Compounding using previous row values
Greetings!
Im trying to write SQL in Oracle 10.2.0.4. I have no success using OVER , PARTITION analytic functions. Can someone help with this requirement..
Data :
with t as (select 1 AS Calc_Seq, 700 as erncd, 0.05 as pct, 22 Hrly from dual union all
select 2, 701, 0.05 , 22 from dual union all
select 3, 702, 0.075 , 22 from dual)
SELECT T.Calc_Seq,T.erncd,T.pct,T.hrly FROM T;
Output required as follows ...
Need compounded rate field, as PCT rate of previous row "Compounded rate" field value.
CALC_SEQ ERNCD PCT HRLY Compounded rate
1 700 0.05 22 23.1
2 701 0.05 22 24.255
3 702 0.075 22 26.074125Thanks in advance.
Edited by: Rama on Jan 9, 2013 11:39 AM
Edited by: Rama on Jan 9, 2013 11:50 AM
Hi,
Try this:
WITH
t AS
SELECT 1 AS Calc_Seq ,700 AS erncd ,0.05 AS pct ,22 Hrly FROM dual UNION ALL
SELECT 2 ,701 ,0.05 ,22 FROM dual UNION ALL
SELECT 3 ,702 ,0.075 ,22 FROM dual
SELECT
FROM
T
model
dimension by (calc_seq)
measures (erncd, pct, hrly, 0 cr)
rules
cr[any] order by calc_seq = case when cv(calc_seq) = 1 then hrly[cv(calc_seq) ] * (1 +pct[cv(calc_seq)])
else cr[ cv() -1] * (1 + pct[cv()]) end
CALC_SEQ ERNCD PCT HRLY CR
1 700 0.05 22 23.1
2 701 0.05 22 24.255
3 702 0.075 22 26.074125 Regards,
Peter
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GO
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http://sqlblog.com/blogs/uri_dimant/
MS SQL optimization: MS SQL Development and Optimization
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Calculate using previous column and rows
Hello TNO members,
I have a complicated problem I need to solve, how ever I am missing knowledge about calculating using previous rows and columns in current select.
Test data
with t as (
select 1 as box, 1 as box_group, 10 as max_qty from dual union all
select 2, 1, 15 from dual union all
select 3, 1, 40 from dual union all
select 4, 1, 45 from dual union all
select 5, 2, 15 from dual union all
select 6, 2, 20 from dual union all
select 7, 2, 20 from dual union all
select 8, 3, 20 from dual)Expected Output result
box box_group max_qty assigned_from_60
1 1 10 10
2 1 15 15
3 1 40 17
4 1 45 18
5 2 15 0
6 2 20 0
7 2 20 0
8 3 20 0The problem:
In total 60 items are shared among the boxes in the same group, ordered by the lowest max_qty.
10 items can be assign to each box in group 1. (Used items: 40)
The remaining 20 items will be assigned to the other boxes in group 1.
5 more items can be assign to each box in group 1 (Used items: 15)
The remaining 15 items will be assigned to the remaining boxes in group 1.
2 more items can be assign to each box in group 1 (used items: 4)
One item remains. When items cannot be shared equally among the remaining boxes, ordered by the highest max_quantity, assign +1 till no item remains.
My solution in steps:
1. Calculate max_qty difference. How can I calculate the difference between the max_qty from box 1 and 2? Tricky is not to calculate the difference between different groups.
This means output result should be something like
box box_group max_qty qty_dif
1 1 10 10
2 1 15 5
3 1 40 25
4 1 45 5
5 2 15 15
6 2 20 5
7 2 20 0
8 3 20 202. Remaining boxes in the same group. I want to know how many boxes are in the same group. Especially the remaining boxes when the current max_quantity is filled.
