How to get the rows of the longest consecutive same value?
Table Learning:
rowID
values
1
1
2
1
3
0
4
0
5
0
6
1
7
0
8
1
9
1
10
1
Longest consecutive value is 1 (rowID 8-10 as rowID 1-2 is 2 and rowID 6-6 is 1). How to query to get the actual rows of consecutive values (not just rowStart and rowEnd values) like :
rowID
values
8
1
9
1
10
1
And for longest consecutive values of both 1 and 0?
DB Fiddle
I think that the simplest approach is to use a window count to define the islands. Then to get the "longest" island, we just need to aggregate, sort and limit:
select min(valueid) grp_start, max(valueid) grp_end
from (select t.*, sum(value = 0) over(order by valueid) grp from testing t) t
where value = 1
group by grp
order by count(*) desc limit 1
In the DB Fiddle that you provided, the query returns:
grp_start
grp_end
8
10
This is a gaps and islands problem, and one approach is to use the difference in row numbers method:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY rowID) rn1,
ROW_NUMBER() OVER (PARTITION BY values ORDER BY rowID) rn2
FROM yourTable
),
cte2 AS (
SELECT *,
MIN(rowID) OVER (PARTITION BY values, rn1 - rn2) AS minRowID,
MAX(rowID) OVER (PARTITION BY values, rn1 - rn2) AS maxRowID
FROM cte1
),
cte3 AS (
SELECT *, RANK() OVER (PARTITION BY values ORDER BY maxRowID - minRowID DESC) rnk
FROM cte2
)
SELECT rowID, values
FROM cte3
WHERE rnk = 1
ORDER BY values, rowID;
Related
Is it possible to rank item by partition without use CTE method
Expected Table
item
value
ID
A
10
1
A
20
1
B
30
2
B
40
2
C
50
3
C
60
3
A
70
4
A
80
4
By giving id to the partition to allow agitated function to work the way I want.
item
MIN
MAX
ID
A
10
20
1
B
30
40
2
C
50
60
3
A
70
80
4
SQL Version: Microsoft SQL Sever 2017
Assuming that the value column provides the intended ordering of the records which we see in your question above, we can try using the difference in row numbers method here. Your problem is a type of gaps and islands problem.
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY value) rn1,
ROW_NUMBER() OVER (PARTITION BY item ORDER BY value) rn2
FROM yourTable
)
SELECT item, MIN(value) AS [MIN], MAX(value) AS [MAX], MIN(ID) AS ID
FROM cte
GROUP BY item, rn1 - rn2
ORDER BY MIN(value);
Demo
If you don't want to use a CTE here, for whatever reason, you may simply inline the SQL code in the CTE into the bottom query, as a subquery:
SELECT item, MIN(value) AS [MIN], MAX(value) AS [MAX], MIN(ID) AS ID
FROM
(
SELECT *, ROW_NUMBER() OVER (ORDER BY value) rn1,
ROW_NUMBER() OVER (PARTITION BY item ORDER BY value) rn2
FROM yourTable
) t
GROUP BY item, rn1 - rn2
ORDER BY MIN(value);
You can generate group IDs by analyzing the previous row item value that could be obtained with the LAG function and finally use GROUP BY to get the minimum and maximum value in item groups.
SELECT
item,
MIN(value) AS "min",
MAX(value) AS "max",
group_id + 1 AS id
FROM (
SELECT
*,
SUM(CASE WHEN item = prev_item THEN 0 ELSE 1 END) OVER (ORDER BY value) AS group_id
FROM (
SELECT
*,
LAG(item, 1, item) OVER (ORDER BY value) AS prev_item
FROM t
) items
) groups
GROUP BY item, group_id
Query produces output
item
min
max
id
A
10
20
1
B
30
40
2
C
50
60
3
A
70
80
4
You can check a working demo here
If I had a SQL Server query that returns numbers in order like this
1
2
3
5
6
7
9
10
11
how can I remove numbers such that no two adjacent pairs are consecutive by 1? The above should be returned like
3
5
7
9
Is this possible to do?
We can use LEAD and LAG here:
WITH cte AS (
SELECT id, LAG(id) OVER (ORDER BY id) lag_id, LEAD(id) OVER (ORDER BY id) lead_id
FROM yourTable
)
SELECT id
FROM cte
WHERE lag_id <> id - 1 OR lead_id <> id + 1
ORDER BY id;
You can try to use LEAD and LAG window functions and calculation what rows are consecutive by 1.
SELECT Val
FROM (
SELECT *,
LEAD(Val) OVER(ORDER BY Val) - Val gap1,
Val - LAG(Val) OVER(ORDER BY Val) gap2
FROM T
) t1
WHERE gap1 > 1 OR gap2 > 1
The source table:
id num
-------------------
1 1
2 1
3 1
4 2
5 2
6 1
The output:(appear at least 2 times)
num times
--------------
1 3
2 2
Based on the addition logic defined in the comments it appears this is what you're after:
WITH YourTable AS(
SELECT V.id,
V.num
FROM (VALUES(1,1),
(2,1),
(3,1),
(4,2),
(5,2),
(6,1),
(7,1))V(id,num)), --Added extra row due to logic defined in comments
Grps AS(
SELECT YT.id,
YT.num,
ROW_NUMBER() OVER (ORDER BY id) -
ROW_NUMBER() OVER (PARTITION BY Num ORDER BY id) AS Grp
FROM YourTable YT),
Counts AS(
SELECT num,
COUNT(num) AS Times
FROM grps
GROUP BY grp,
num)
SELECT num,
MAX(times) AS times
FROM Counts
GROUP BY num;
This uses a CTE and ROW_NUMBER to define the groups, and then an additional CTE to get the COUNT per group. Finally you can then get the MAX COUNT per num.
I would adress this with a gaps-and-islands technique:
select num, max(cnt)
from (
select num, count(*) cnt
from (
select
id,
num,
row_number() over(order by id) rn1,
row_number() over(partition by num order by id) rn2
from mytable
) t
group by num, rn1 - rn2
) t
group by num
The most inner query computes row numbers over the whole table and within num groups; the difference between the row numbers gives you the group of adjacent records that each record belong to (you can run that subquery independently and follow how the difference evolves to understand more).
Then, the next level count the number of records in each group of adjacent records. The most outer query takes the maximum count of adjacent records in for each num.
Demo on DB Fiddle:
num | (No column name)
--: | ---------------:
1 | 3
2 | 2
this will work for you
select num,count(num) times from Tabl
group by num
I'm having problem with getting only TOP 2 values for each group (groups are in column).
Example :
ID Group Value
1 A 30
2 A 150
3 A 40
4 A 70
5 B 0
6 B 100
7 B 90
I expect my output to be
ID Group Value
1 A 150
2 A 70
3 B 100
4 B 90
Simply, for each group I want just 2 rows with the highest Value
Most databases support the ANSI standard row_number() function. You would use it as:
select group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
To set the id you can use row_number() in the outer query:
select row_number() over (order by group, value) as id,
group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
However, changing the id seems suspicious.
You can use CTE with rank function ROW_NUMBER() .
Here is query to get your result.
;WITH cte AS
( SELECT Group, value,
ROW_NUMBER() OVER (PARTITION BY Group ORDER BY value DESC) AS rn
FROM test
)
SELECT Group, value FROM cte
WHERE rn <= 2
ORDER BY value
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