Teradata 划分错误？

Question

我有一位同事不想在百分等级中包含空行。默认的 Teradata 函数似乎只是将 null 视为集合中的最小数字，因此我决定手动进行数学运算。我开始使用下面的查询来测试我的方程式

drop table tmp;

create multiset volatile table tmp (
  num byteint
) primary index (num)
  on commit preserve rows
;

insert into tmp
values (1)
;insert into tmp
values (2)
;insert into tmp
values (1)
;insert into tmp
values (4)
;insert into tmp
values (null)
;insert into tmp
values (4)
;insert into tmp
values (null)
;insert into tmp
values (2)
;insert into tmp
values (9)
;insert into tmp
values (null)
;insert into tmp
values (10)
;insert into tmp
values (10)
;insert into tmp
values (11)
;

select
  num,
  case
    when num is null then 0
    else cast(dense_rank() over (partition by case when num is not null then 1 else 2 end order by num) as number)
  end as str_rnk,
  q.nn,
  str_rnk/q.nn as pct_rnk
from tmp
cross join (
    select cast(count(num) as number) as nn from tmp
) q
order by num
;

所以我希望在结果集中看到的是：

num   str_rnk  nn  pct_rnk
null        0  10        0
null        0  10        0
null        0  10        0
   1        1  10      0.1
   1        1  10      0.1
   2        2  10      0.2
   2        2  10      0.2
   4        3  10      0.3
   4        3  10      0.3
   9        4  10      0.4
  10        5  10      0.5
  10        5  10      0.5

但我得到的结果看起来像是常规的rank 而不是dense_rank，如下所示：

num   str_rnk  nn  pct_rnk
null        0  10        0
null        0  10        0
null        0  10        0
   1        1  10      0.1
   1        1  10      0.1
   2        2  10      0.3
   2        2  10      0.3
   4        3  10      0.5
   4        3  10      0.5
   9        4  10      0.7
  10        5  10      0.8
  10        5  10      0.8

我知道我可以在子查询中设置排名，它会按照我期望的方式进行计算，但为什么不按照我现在的方式进行呢？

Answer 1

虽然这不能回答您的问题。这不是一个问题，这似乎是一个奇怪的问题，在同一个 SELECT 中运行 CAST 和 Dense_Rank 两次。

考虑：

select
  num,
  case
    when num is null then 0
    else cast(dense_rank() over (partition by case when num is not null then 1 else 2 end order by num) as number)
  end as str_rnk,

  case
    when num is null then 0
    else cast(dense_rank() over (partition by case when num is not null then 1 else 2 end order by num) as number)
  end as str_rnk2
from tmp
cross join (
    select cast(count(num) as number) as nn from tmp
) q;


+--------+---------+----------+
|  num   | str_rnk | str_rnk2 |
+--------+---------+----------+
| 1      |       1 |        1 |
| 1      |       1 |        1 |
| 2      |       2 |        3 |
| 2      |       2 |        3 |
| 4      |       3 |        5 |
| 4      |       3 |        5 |
| 9      |       4 |        7 |
| 10     |       5 |        8 |
| 10     |       5 |        8 |
| 11     |       6 |       10 |
| <null> |       0 |        0 |
| <null> |       0 |        0 |
| <null> |       0 |        0 |
+--------+---------+----------+

因为这里不需要 CAST：

select
  num,

  case
    when num is null then 0
    else dense_rank() over (partition by case when num is not null then 1 else 2 END order by num) 
  end as str_rnk,

  case
    when num is null then 0
    else dense_rank() over (partition by case when num is not null then 1 else 2 END order by num) 
  end as str_rnk2
from tmp
cross join (
    select cast(count(num) as number) as nn from tmp
) q;

+--------+---------+----------+
|  num   | str_rnk | str_rnk2 |
+--------+---------+----------+
| 1      |       1 |        1 |
| 1      |       1 |        1 |
| 2      |       2 |        2 |
| 2      |       2 |        2 |
| 4      |       3 |        3 |
| 4      |       3 |        3 |
| 9      |       4 |        4 |
| 10     |       5 |        5 |
| 10     |       5 |        5 |
| 11     |       6 |        6 |
| <null> |       0 |        0 |
| <null> |       0 |        0 |
| <null> |       0 |        0 |
+--------+---------+----------+

您的查询，快速重写：

select
  num,
  case
    when num is null then 0
    else dense_rank() over (partition by num * 0 order by num) 
    end as str_rnk, 
  str_rnk * 1.0/COUNT(*) OVER (PARTITION BY num * 0) as pct_rnk
from tmp
order by num
;

+--------+---------+---------+
|  num   | str_rnk | pct_rnk |
+--------+---------+---------+
| <null> |       0 |     0.0 |
| <null> |       0 |     0.0 |
| <null> |       0 |     0.0 |
| 1      |       1 |     0.1 |
| 1      |       1 |     0.1 |
| 2      |       2 |     0.2 |
| 2      |       2 |     0.2 |
| 4      |       3 |     0.3 |
| 4      |       3 |     0.3 |
| 9      |       4 |     0.4 |
| 10     |       5 |     0.5 |
| 10     |       5 |     0.5 |
| 11     |       6 |     0.6 |
+--------+---------+---------+

或者，如果您想完全摆脱 CASE 语句：

select
  num,
  dense_rank() over (partition by num * 0 order by num) * (num * 0 + 1.0) as str_rnk,  
  str_rnk/COUNT(*) OVER (PARTITION BY num * 0) as pct_rnk
from tmp
order by num;

Answer 2

正如 JNevill 指出的那样，这是一个错误，您应该使用 Teradata 支持打开一个事件：

SELECT
   num,
   -- cast to FLOAT or DECIMAL works as expected
   Cast(Dense_Rank() Over (ORDER BY num) AS NUMBER) AS a,
   a AS b
FROM tmp

 num    a    b
----  ---  ---
   ?    1    1
   ?    1    1
   ?    1    1
   1    2    4
   1    2    4
   2    3    6
   2    3    6
   4    4    8
   4    4    8
   9    5   10
  10    6   11
  10    6   11
  11    7   13

但是添加QUALIFY a<>b 会返回一个空结果:-)

PERCENT_RANK的原始计算是基于

Cast(Rank() Over (ORDER BY num) -1 AS DEC(18,6)) / Count(*) Over ()

如果要排除 NULL，可以切换到 Count(num) 和 NULLS LAST：

SELECT
   num,
   CASE
      WHEN num IS NOT NULL 
      THEN Cast(Dense_Rank() Over (ORDER BY num NULLS LAST) AS DECIMAL(18,6)) 
      ELSE 0
   END AS str_rnk,
   str_rnk / Count(num) Over ()
FROM tmp

或者使用巧妙的num * 0技巧：

SELECT
   num,
   Coalesce(Dense_Rank()
            Over (ORDER BY num NULLS LAST) 
             * (num * 0 +1.000000), 0) AS str_rnk,
   str_rnk / Count(num) Over ()
FROM tmp