Question

我有一个包含以下列的测试表：

Transaction_date  date
account_number    Number(10)
transaction_ammount  number (10,5)


account_number  transaction_amount    transaction date
111              10000                 10-OCT-2014
111              20000                 10-OCT-2014
111              50000                 08-OCT-2014
111              30000                 06-OCT-2014
222              60000                 10-OCT-2014
222              50000                 10-OCT-2014
222              30000                 08-OCT-2014

我需要根据以下公式计算评分：

评级=（（1天的总交易额） - （最近10个日历天的平均总值））/ 最后10天的标准偏差（STDDEV）。

我已经使用STDDEV函数来计算它。但它给我的输出不正确。

account_number 111的要求是：

排名第10 OCT =第10个OCt-Avg的所有金额为111（过去10天）/ STDEV为帐户111。

排名第10 OCT =（10000 + 20000） - （（10000 + 2000 + 50000 + 30000）/ 10）/ STDEV（10000,2000,50000,30000）

请建议我如何在sql query的帮助下实现这一目标。谢谢！

Answer 1

with t1 as (
select '111' account_number, 10000 transaction_amount, to_date('10-10-2014', 'DD-MM-YYYY') transaction_date from dual
union all select '111', 20000, to_date('10-10-2014', 'DD-MM-YYYY') from dual
union all select '111', 50000, to_date('08-10-2014', 'DD-MM-YYYY') from dual
union all select '111', 30000, to_date('06-10-2014', 'DD-MM-YYYY') from dual
union all select '222', 60000, to_date('10-10-2014', 'DD-MM-YYYY') from dual
union all select '222', 50000, to_date('10-10-2014', 'DD-MM-YYYY') from dual
union all select '222', 30000, to_date('08-10-2014', 'DD-MM-YYYY') from dual
),
t2 as (
select account_number, transaction_amount, transaction_date,
       -- aggregate transaction_amount for 1 day
       sum(transaction_amount) over (partition by account_number order by transaction_date range 0 preceding) aggr_amount,

       -- Avg aggregate value for last 10 calender days
       avg(transaction_amount) over (partition by account_number order by transaction_date range 10 preceding) aggr_avg,

       -- Standared Deviation(STDDEV) for last 10 days
       stddev(transaction_amount) over (partition by account_number order by transaction_date range 10 preceding) aggr_stddev
from t1)      
select account_number, transaction_amount, transaction_date,
       aggr_amount - aggr_avg / nullif(aggr_stddev, 0)
from t2;

计算Oracle中的标准差

1 个答案: