PostgreSQL:如何使用n步骤对组中的记录进行排序

时间:2019-05-22 01:53:18

标签: sql postgresql

我的表格中的行按排名排序。所有 rank 值均不同。另外,我还有颜色列,其中包含可能的值(红色,蓝色,绿色)。我需要遍历earch行(例如,将其命名为 X )并检查下一个 n 行中是否存在行(例如,将其命名为 Y )其中 X.color == Y.color 然后需要移动 Y 行并将其插入 X 行之后。 例如 n = 5 和起始表:

id  rank    color
1   0.8     red                                               
3   0.76    blue                                              
4   0.75    blue                                              
7   0.743   green                                             
5   0.742   red                                               
6   0.741   green                                             
8   0.6     blue                                              
9   0.5     blue                                              
11  0.47    green                                             
12  0.45    blue                                              
10  0.4     red 

1进行迭代,在接下来的 n 行中,我们有颜色为红色的行,然后:

  id    rank    color
    1   0.8     red
    5   0.742   red                                                                                              
    3   0.76    blue                                              
    4   0.75    blue                                              
    7   0.743   green                                             
    6   0.741   green                                             
    8   0.6     blue                                              
    9   0.5     blue                                              
    11  0.47    green                                             
    12  0.45    blue                                              
    10  0.4     red

2进行迭代,接​​下来的 n 行中没有红色,因此没有更改

3进行迭代,在接下来的 n 行中有 blue ,但是在当前行之后已经有了下一个蓝色行。

4迭代,接下来的 n 行中,我们有蓝色,然后:

  id    rank    color
    1   0.8     red
    5   0.742   red                                                                                              
    3   0.76    blue                                              
    4   0.75    blue
    8   0.6     blue                                                                                            
    7   0.743   green                                             
    6   0.741   green                                             
    9   0.5     blue                                              
    11  0.47    green                                             
    12  0.45    blue                                              
    10  0.4     red

5进行迭代,我们在接下来的 n 行中添加蓝色,然后:

  id    rank    color
    1   0.8     red
    5   0.742   red                                                                                              
    3   0.76    blue                                              
    4   0.75    blue
    8   0.6     blue
    9   0.5     blue                                                                                                                                          
    7   0.743   green                                             
    6   0.741   green                                             
    11  0.47    green                                             
    12  0.45    blue                                              
    10  0.4     red

6迭代,接下来的 n 行中,我们有蓝色,然后:

  id    rank    color
    1   0.8     red
    5   0.742   red                                                                                              
    3   0.76    blue                                              
    4   0.75    blue
    8   0.6     blue
    9   0.5     blue
    12  0.45    blue                                                                                                                                                                                        
    7   0.743   green                                             
    6   0.741   green                                             
    11  0.47    green                                             
    10  0.4     red

7进行迭代,在接下来的 n 行中有 green ,但是在当前行之后已经有下一个绿色行。

8进行迭代,在接下来的 n 行中有 green ,但是在当前行之后已经有下一个绿色行。

9进行迭代,在接下来的 n 行中有 green ,但是在当前行之后已经有下一个绿色行。

10不进行任何更改。

有人对如何解决这个问题有一个想法,我试图固定窗口功能,但尚不清楚是否可以借助它们来完成,也许还需要其他机制吗?

1 个答案:

答案 0 :(得分:1)

对于刚接触SQL的人们来说,遇到诸如此类的问题时考虑迭代确实是很常见的做法,但是无需任何迭代就可以做到这一点。让我们分解一下。首先,让我们根据排名列找到所有数据的row_id:

