I have the following table:
Name Field
A 1
B 1
C 1
D 1
E 1
F 1
G 1
H 2
I 2
J 2
K 3
L 3
M 3
N 3
O 3
P 3
Q 3
R 3
S 3
T 3
I need a SQL query which will generate me a set with 5 random rows for each distinct value on column Field.
For example, results expected:
Name Field
A 1
B 1
D 1
E 1
G 1
J 2
I 2
H 2
M 3
Q 3
T 3
S 3
P 3
Is there an easy way to do this? Or should i split that table into more tables and generate random for each table then union them?
You can do this with a CTE using a ROW_NUMBER() whilst PARTITIONing on the Field:
;With Cte As
(
Select Name, Field,
Row_Number() Over (Partition By Field Order By NewId()) RN
From YourTable
)
Select Name, Field
From Cte
Where RN <= 5
SQL Fiddle
You can readily do this with row_number():
select name, field
from (select t.*,
row_number() over (partition by field order by newid()) as seqnum
from t
) t
where seqnum <= 5;
An enhancement to Gordon Linoff's code, This code really helped me if you need criteria in your query.
select *
from (select t.*,
row_number() over (partition by region order by newid()) as seqnum
from MyTable t
WHERE t.program = 'ACME'
) t
where seqnum <= 1500;
Related
I have a table like this:
D
S
2
1
2
3
4
2
4
3
4
5
6
1
in which the code of symptoms(S) of three diseases(D) are shown. I want to rearrange this table (D-S) such that the diseases with more symptoms come up i.e. order it by decreasing the numbers of symptoms as below:
D
S
4
2
4
3
4
5
2
1
2
3
6
1
Can anyone help me to write a SQL code for it in SQL server?
I had tried to do this as the following but this doesn't work:
SELECT *
FROM (
select D, Count(S) cnt
from [D-S]
group by D
) Q
order by Q.cnt desc
select
D,
S
from
D-S
order by
count(*) over(partition by D) desc,
D,
S;
Two easy ways to approach this:
--==== Sample Data
DECLARE #t TABLE (D INT, S INT);
INSERT #t VALUES(2,1),(2,3),(4,2),(4,3),(4,5),(6,1);
--==== Using Window Function
SELECT t.D, t.S
FROM (SELECT t.*, Rnk = COUNT(*) OVER (PARTITION BY t.D) FROM #t AS t) AS t
ORDER BY t.Rnk DESC;
--==== Using standard GROUP BY
SELECT t.*
FROM #t AS t
JOIN
(
SELECT t2.D, Cnt = COUNT(*)
FROM #t AS t2
GROUP BY t2.D
) AS t2 ON t.D = t2.D
ORDER BY t2.Cnt DESC;
Results:
D S
----------- -----------
4 2
4 3
4 5
2 1
2 3
6 1
I am using PostgreSQL.
I need stored procedure using PLPGSQL language that will return table (SET OF RECORDS) containing count of top 2 and bottom 2 results from my_table.
For example:
my_table
id value
1 a
2 a
3 a
4 b
5 b
6 c
7 c
8 e
9 f
10 g
11 g
12 g
13 g
14 h
15 h
Returns:
count value
4 g
3 a
1 e
1 f
Thank you
You can use window functions with aggration
select v.value, v.cnt
from (select value, count(*) as cnt,
row_number() over (order by count(*) desc) as seqnum_desc,
row_number() over (order by count(*) asc) as seqnum_asc
from t
group by value
) v
where seqnum_desc <= 2 or seqnum_asc <= 2;
Note: In the case of ties -- particularly likely at the bottom end -- this returns arbitrary values with the same count. You can adjust for this using rank() or dense_rank(), depending on what you want in this case.
I have a set of numbers like this
ID
===
1
2
3
1
2
1
1
2
3
4
5
...
I want to select a new row that increase when fetch next 1 like this
ID number
=== ========
1 1
2 1
3 1
1 2
2 2
1 3
1 4
2 4
3 4
4 4
5 4
Any suggestion ?
