我想用平均值计算库存成本，我有点卡在这里......
考虑一个简单的交易表tr :( ids是自动增量，负数量表示卖出交易）
order_id | volume | price | type
1 | 1000 | 100 | B
2 | -500 | 110 | S
3 | 1500 | 80 | B
4 | -100 | 150 | S
5 | -600 | 110 | S
6 | 700 | 105 | B
现在我想知道每笔交易后的总交易量和总成本。困难在于销售正确。卖出总是以此时的平均成本计算（即卖出价格在这里实际上并不相关），因此交易订单在这里很重要。
最理想的结果是：
order_id | volume | price | total_vol | total_costs | unit_costs
1 | 1000 | 100 | 1000 | 100000 | 100
2 | -500 | 110 | 500 | 50000 | 100
3 | 1500 | 80 | 2000 | 170000 | 85
4 | -100 | 150 | 1900 | 161500 | 85
5 | -600 | 110 | 1300 | 110500 | 85
6 | 700 | 105 | 2000 | 184000 | 92
现在，另一方面，total_vol很容易sum(volume) over (...)，总费用。我玩过窗口函数，但除非我遗漏了一些非常明显（或非常聪明）的东西，否则我认为不能单独使用窗口函数...
任何帮助将不胜感激。 ：）
更新
这是我最终使用的代码，两个答案的组合（数据模型比上面的简化示例复杂一点，但你明白了）：
select ser_num
, tr_id
, tr_date
, action_typ
, volume
, price
, total_vol
, trunc(total_costs,0) total_costs
, trunc(unit_costs,4) unit_costs
from itt
model
partition by (ser_num)
dimension by (row_number() over (partition by ser_num order by tr_date, tr_id) rn)
measures (tr_id, tr_date, volume, price, action_typ, 0 total_vol, 0 total_costs, 0 unit_costs)
rules automatic order
( total_vol[ANY] order by rn
= nvl(total_vol[cv()-1],0) +
decode(action_typ[cv()], 'Buy', 1, 'Sell', -1) * volume[cv()]
, total_costs[ANY] order by rn
= case action_typ[cv()]
when 'Buy' then volume[cv()] * price[cv()] + nvl(total_costs[cv()-1],0)
when 'Sell' then total_vol[cv()] * nvl(unit_costs[cv()-1],price[cv()])
end
, unit_costs[ANY] order by rn
= decode(total_vol[cv()], 0, unit_costs[cv()-1],
total_costs[cv()] / total_vol[cv()])
)
order by ser_num, tr_date, tr_id
一些观察结果：
cv()-1）的引用时，维度必须以与整个模型子句相同的方式进行分区（这也是为什么使用iteration_number可能会很棘手）order by rn Automatic order自动执行此操作，此处不需要迭代）答案 0 :(得分：15)
您可以使用MODEL子句执行此递归计算
创建样本表并插入数据
create table costs (order_id int, volume int, price numeric(16,4), type char(1));
insert into costs (order_id, volume, price) values (1,1000,100);
insert into costs (order_id, volume, price) values (2,-500,110);
insert into costs (order_id, volume, price) values (3,1500,80);
insert into costs (order_id, volume, price) values (4,-100,150);
insert into costs (order_id, volume, price) values (5,-600,110);
insert into costs (order_id, volume, price) values (6,700,105);
查询（已编辑更改rules iterate(1000)至rules automatic order实现了MODEL子句，因为它打算正常运行，即从上到下顺序执行。查询从0.44s到0.01s！）
select order_id, volume, price, total_vol, total_costs, unit_costs
from (select order_id, volume, price,
volume total_vol,
