Question

给出下表

  grp |   ind |   val
----------------------
    a |     1 |     1
    a |     2 |     1
    a |     3 |     1
    a |     4 |     2
    a |     5 |     2
    a |     6 |     4
    a |     7 |     2
    b |     1 |     1
    b |     2 |     1
    b |     3 |     1
    b |     4 |     3
    b |     5 |     3
    b |     6 |     4

我需要选择以下内容：

  grp |   ind |   val
----------------------
    a |     1 |     1
    a |     4 |     2
    a |     6 |     4
    a |     7 |     2
    b |     1 |     1
    b |     4 |     3
    b |     6 |     4

对于每个'grp'，每个记录的'val'与前面的'val'不同（按'index'排序）所以每个记录中'value'“step”。

实现这一目标的最有效方法是什么？

感谢。

以下是创建测试用例的脚本：

create temp table test_table
(
    grp character varying,
    ind numeric,
    val numeric
);
insert into test_table values
    ('a', 1 , 1),
    ('a', 2 , 1),
    ('a', 3 , 1),
    ('a', 4 , 2),
    ('a', 5 , 2),
    ('a', 6 , 4),
    ('a', 7 , 2),
    ('b', 1 , 1),
    ('b', 2 , 1),
    ('b', 3 , 1),
    ('b', 4 , 3),
    ('b', 5 , 3),
    ('b', 6 , 4);

Answer 1

DROP SCHEMA tmp CASCADE;
CREATE SCHEMA tmp ;
SET search_path=tmp;

CREATE TABLE ztable
        ( zgroup CHAR(1)
        , zindex int
        , zvalue INTEGER
        );
INSERT INTO ztable(zgroup,zindex,zvalue) VALUES
    ('a',     1,      1)
    ,('a',     2,      1)
    ,('a',     3,      1)
    ,('a',     4,      2)
    ,('a',     5,      2)
    ,('a',     6,      4)
    ,('b',     1,      1)
    ,('b',     2,      1)
    ,('b',     3,      1)
    ,('b',     4,      3)
    ,('b',     5,      3)
    ,('b',     6,      4)
        ;

WITH agg AS (
        SELECT zgroup
        , zindex
        , zvalue
        , row_number() OVER (PARTITION BY zgroup ORDER BY zindex) AS zrank
        FROM ztable
        )
SELECT t1.zgroup,t1.zindex,t1.zvalue
FROM agg t1
LEFT JOIN agg t0 ON t0.zgroup = t1.zgroup AND 1+t0.zrank = t1.zrank
WHERE t0.zvalue <> t1.zvalue OR t0.zrank IS NULL
        ;

Answer 2

select grp,
       ind,
       val
from (
   select grp, 
          ind, 
          val,
          lag(val,1,0::numeric) over (partition by grp order by ind) - val as diff
   from test_table
) t
where diff <> 0;

Answer 3

select group,min(index) as index,value from table
group by group,value

选择“步骤”记录

3 个答案: