Displays data divided into columns - sql

I have a list of data and I want to display them no matter the order but by a particular order of columns for example 3
id
1
2
3
4
5
6
7
8
9
10
11
12
13
14
id1 id2 id3
1 6 11
2 7 12
3 8 13
4 9 14
5 10 NULL
Final result

This is tricky. One approach uses two calls to ROW_NUMBER, one for the row position and the other for the column position. The logic here is that each row belongs to a group which rolls over as the id increases by multiples of 5, while the column is determined by how many multiples of 5 the id value represents.
WITH cte AS (
SELECT id,
(ROW_NUMBER() OVER (ORDER BY id) - 1) / 5 AS rn1,
(ROW_NUMBER() OVER (ORDER BY id) - 1) % 5 AS rn2
FROM yourTable
)
SELECT
rn2,
MAX(CASE WHEN rn1 = 0 THEN id END) AS id1,
MAX(CASE WHEN rn1 = 1 THEN id END) AS id2,
MAX(CASE WHEN rn1 = 2 THEN id END) AS id3
FROM cte
GROUP BY
rn2
ORDER BY
rn2;
Demo

You can use window functions and conditional aggregation:
select max(case when seqnum % 3 = 1 then id end),
max(case when seqnum % 3 = 2 then id end),
max(case when seqnum % 3 = 0 then id end)
from (select t.*,
row_number() over (order by (select null)) as seqnum
from t
) t
group by (seqnum - 1) / 3

Related

Can I start a new group when value changes from 0 to 1?

Can I somehow assign a new group to a row when a value in a column changes in T-SQL?
I would be grateful if you can provide solution that will work on unlimited repeating numbers without CTE and functions. I made a solution that work in sutuation with 100 consecutive identical numbers(with
coalesce(lag()over(), lag() over(), lag() over() ) - it is too bulky
but can not make a solution for a case with unlimited number of consecutive identical numbers.
Data
id somevalue
1 0
2 1
3 1
4 0
5 0
6 1
7 1
8 1
9 0
10 0
11 1
12 0
13 1
14 1
15 0
16 0
Expected
id somevalue group
1 0 1
2 1 2
3 1 2
4 0 3
5 0 3
6 1 4
7 1 4
8 1 4
9 0 5
10 0 5
11 1 6
12 0 7
13 1 8
14 1 8
15 0 9
16 0 9
If you just want a group identifier, you can use:
select t.*,
min(id) over (partition by some_value, seqnum - seqnum_1) as grp
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by somevalue order by id) as sequm_1
from t
) t;
If you want them enumerated . . . well, you can enumerate the id above using dense_rank(). Or you can use lag() and a cumulative sum:
select t.*,
sum(case when some_value = prev_sv then 0 else 1 end) over (order by id) as grp
from (select t.*,
lag(somevalue) over (order by id) as prev_sv
from t
) t;
Here's a different approach:
First I created a view to provide the group increment on each row:
create view increments as
select
n2.id,n2.somevalue,
case when n1.somevalue=n2.somevalue then 0 else 1 end as increment
from
(select 0 as id,1 as somevalue union all select * from mytable) n1
join mytable n2
on n2.id = n1.id+1
Then I used this view to produce the group values as cumulative sums of the increments:
select id, somevalue,
(select sum(increment) from increments i1 where i1.id <= i2.id)
from increments i2

How to Generate Row number Partition by two column match in sql

Tbl1
---------------------------------------------------------
Id Date Qty ReOrder
---------------------------------------------------------
1 1-1-18 1 3
2 2-1-18 0 3
3 3-1-18 2 3
4 4-1-18 3< >3
5 5-1-18 2 3
6 6-1-18 0 3
7 7-1-18 1 3
8 8-1-18 0 3
---------------------------------------------------------
I want the result like below
---------------------------------------------------------
Id Date Qty ReOrder
---------------------------------------------------------
1 1-1-18 1 3
5 5-1-18 2 3
---------------------------------------------------------
if ReOrder not same with Qty then date will be same upto after reorder=Qty
You can use cumulative approach with row_number() function :
select top (1) with ties *
from (select *, max(case when qty = reorder then 'v' end) over (order by id desc) grp
from table
) t
order by row_number() over(partition by grp order by id);
Unfortunately this will require SQL Server, But you can also do:
select *
from (select *, row_number() over(partition by grp order by id) seq
from (select *, max(case when qty = reorder then 'v' end) over (order by id desc) grp
from table
) t
) t
where seq = 1;

