count(distinct) over (partition by... 在 Oracle SQL 中不起作用

Question

我想统计过去 30 天的 distinct day_number。但是，distinct 函数不能与 over

一起使用

如果我删除 distinct，它会给我 day_number 的总数，但是 day_number 可能有很多重复项。所以这就是为什么我要添加 distinct.

select tr.*,
       count( distinct day_number) OVER (PARTITION BY ACCOUNT ORDER BY DAY_number range 29 PRECEDING) as result
from table tr;

任何人都可以告诉我如何计算 over(partition by..) 语句中的不同数字吗？提前致谢。

Answer 1

您可以先创建一个只列出每个 ID 一次的列，然后对该列进行范围计数，例如：

WITH sd AS (SELECT 1 ID, 10 val FROM dual UNION ALL
            SELECT 1 ID, 20 val FROM dual UNION ALL
            SELECT 2 ID, 30 val FROM dual UNION ALL
            SELECT 2 ID, 40 val FROM dual UNION ALL
            SELECT 4 ID, 50 val FROM dual UNION ALL
            SELECT 4 ID, 60 val FROM dual UNION ALL
            SELECT 6 ID, 70 val FROM dual)
SELECT ID,
       val,
       COUNT(id_distinct) OVER (ORDER BY ID RANGE 3 PRECEDING) cnt_disinct_ids
FROM   (SELECT ID,
               val,
               CASE WHEN row_number() OVER (PARTITION BY ID ORDER BY val) = 1 THEN ID END id_distinct
        FROM   sd);

        ID        VAL CNT_DISINCT_IDS
---------- ---------- ---------------
         1         10               1
         1         20               1
         2         30               2
         2         40               2
         4         50               3
         4         60               3
         6         70               2

ETA：证明上述技术适用于您的数据：

WITH your_table AS (SELECT 'ABCDE' account_sk, 23 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 23 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 24 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 25 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 53 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 53 day_sk FROM dual UNION ALL
                    SELECT 'ABCDE' account_sk, 55 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 10 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 12 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 40 day_sk FROM dual UNION ALL
                    SELECT 'VWXYZ' account_sk, 40 day_sk FROM dual)
SELECT account_sk,
       day_sk,
       COUNT(day_sk_distinct) OVER (PARTITION BY account_sk ORDER BY day_sk RANGE BETWEEN 29 PRECEDING AND CURRENT ROW) count_distinct_day_sk
FROM   (SELECT account_sk,
               day_sk,
               CASE WHEN row_number() OVER (PARTITION BY account_sk, day_sk ORDER BY day_sk) = 1 THEN day_sk END day_sk_distinct
        FROM   your_table);

ACCOUNT_SK     DAY_SK COUNT_DISTINCT_DAY_SK
---------- ---------- ---------------------
ABCDE              23                     1
ABCDE              23                     1
ABCDE              24                     2
ABCDE              25                     3
ABCDE              53                     3
ABCDE              53                     3
ABCDE              55                     2
VWXYZ              10                     1
VWXYZ              12                     2
VWXYZ              40                     2
VWXYZ              40                     2

Answer 2

count(distinct ...) 与 over 子句配合使用效果很好，主要问题是排序依据。您不能执行 count (distinct ..) over (partition by ... order by ...)，因为 DISTINCT 函数和 RATIO_TO_REPORT 不能有 ORDER BY。所以我这样做了：

select tr.*, count (distinct day_number) over (partition by account)
from (select t.*, row_number() over (partition by account order by day_number) row_number from table t) tr
where row_number < 30;

我在人力资源部的employees scheme中测试过（免费的oracle scheme随处可见）我不确定它是否适用于您的架构，因为我没有它的副本，但如果没有，它应该会给您一些想法：

 select count (distinct manager_id) over (partition by department_id), department_id, manager_id
from (select e.*, row_number() over (partition by department_id order by employee_id) row_number from employees e)
where row_number < 30;

Answer 3

您可以通过创建一个列来模拟这一点，当“新中断”出现在有序列表中时该列为 1，否则为 null。然后你只需要计算或总结这些突破指标。 count() 和 sum() 都支持“over(order by...)”。在您的示例中，它将是：

with TR as
(select 1 as PK, 'ABCDE' as ACCOUNT, 23 as DAY_number from dual
union all select 2, 'ABCDE', 23 from dual
union all select 3, 'ABCDE', 24 from dual
union all select 4, 'ABCDE', 25 from dual
)
select tr.*
    ,  count( /*distinct*/ day_number) OVER (PARTITION BY ACCOUNT ORDER BY DAY_number range 29 PRECEDING) as wrong_result
    ,  count(IS_NEW_BREAK) over(PARTITION BY ACCOUNT order by day_number range 29 PRECEDING) as desired_output
    from
    (select tr.*
        ,   case min(case when day_number is not null then PK end) over(PARTITION BY ACCOUNT, day_number) when PK then 1 end as IS_NEW_BREAK
        from tr
    ) tr;

count(distinct) over (partition by... 在 Oracle SQL 中不起作用

count(distinct) over (partition by... doesn't work in Oracle SQL

sql

oracle

distinct

window-functions