Using the following code does not result in the correct output, what is wrong or missing here?
count(*) over(partition by box_group order by max_qty asc range between current row and unbounded following) This means output result should be something like
box box_group max_qty qty_dif rem_boxes
1 1 10 10 4
2 1 15 5 3
3 1 40 25 2
4 1 45 5 1
5 2 15 15 3
6 2 20 5 2
7 2 20 0 1
8 3 20 20 13. Calculate costs. This one is faily easy rem_boxes * qty_dif (*per row*)
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost
1 1 10 10 4 40
2 1 15 5 3 15
3 1 40 25 2 50
4 1 45 5 1 5
5 2 15 15 3 45
6 2 20 5 2 10
7 2 20 0 1 0
8 3 20 20 1 204. Calculate rem_items. 60 - (rem_boxes * qty_dif of box 1) - (rem_boxes * qty_dif of box 2) - (rem_boxes * qty_dif
of box n). How can I calculate using results of previous rows? (*all, not per group*)
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost rem_items
1 1 10 10 4 40 20
2 1 15 5 3 15 5
3 1 40 25 2 50 -45
4 1 45 5 1 5 -50
5 2 15 15 3 45 -95
6 2 20 5 2 10 -105
7 2 20 0 1 0 -105
8 3 20 20 1 20 -1255. Assign full quantity. For each row check if rem_items > 0 then 1 else 0
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost rem_items assign
1 1 10 10 4 40 20 1
2 1 15 5 3 15 5 1
3 1 40 25 2 50 -45 0
4 1 45 5 1 5 -50 0
5 2 15 15 3 45 -95 0
6 2 20 5 2 10 -105 0
7 2 20 0 1 0 -105 0
8 3 20 20 1 20 -125 06. Calculate assign quantity attemp 1. Calculate assign quantity of remaining boxes per group
When assign = 1 then max_qty else pervious a_qty (*within same group*)
This means output result should be something like
box box_group max_qty qty_dif rem_boxes cost rem_items assign a_qty
1 1 10 10 4 40 20 1 10
2 1 15 5 3 15 5 1 15
3 1 40 25 2 50 -45 0 15
4 1 45 5 1 5 -50 0 15
5 2 15 15 3 45 -95 0 0
6 2 20 5 2 10 -105 0 0
7 2 20 0 1 0 -105 0 0
8 3 20 20 1 20 -125 0 0How to solve the rest, I do not know yet. Any other suggestion to solve this problem, is welcome.
Since I'm not really a professional this is what I tried till now
with z as (
select 1 as box, 1 as box_group, 10 as max_qty from dual union all
select 2, 1, 15 from dual union all
select 3, 1, 40 from dual union all
select 4, 1, 45 from dual union all
select 5, 2, 15 from dual union all
select 6, 2, 20 from dual union all
select 7, 2, 20 from dual union all
select 8, 3, 20 from dual)
select u.*,
case
when u.assign = 2 then u.max_qty
when u.assign = 1 then 0
when u.assign = 0 then 0
end as assigned_qty
from
select v.*,
case
when 60 - sum(v.max_qty) over (order by v.box_group, v.max_qty, v.box) >= 0
and v.rem_items_before >= 0 then 2
when 60 - sum(v.max_qty) over (order by v.box_group, v.max_qty, v.box) < 0
and v.rem_items_before > 0 then 1 else 0
end as assign
from
select w.*,
w.rem_items_after + w.max_qty as rem_items_before
from
select x.*,
60 - x.qty_assigned as rem_items_after
from
select y.*,
y.max_qty * y.rem_boxes as total_cost,
sum(y.max_qty) over (order by y.box_group, y.max_qty, y.box) as qty_assigned
from
select z.*,
count(*) over (partition by z.box_group order by z.max_qty, z.box asc range between current row and unbounded following) as rem_boxes
from z
) y
) x
) w
) v
) uKind regards,
Metro
Edited by: 858378 on 30-mei-2011 4:39
Edited by: 858378 on 30-mei-2011 5:05Hi, Metro,
858378 wrote:
Hello, thanks for your help so far.
The course I am in, teaches you how to use the basic pl sql language such are selecting from tables.Is it about PL/SQL or SQL?