0.0 total_costs,
0.0 unit_costs,
row_number() over (order by order_id) rn
from costs order by order_id)
model
dimension by (order_id)
measures (volume, price, total_vol, total_costs, unit_costs)
rules automatic order -- iterate(1000)
( total_vol[any] = volume[cv()] + nvl(total_vol[cv()-1],0.0),
total_costs[any] =
case SIGN(volume[cv()])
when -1 then total_vol[cv()] * nvl(unit_costs[cv()-1],0.0)
else volume[cv()] * price[cv()] + nvl(total_costs[cv()-1],0.0)
end,
unit_costs[any] = total_costs[cv()] / total_vol[cv()]
)
order by order_id
输出
ORDER_ID VOLUME PRICE TOTAL_VOL TOTAL_COSTS UNIT_COSTS
1 1000 100 1000 100000 100
2 -500 110 500 50000 100
3 1500 80 2000 170000 85
4 -100 150 1900 161500 85
5 -600 110 1300 110500 85
6 700 105 2000 184000 92
本网站有一个关于MODEL子句的好教程
<小时/> 上面数据的EXCEL表格如下所示，公式向下延伸
A B C D E F
---------------------------------------------------------------------------
1| order_id volume price total_vol total_costs unit_costs
2| 0 0 0
3| 1 1000 100 =C4+E3 =IF(C4<0,G3*E4,F3+C4*D4) =F4/E4
4| 2 -500 110 =C5+E4 =IF(C5<0,G4*E5,F4+C5*D5) =F5/E5
5| 3 1500 80 =C6+E5 =IF(C6<0,G5*E6,F5+C6*D6) =F6/E6
6| 4 -100 150 =C7+E6 =IF(C7<0,G6*E7,F6+C7*D7) =F7/E7
7| 5 -600 110 =C8+E7 =IF(C8<0,G7*E8,F7+C8*D8) =F8/E8
8| 6 700 105 =C9+E8 =IF(C9<0,G8*E9,F8+C9*D9) =F9/E9
答案 1 :(得分：6)
Richard的模型子句查询存在问题。它在没有UNTIL子句的情况下进行了1000次迭代。经过四次迭代后，最终结果已经实现。接下来的996次迭代消耗CPU功率，但什么都不做。
在这里，您可以看到查询在使用当前数据集进行4次迭代后完成处理：
SQL> select order_id
2 , volume
3 , price
4 , total_vol
5 , total_costs
6 , unit_costs
7 from ( select order_id
8 , volume
9 , price
10 , volume total_vol
11 , 0.0 total_costs
12 , 0.0 unit_costs
13 , row_number() over (order by order_id) rn
14 from costs
15 order by order_id
16 )
17 model
18 dimension by (order_id)
19 measures (volume, price, total_vol, total_costs, unit_costs)
20 rules iterate (4)
21 ( total_vol[any] = volume[cv()] + nvl(total_vol[cv()-1],0.0)
22 , total_costs[any]
23 = case SIGN(volume[cv()])
24 when -1 then total_vol[cv()] * nvl(unit_costs[cv()-1],0.0)
25 else volume[cv()] * price[cv()] + nvl(total_costs[cv()-1],0.0)
26 end
27 , unit_costs[any] = total_costs[cv()] / total_vol[cv()]
28 )
29 order by order_id
30 /
ORDER_ID VOLUME PRICE TOTAL_VOL TOTAL_COSTS UNIT_COSTS
---------- ---------- ---------- ---------- ----------- ----------
1 1000 100 1000 100000 100
2 -500 110 500 50000 100
3 1500 80 2000 170000 85
4 -100 150 1900 161500 85
5 -600 110 1300 110500 85
6 700 105 2000 184000 92
6 rows selected.
