如何将每一行除以这些行的总和？

Posted 2023-03-23

【中文标题】如何将每一行除以这些行的总和？

【英文标题】：how to divide each row by the sum of those rows?

【发布时间】：2021-12-30 04:51:38

【问题描述】：

我有一张表 CREDITS 包含这些列：

DAY_OPERATIONAL TOTAL_LOAN CREDITS_PERCENT

我应该将每一行的TOTAL_LOAN 和CREDITS_PERCENT 相乘，然后除以SUM(TOTAL_LOAN) 中的DAY_OPERATIONAL。

我的代码不起作用。

我怎样才能做到这一点？如何正确使用 group by？

SELECT 
    (total_loan * credits_percent) / SUM(total_loan) 
FROM
    credits
WHERE
    day_operational = '29.12.2021'

【问题讨论】：

你能分享样本数据集和想要的结果吗？

【参考方案1】：

这就是我对问题的理解。

tc 和 tl 在这里只是为了显示用于计算最终 result 的值。

SQL> with credits (day_operational, total_loan, credits_percent) as
  2    (select date '2021-12-01', 100, 1 from dual union all
  3     select date '2021-12-01', 200, 4 from dual union all
  4     --
  5     select date '2021-12-02', 500, 5 from dual
  6    )
  7  select day_operational,
  8         sum(total_loan * credits_percent) tc,
  9         sum(total_loan) tl,
 10         --
 11         sum(total_loan * credits_percent) / sum(total_loan) result
 12  from credits
 13  group by day_operational
 14  order by day_operational;

DAY_OPERAT         TC         TL     RESULT
---------- ---------- ---------- ----------
01.12.2021        900        300          3
02.12.2021       2500        500          5

SQL>

所以：

01.12.2021: 100 * 1 + 200 * 4 = 100 + 800 =  900
            100 + 200                     =  300  -->  900 / 300 = 3

02.12.2021: 500 * 5                       = 2500
            500                           =  500  --> 2500 / 500 = 5

---

截至day_operational：如果它的数据类型是date（应该是！），那么不要将它与字符串进行比较。 '29.12.2021' 是一个字符串。使用

日期文字（和我一样）：date '2021-12-29' 或

to_date 具有适当格式掩码的函数：

to_date('29.12.2021', 'dd.mm.yyyy')

【参考方案2】：

TOTAL_LOAN * CREDITS_PERCENT / SUM(TOTAL_LOAN)

=

CREDITS_PERCENT * (TOTAL_LOAN / SUM(TOTAL_LOAN)

因此

SELECT
  day_operational
, credits_percent * RATIO_TO_REPORT(total_loan) OVER (PARTITION BY day_operational)
FROM credits

见RATIO_TO_REPORT