I have learned things about
- SELECT
- FROM
- WHERE
- GROUP BY
- ORDER BY
- SUB SELECTION IN SELECT
- SUB SELECTION IN FROM
- SUM, COUNT, MIN, MAX
- CASES
- INNER, OUTER, LEFT, FULL, CROSS JOINSAll of these are parts of the SQL language, not PL/SQL.
We are now at partitioning.Are you specifically at partitioning, or iare you at a point where the book talks about analytic functions, which sometimes, but not always, have a PARTTION BY clause?
It's written in Dutch, I tried to translate it, so it might not be well written English.
Excercise 192
Distribution center 'One by one' wants to automate the distribution process.
Items are distributed to the boxes one by one. Sorry, I can't figure out what Exercise 192 is, based on just that.
It is similar to the next excercise, but this above was a lot more straight to the point.
*Excercise 193*
Distribution center 'All in one box' wants to automate the distribution process.
One of the major changes in this process is to distribute items equally to all boxes in the same group.
This means starting from the lowest quantity, assign the lowest quantity to all boxes in the same group if possible.
If this is not possible distribute the amount of items divided by the number of boxes in the same group. If the amount of items per box is lower than 1 and not 0.
Divide the remaining items per box ordered by the box with the highest quantity, till there are no items left.
When it is possible to distribute the lowest quantity to all boxes, move up to the next box in the same group.
When all boxes in the same group are filled to their maximum quantity, move up to the next group and repeat this process. ...
So Exercise 193 is what you asked yesterday, right?
A) Order the following information by box_group and max_qty as described.
B) Calculate the distribution results for 60, 120, 170 items.
When I have 60 items, the following output result should be
box box_group max_qty assigned_from_60
1 1 10 10
2 1 15 15
3 1 40 17
4 1 45 18
5 2 15 0
6 2 20 0
7 2 20 0
8 3 20 0Box 1 and 2 can be filled completely as you can fill atleast 10 to box 1,2,3,4 and an additional 5 to box 2, 3 and 4.
The last item goes to the box 4.
when I have 120 items
box box_group max_qty assigned_from_120
1 1 10 10
2 1 15 15
3 1 40 40
4 1 45 45
5 2 15 3
6 2 20 3
7 2 20 4
8 3 20 0
Based on what you posted, it seems like the following should be equally acceptable:
box box_group max_qty assigned_from_120
1 1 10 10
2 1 15 15
3 1 40 40
4 1 45 45
5 2 15 3
6 2 20 4
7 2 20 3
8 3 20 0That is, the last 10 items have to be distributed among the 3 boxes in box_group=2 as equally as possible. So one box will get 4 items and the others will get 3. The extra item will go to the box with the highest max_qty, but in this case, there is a tie: box 6 has just as much of a claim to having the highest max_qty as box 7. The line marked "***** Add if needed *****" in the query blow guarantees that, in case of a tie like this, the box with the higher box value will be considered "larger" than another box with the same max_qty.
when I have 170 items
box box_group max_qty assigned_from_170
1 1 10 10
2 1 15 15
3 1 40 40
4 1 45 45
5 2 15 15
6 2 20 20
7 2 20 20
8 3 20 5
I accidentally posted the wrong query yesterday. This is what I should have posted:
WITH cntr AS
SELECT LEVEL AS n
FROM ( SELECT MAX (max_qty) AS max_max_qty
FROM z
-- WHERE box_group = :target_box_group -- ***** Removed *****
CONNECT BY LEVEL <= max_max_qty
, got_r_num AS
SELECT z.box
, c.n
, ROW_NUMBER () OVER ( ORDER BY box_group -- ***** Added *****
, c.n
, z.max_qty DESC
, box DESC -- ***** Add if needed *****
) AS r_num
FROM cntr c
JOIN z ON c.n <= z.max_qty
-- WHERE z.box_group = :target_box_group -- ***** Removed *****
, got_assigned AS
SELECT box
, COUNT (*) AS assigned
FROM got_r_num
WHERE r_num <= :total_items
GROUP BY box
SELECT z.*, NVL (a.assigned, 0) AS assigned
FROM z
LEFT OUTER JOIN got_assigned a ON z.box = a.box
ORDER BY z.box_group
, z.max_qty
;Yesterday, I described how you need to remove 2 lines and add 1, but the code I posted was the unchanged query. The query above is what I should have posted then. Look for comments beginning "*****" above for the changes. I apologize for my mistake.