它需要4次迭代而不是6次，因为使用了自动顺序，每次迭代都会尝试调整所有6行。
如果您使用与行数一样多的迭代并且每次迭代只调整一行，那么您的性能会更高。您也可以跳过子查询，然后最终查询变为：
SQL> select order_id
2 , volume
3 , price
4 , total_vol
5 , total_costs
6 , unit_costs
7 from costs
8 model
9 dimension by (row_number() over (order by order_id) rn)
10 measures (order_id, volume, price, type, 0 total_vol, 0 total_costs, 0 unit_costs)
11 rules iterate (1000) until (order_id[iteration_number+2] is null)
12 ( total_vol[iteration_number+1]
13 = nvl(total_vol[iteration_number],0) + volume[iteration_number+1]
14 , total_costs[iteration_number+1]
15 = case type[iteration_number+1]
16 when 'B' then volume[iteration_number+1] * price[iteration_number+1] + nvl(total_costs[iteration_number],0)
17 when 'S' then total_vol[iteration_number+1] * nvl(unit_costs[iteration_number],0)
18 end
19 , unit_costs[iteration_number+1]
20 = total_costs[iteration_number+1] / total_vol[iteration_number+1]
21 )
22 order by order_id
23 /
ORDER_ID VOLUME PRICE TOTAL_VOL TOTAL_COSTS UNIT_COSTS
---------- ---------- ---------- ---------- ----------- ----------
1 1000 100 1000 100000 100
2 -500 110 500 50000 100
3 1500 80 2000 170000 85
4 -100 150 1900 161500 85
5 -600 110 1300 110500 85
6 700 105 2000 184000 92
6 rows selected.
希望这有帮助。
的问候，
罗布。
EDIT 一些证据可以支持我的声明：
SQL> create procedure p1 (p_number_of_iterations in number)
2 is
3 begin
4 for x in 1 .. p_number_of_iterations
5 loop
6 for r in
7 ( select order_id
8 , volume
9 , price
10 , total_vol
11 , total_costs
12 , unit_costs
13 from ( select order_id
14 , volume
15 , price
16 , volume total_vol
17 , 0.0 total_costs
18 , 0.0 unit_costs
19 , row_number() over (order by order_id) rn
20 from costs
21 order by order_id
22 )
23 model
24 dimension by (order_id)
25 measures (volume, price, total_vol, total_costs, unit_costs)
26 rules iterate (4)
27 ( total_vol[any] = volume[cv()] + nvl(total_vol[cv()-1],0.0)
28 , total_costs[any]
29 = case SIGN(volume[cv()])
30 when -1 then total_vol[cv()] * nvl(unit_costs[cv()-1],0.0)
31 else volume[cv()] * price[cv()] + nvl(total_costs[cv()-1],0.0)
32 end
33 , unit_costs[any] = total_costs[cv()] / total_vol[cv()]
34 )
35 order by order_id
36 )
37 loop
38 null;
39 end loop;
40 end loop;
41 end p1;
42 /
Procedure created.
SQL> create procedure p2 (p_number_of_iterations in number)
2 is
3 begin
4 for x in 1 .. p_number_of_iterations
5 loop
6 for r in
7 ( select order_id
8 , volume
9 , price
10 , total_vol
11 , total_costs
12 , unit_costs
13 from costs
14 model
15 dimension by (row_number() over (order by order_id) rn)
16 measures (order_id, volume, price, type, 0 total_vol, 0 total_costs, 0 unit_costs)
17 rules iterate (1000) until (order_id[iteration_number+2] is null)
18 ( total_vol[iteration_number+1]
19 = nvl(total_vol[iteration_number],0) + volume[iteration_number+1]
20 , total_costs[iteration_number+1]
21 = case type[iteration_number+1]
22 when 'B' then volume[iteration_number+1] * price[iteration_number+1] + nvl(total_costs[iteration_number],0)
23 when 'S' then total_vol[iteration_number+1] * nvl(unit_costs[iteration_number],0)
24 end
25 , unit_costs[iteration_number+1]
26 = total_costs[iteration_number+1] / total_vol[iteration_number+1]
27 )
28 order by order_id
29 )
30 loop
31 null;
32 end loop;
33 end loop;
34 end p2;
35 /
Procedure created.
SQL> set timing on
SQL> exec p1(1000)
PL/SQL procedure successfully completed.
Elapsed: 00:00:01.32
SQL> exec p2(1000)
PL/SQL procedure successfully completed.
Elapsed: 00:00:00.45
SQL> exec p1(1000)
PL/SQL procedure successfully completed.
Elapsed: 00:00:01.28
SQL> exec p2(1000)
PL/SQL procedure successfully completed.
Elapsed: 00:00:00.43