', 110, 210, to_date('20160119', 'YYYYMMDD')
    from dual
  union all
  select 1, 'Mark', 106, 205, to_date('20160115', 'YYYYMMDD')
    from dual
  union all
  select 1, 'Mark', 103, 201, to_date('20160112', 'YYYYMMDD')
    from dual
  union all
  select 2, 'Steve', 120, 220, to_date('20151218', 'YYYYMMDD')
    from dual
  union all
  select 2, 'Steve', 111, 210, to_date('20151210', 'YYYYMMDD')
    from dual
  union all
  select 2, 'Steve', 104, 206, to_date('20151203', 'YYYYMMDD') from dual),

-- first transform value1, value2 to value_id (1,2), value
data2 as
 (select d.rn, d.name, 1 as val_id, d.value1 as value, d.timestamp
    from data d
  union all
  select d.rn, d.name, 2 as val_id, d.value2 as value, d.timestamp
    from data d)

select *  -- find previous row P of row D, evaluate difference and build column name as desired
  from (select d.rn,
               d.name,
               d.value - p.value as value,
               'value' || d.val_id || 'Lag' || row_number() over(partition by d.rn, d.val_id order by d.timestamp desc) as col
          from data2 p, data2 d
         where p.rn = d.rn
           and p.val_id = d.val_id
           and p.timestamp =
               (select max(pp.timestamp)
                  from data2 pp
                 where pp.rn = p.rn
                   and pp.val_id = p.val_id
                   and pp.timestamp < d.timestamp))
       -- pivot
       pivot(sum(value) for col in('value1Lag1',
                                   'value1Lag2',
                                   'value2Lag1',
                                   'value2Lag2'));
于 2016-02-02T08:31:14.373 回答
0

如果只是在这种情况下，这并不难。你需要2个步骤

  1. 自加入并得到减号的结果

    select t1.RN,
       t1.Name,
       t1.rm,
       t2.value1-t1.value1 as value1, 
       t2.value2-t1.value2 as value2
from 
(select RN,Name,value1,value2,
        row_number(partition by Name order by Timestamp desc) as rm from table)t1 
left join
(select RN,Name,value1,value2,
        row_number(partition by Name order by Timestamp desc) as rm from table) t2 
on t1.rm = t2.rm-1
where t2.RN is not null.

您将其设置为表格，假设为 table3。

2.你转动它

select * from (
  select t3.RN, t3.Name,t3.rm,t3.value1,t3.value2 from table3 t3
               )
pivot 
 (
   max(value1)
    for rm in ('1','2')
  )v1

3.你得到了 value1 和 value2 的 2 个数据透视表，将它们连接在一起以获得结果。

但我认为可能有更好的方法，我不确定我们是否可以在旋转它时加入枢轴，所以我将在获得枢轴结果后使用加入，这将再制作 2 个表。它不好，但我能做的最好

于 2016-02-02T06:50:51.250 回答