with data as (select * from (values
(1,0.8 ,'red'),
(3,0.76 ,'blue'),
(4,0.75 ,'blue'),
(7,0.743 ,'green'),
(5,0.742 ,'red'),
(6,0.741 ,'green'),
(8,0.6 ,'blue'),
(9,0.5 ,'blue'),
(11,0.47 ,'green'),
(12,0.45 ,'blue'),
(10,0.4 ,'red')
) v(id, rank, color)
)
select id, rank, color,
  row_number() over (order by rank desc) as row_id
FROM data
ORDER BY rank desc;
 id | rank  | color | row_id
----+-------+-------+--------
  1 |   0.8 | red   |      1
  3 |  0.76 | blue  |      2
  4 |  0.75 | blue  |      3
  7 | 0.743 | green |      4
  5 | 0.742 | red   |      5
  6 | 0.741 | green |      6
  8 |   0.6 | blue  |      7
  9 |   0.5 | blue  |      8
 11 |  0.47 | green |      9
 12 |  0.45 | blue  |     10
 10 |   0.4 | red   |     11
(11 rows)

从那里,我们可以找出前五行中具有相同颜色的最小行:

with data as (select * from (values
(1,0.8 ,'red'),
(3,0.76 ,'blue'),
(4,0.75 ,'blue'),
(7,0.743 ,'green'),
(5,0.742 ,'red'),
(6,0.741 ,'green'),
(8,0.6 ,'blue'),
(9,0.5 ,'blue'),
(11,0.47 ,'green'),
(12,0.45 ,'blue'),
(10,0.4 ,'red')
) v(id, rank, color)
), all_rows as (
select id, rank, color,
  row_number() over (order by rank desc) as row_id
FROM data
ORDER BY rank desc
)
select id, rank, color, row_id,
  first_value(row_id) over (partition by color order by row_id range between 5 
  preceding and current row)
from all_rows
order by 5, rank desc;
 id | rank  | color | row_id | first_value
----+-------+-------+--------+-------------
  1 |   0.8 | red   |      1 |           1
  5 | 0.742 | red   |      5 |           1
  3 |  0.76 | blue  |      2 |           2
  4 |  0.75 | blue  |      3 |           2
  8 |   0.6 | blue  |      7 |           2
  9 |   0.5 | blue  |      8 |           3
  7 | 0.743 | green |      4 |           4
  6 | 0.741 | green |      6 |           4
 11 |  0.47 | green |      9 |           4
 12 |  0.45 | blue  |     10 |           7
 10 |   0.4 | red   |     11 |          11
(11 rows)

这很接近,但是不太正确。我们需要从每种颜色的最后5行中获取最小值。我们还需要一步:

with data as (select * from (values
(1,0.8 ,'red'),
(3,0.76 ,'blue'),
(4,0.75 ,'blue'),
(7,0.743 ,'green'),
(5,0.742 ,'red'),
(6,0.741 ,'green'),
(8,0.6 ,'blue'),
(9,0.5 ,'blue'),
(11,0.47 ,'green'),
(12,0.45 ,'blue'),
(10,0.4 ,'red')
) v(id, rank, color)
), all_rows as (
select id, rank, color,
  row_number() over (order by rank desc) as row_id
FROM data
ORDER BY rank desc
), first_values as (
select id, rank, color, row_id,
  first_value(row_id) over (partition by color order by row_id range between 5 
  preceding and current row)
from all_rows
order by 5, rank desc
)
select id, rank, color, row_id,
first_value(first_value) over (partition by color order by row_id range between 5 preceding and current row) as overall_rank
FROM first_values
ORDER BY overall_rank, rank desc
;
 id | rank  | color | row_id | overall_rank
----+-------+-------+--------+--------------
  1 |   0.8 | red   |      1 |            1
  5 | 0.742 | red   |      5 |            1
  3 |  0.76 | blue  |      2 |            2
  4 |  0.75 | blue  |      3 |            2
  8 |   0.6 | blue  |      7 |            2
  9 |   0.5 | blue  |      8 |            2
 12 |  0.45 | blue  |     10 |            2
  7 | 0.743 | green |      4 |            4
  6 | 0.741 | green |      6 |            4
 11 |  0.47 | green |      9 |            4
 10 |   0.4 | red   |     11 |           11
(11 rows)