oracle sql - numbering group of rows - sql

Question

i have the following table with different prices in every week and need a numbering like in the last column. consecutive rows with same prices should have the same number like in weeks 11/12 or 18/19. but on the other side weeks 2 and 16 have the same prices but are not consecutive so they should get a different number.
w | price | r1 | need
===========================
1 167,93 1 1
2 180 1 2
3 164,72 1 3
4 147,42 1 4
5 133,46 1 5
6 145,43 1 6
7 147 1 7
8 147,57 1 8
9 150,95 1 9
10 158,14 1 10
11 170 1 11
12 170 2 11
13 166,59 1 12
14 161,06 1 13
15 162,88 1 14
16 180 2 15
17 183,15 1 16
18 195 1 17
19 195 2 17
i have already experimented with the analytics functions (row_number, rank, dens_rank), but didn't found a solution for this problem so far.
(oracle sql 10,11)
does anyone have a hint? thanks.

Answer 1

Simulating your table first:
SQL> create table mytable (w,price,r1)
2 as
3 select 1 , 167.93, 1 from dual union all
4 select 2 , 180 , 1 from dual union all
5 select 3 , 164.72, 1 from dual union all
6 select 4 , 147.42, 1 from dual union all
7 select 5 , 133.46, 1 from dual union all
8 select 6 , 145.43, 1 from dual union all
9 select 7 , 147 , 1 from dual union all
10 select 8 , 147.57, 1 from dual union all
11 select 9 , 150.95, 1 from dual union all
12 select 10, 158.14, 1 from dual union all
13 select 11, 170 , 1 from dual union all
14 select 12, 170 , 2 from dual union all
15 select 13, 166.59, 1 from dual union all
16 select 14, 161.06, 1 from dual union all
17 select 15, 162.88, 1 from dual union all
18 select 16, 180 , 2 from dual union all
19 select 17, 183.15, 1 from dual union all
20 select 18, 195 , 1 from dual union all
21 select 19, 195 , 2 from dual
22 /
Table created.
Your need column is calculated in two parts: first compute a delta column which denotes whether the previous price-column differs from the current rows price column. If you have that delta column, the second part is easy by computing the sum of those deltas.
SQL> with x as
2 ( select w
3 , price
4 , r1
5 , case lag(price,1,-1) over (order by w)
6 when price then 0
7 else 1
8 end delta
9 from mytable
10 )
11 select w
12 , price
13 , r1
14 , sum(delta) over (order by w) need
15 from x
16 /
W PRICE R1 NEED
---------- ---------- ---------- ----------
1 167.93 1 1
2 180 1 2
3 164.72 1 3
4 147.42 1 4
5 133.46 1 5
6 145.43 1 6
7 147 1 7
8 147.57 1 8
9 150.95 1 9
10 158.14 1 10
11 170 1 11
12 170 2 11
13 166.59 1 12
14 161.06 1 13
15 162.88 1 14
16 180 2 15
17 183.15 1 16
18 195 1 17
19 195 2 17
19 rows selected.

Answer 2

You can nest your analytic functions using inline views, so you first group the consecutive weeks with same prices and then dense_rank using those groups:
select w
, price
, r1
, dense_rank() over (
order by first_w_same_price
) drank
from (
select w
, price
, r1
, last_value(w_start_same_price) ignore nulls over (
order by w
rows between unbounded preceding and current row
) first_w_same_price
from (
select w
, price
, r1
, case lag(price) over (order by w)
when price then null
else w
end w_start_same_price
from your_table
)
)
order by w
The innermost inline view with LAG function lets the starting week of every consecutive group get it's own week number, but every consecutive week with same price gets null (weeks 12 and 19 in your data.)
The middle inline view with LAST_VALUE function then use the IGNORE NULLS feature to give the consecutive weeks the same value as the first week within each group. So week 11 and 12 both gets 11 in first_w_same_price and week 18 and 19 both gets 18 in first_w_same_price.
And finally the outer query use DENSE_RANK to give the desired result.

Answer 3

For each row you should count previous rows where (w-1) row price isn't the same as (w) price:
select T1.*,
(SELECT count(*)
FROM T T2
JOIN T T3 ON T2.w-1=T3.w
WHERE T2.Price<>T3.Price
AND T2.W<=T1.W)+1 rn
from t T1
SQLFiddle demo

Answer 4

Try this:
with tt as (
select t.*, decode(lag(price) over(order by w) - price, 0, 1, 0) diff
from t
)
select w
, price
, r1
, row_number() over (order by w) - sum(diff) over(order by w rows between UNBOUNDED PRECEDING and current row) need
from tt

Answer 5

SELECT w, price, r1,
ROW_NUMBER () OVER (PARTITION BY price ORDER BY price) row_column
FROM TABLE