This query gets the results you posted for all 3 values of :total_items that you posted. If it doesn't work for some other value, or some other data, post the new values and the correct results you want from them, and point out where the query above is wrong.
C) Sum the maximum quantities per group
D) Get the amount of boxes in each group
E) Create a column for remaining boxes per group
F) For each distinct quantity, count the amount of boxes with this quantity
G) Calculate how many items are required to fill all boxes in the same group
H) Create a plan how to solve this distribution problem described in the introduction?
I) Solve this problem using your plan.Are these the steps that the book suggests using?
I don't understand that approach. It might be a good way to solve the problem without using a computer. It might be a good way to solve the problem using a procedural language, such as PL/SQL. It might be one way of solving the problem in SQL, but I think it will be more complicated and less efficient than what I psoted.
The approach above is iterative; that is, you repeat certain steps, with different values. For example, you distribute a certain number of items to all boxes in a box_group. The you remove the smallest box(es) from the group, and repeat, distributing the remaining items among the remianing boxes. That's not hard to do in a language like PL/SQL, where you have loops and variables. In SQL, the closest thing to that is the MODEL clause. I'm sure you could write a MODEL solution to this problem, but, if your book hasn't mentioned MODEL yet, then that's certainly not what it's expecting you to do.
Even using the approach in steps A) trhough G) above, I don't see how a PARTITION BY would help. -
Derived logical column using Row values in OBIEE11g
Hi Experts
I have a requirement whereby I need to create say 100 new logical column in obiee using the row values of one dimension.
For eg. say I have dimension values like A and B in one of the table , I need to create logical columns like C=A+B and D = A/B and E = AxB as such when the user drags this to the analysis the calculation should happen automatically.
The problem is they are in the rows and not the coulmn , so I can't do plain calculation like sales per Unit = Sales/Units - as Sales and Units are two differnet measure column.
How to achieve this in RPD.
RegdsHi
Pls. find below the examples :
You have a dimesion called KPI which has soem KPIs.
KPI Measure Values
KPI1 10
KPI2 20
KPI3 50
etc
Now I need to calulate a driven KPI i.e KPI10 which is KPI10 = KPI2/KPI3 , and I need to do that in RPD.
I know I can do it in Front end analaytics by using Pivot or Selection steps . But I want to do it in RPD.
Regds -
Using the Previous Row in a Calculator Formula
Hello. I am using the calculator to create a new channel that is a custom filtered version of an existing channel. Unfortunately, I have not been able to identify the previous row entry in the channel. To make things easier, let's say my filter is a simple averager. Here is simple pseudocode for what I want to do:
New_Channel = (Existing_Channel[n] + Existing_Channel[n-1]) / 2
where n is the current row.
This is the calculator formula I have come up with:
ch("[1]/New_Channel") = (CHD(chnRow - 1, "[1]/Existing_Channel") + ch("[1]/Existing_Channel")) / 2 + CTNV(chnRow > 1)
Unfortunately, I think chnRow only returns 0.
Obviously, I am a beginner. I am open to other approaches to creating a new channel with a custom filter.
Thanks!
Solved!
Go to Solution.Hi jbuttron,
What you need to do is to copy the channel and delete the first value from the copied channel-- then the Nth row of the original channel will line up with the (N-1)th row of the copied channel. You should also add the last value of the copied channel to the end of that channel as a new value so both channels end up with the same channel length. Now you can reference the channels with Ch("[1]/old") and Ch("[1]/new") in the channel calculator expression, assuming that the channel names are "old" and "new" respectively and that both are in the first group. You don't need a row variable in the expression now, which is good because there's no way to iterate a row variable in a channel calculator expression. The row iteration is implicit in the channel referencing Ch("[1]/new").
Brad Turpin
DIAdem Product Support Engineer
National Instruments
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