COUNT() OVER possible using DISTINCT and WINDOWING IN HIVE

I want to calculate the number of distinct port numbers that exist between the current row and the X previous rows (sliding window), where x can be any integer number.
For instance,
If the input is:
ID PORT
1 21
2 22
3 23
4 25
5 25
6 21
The output should be:
ID PORT COUNT
1 21 1
2 22 2
3 23 3
4 25 4
5 25 4
6 21 4
I am using Hive, over RapidMiner and I have tried the following:
select id, port,
count (*) over (partition by srcport order by id rows between 5 preceding and current row)
This must work for big data and when X is big integer number.
Any feedback would be appreciated.
I don't think there is an easy way. One method uses lag():
select ( (case when port_5 is not null then 1 else 0 end) +
(case when port_4 is not null and port_4 not in (port_5) then 1 else 0 end) +
(case when port_3 is not null and port_3 not in (port_5, port_4) then 1 else 0 end) +
(case when port_2 is not null and port_2 not in (port_5, port_4, port_3) then 1 else 0 end) +
(case when port_1 is not null and port_1 not in (port_5, port_4, port_3, port_2) then 1 else 0 end) +
(case when port is not null and port not in (port_5, port_4, port_3, port_2, port_2) then 1 else 0 end)
) as cumulative_distinct_count
from (select t.*,
lag(port, 5) over (partition by srcport order by id rows) as port_5,
lag(port, 4) over (partition by srcport order by id rows) as port_4,
lag(port, 3) over (partition by srcport order by id rows) as port_3,
lag(port, 2) over (partition by srcport order by id rows) as port_2,
lag(port, 1) over (partition by srcport order by id rows) as port_1
from t
) t
This is a complicated query, but the performance should be ok.
Note: port and srcport I assume are the same thing, but this borrows from your query.
One way to do it is with a self join as distinct isn't supported in window functions.
select t1.id,count(distinct t2.port) as cnt
from tbl t1
join tbl t2 on t1.id-t2.id>=0 and t1.id-t2.id<=5 --change this number per requirements
group by t1.id
order by t1.id
This assumes id's are in sequential order.
If not, first get the row numbers and use the logic from above. It would be like
with rownums as (select id,port,row_number() over(order by id) as rnum
from tbl)
select r1.id,count(distinct r2.port)
from rownums r1
join rownums r2 on r1.rnum-r2.rnum>=0 and r1.rnum-r2.rnum<=5
group by r1.id

Count consecutive duplicate values in SQL

I have a table like so
ID OrdID Value
1 1 0
2 2 0
3 1 1
4 2 1
5 1 1
6 2 0
7 1 0
8 2 0
9 2 1
10 1 0
11 2 0
I want to get the count of consecutive value where the value is 0. Using the example above the result will be 3 (Rows 6, 7 and 8). I am using sql server 2008 r2.
I am going to presume that id is unique and increasing. You can get counts of consecutive values by using the different of row numbers. The following counts all sequences:
select grp, value, min(id), max(id), count(*) as cnt
from (select t.*,
(row_number() over (order by id) - row_number() over (partition by value order by id)
) as grp
from table t
) t
group by grp, value;
If you want the longest sequence of 0s:
select top 1 grp, value, min(id), max(id), count(*) as cnt
from (select t.*,
(row_number() over (order by id) - row_number() over (partition by value order by id)
) as grp
from table t
) t
group by grp, value
having value = 0
order by count(*) desc
A query using not exists to find consecutive 0s
select top 1 min(t2.id), max(t2.id), count(*)
from mytable t
join mytable t2 on t2.id <= t.id
where not exists (
select 1 from mytable t3
where t3.id between t2.id and t.id
and t3.value <> 0
)
group by t.id
order by count(*) desc
http://sqlfiddle.com/#!3/52989/3

Aggregate within a group of unchanged values

I have sample data:
RowId TypeId Value
1 1 34
2 1 53
3 1 34
4 2 43
5 2 65
6 16 54
7 16 34
8 1 45
9 6 43
10 6 34
11 16 64
12 16 63
I want to count row for each type (The Value does not matter to me), but only for... neighbor TypeId
TypeId Count
1 3
2 2
16 2
1 1
6 2
16 2
How to achieve this result?
This should give you COUNT of rows within a group of unchanged values:
SELECT TypeId, grp, COUNT(*) FROM (
SELECT RowId, TypeId , Value, gap, SUM(gap) over (ORDER BY RowId ) grp
FROM (SELECT RowId, TypeId , Value,
CASE WHEN TypeId = lag(TypeId) over (ORDER BY RowId )
THEN 0
ELSE 1
END gap
FROM dummy
) t
) tt
GROUP BY TypeId, grp;
If you prefer WITH over endless sub-query inclusions:
WITH dummy_with_groups AS (
SELECT RowId, TypeId , Value, SUM(gap) OVER (ORDER BY RowId) grp
FROM (SELECT RowId, TypeId , Value,
CASE WHEN TypeId = lag(TypeId) OVER (ORDER BY RowId)
THEN 0 ELSE 1 END gap
FROM dummy) t
)
SELECT TypeId, COUNT(*) as Result
FROM dummy_with_groups
GROUP BY TypeId, grp;
http://www.sqlfiddle.com/#!6/f16e9/34
Check this fiddle demo. I have renamed your columns a little.
WITH myCTE AS
(SELECT row_id,
type_id,
ROW_NUMBER () OVER (PARTITION BY type_id ORDER BY row_id)
AS cnt,
CASE LEAD (type_id) OVER (ORDER BY row_id)
WHEN type_id THEN 0
ELSE 1
END
AS show
FROM dummy),
innerQuery AS
(SELECT row_id, type_id, cnt
FROM myCTE
WHERE show = 1)
SELECT iq1.type_id, iq1.cnt - ISNULL (iq2.cnt, 0) CNT
FROM innerQuery iq1
LEFT OUTER JOIN innerQuery iq2
ON iq1.type_id = iq2.type_id
AND EXISTS
(SELECT 1
FROM innerQuery iq3
WHERE iq3.type_id = iq1.type_id
AND iq3.row_id < iq1.row_id
HAVING MAX (iq3.row_id) = iq2.row_id)
The output is exactly as expected.