Assuming that you have a column o which specify the ordering then you can use a self-join like this:
select d1.o, d1.id, count(*)
from data d1
join data d2 on d1.o >= d2.o and d2.id = 1
group by d1.o, d1.id
DBFiddle DEMO
You can solve this with use of cte and window functions, as follows:
DECLARE #t TABLE (ID INT);
INSERT INTO #t VALUES (1),(2),(3),(1),(2),(1),(1),(2),(3),(4),(5);
WITH cte AS(
SELECT ID, ROW_NUMBER() OVER (ORDER BY (SELECT 1)) rn
FROM #t
),
cte1 AS(
SELECT ID, rn, ROW_NUMBER() OVER (ORDER BY rn) rn2
FROM cte
WHERE ID = 1
)
SELECT c.ID, MAX(rn2) OVER (ORDER BY c.rn) rn
FROM cte c
LEFT JOIN cte1 c1 ON c1.rn = c.rn
ORDER BY c.rn
I have following table
ID Name Stage
1 A 1
1 B 2
1 C 3
1 A 4
1 N 5
1 B 6
1 J 7
1 C 8
1 D 9
1 E 10
I need output as below with parameters A and N need to select closest rows where difference between stage is smallest
ID Name Stage
1 A 4
1 N 5
I need to select rows where difference between stage is smallest
This query can make use of an index on (name, stage) efficiently:
WITH cte AS (
SELECT TOP 1
a.id AS a_id, a.name AS a_name, a.stage AS a_stage
, n.id AS n_id, n.name AS n_name, n.stage AS n_stage
FROM tbl a
CROSS APPLY (
SELECT TOP 1 *, stage - a.stage AS diff
FROM tbl
WHERE name = 'N'
AND stage >= a.stage
ORDER BY stage
UNION ALL
SELECT TOP 1 *, a.stage - stage AS diff
FROM tbl
WHERE name = 'N'
AND stage < a.stage
ORDER BY stage DESC
) n
WHERE a.name = 'A'
ORDER BY diff
)
SELECT a_id AS id, a_name AS name, a_stage AS stage FROM cte
UNION ALL
SELECT n_id, n_name, n_stage FROM cte;
SQL Server uses CROSS APPLY in place of standard-SQL LATERAL.
In case of ties (equal difference) the winner is arbitrary, unless you add more ORDER BY expressions as tiebreaker.
dbfiddle here
This solution works, if u know the minimum difference is always 1
SELECT *
FROM myTable as a
CROSS JOIN myTable as b
where a.stage-b.stage=1;
a.ID a.Name a.Stage b.ID b.Name b.Stage
1 A 4 1 N 5
Or simpler if u don't know the minimum
SELECT *
FROM myTable as a
CROSS JOIN myTable as b
where a.stage-b.stage in (SELECT min (a.stage-b.stage)
FROM myTable as a
CROSS JOIN myTable as b)
I need a little help on a SQL query. I could not get the result that I wanted.
ID I10 H 10NS HNS CC NSCC
0 1 1 1 1 14 14
1 0 1 0 1 6 2
1 0 2 0 2 12 2
1 0 3 0 3 17 4
1 0 3 0 3 18 4
1 0 3 0 3 19 4
1 0 3 0 3 20 4
What I want to have is one from each ID with highest CC
For example,
ID I10 H 10NS HNS CC NSCC
0 1 1 1 1 14 14
1 0 3 0 3 20 4
I tried with this code:
SELECT a.ID, b.name, a.i10 as[i-10-index], a.h as[h-index], 10ns as[i-10-index based on non-self-citation], a.hns as [h-index based on non-self-citation],
max(a.[Citation Count]), (a.[Non-Self-Citation Count])
FROM tbl_lpNumerical as a
join tbl_lpAcademician as b
on a.ID= (b.ID-1)
GROUP BY a.ID, b.name, a.i10, a.h, a.10ns, a.hns,
a.[Non-Self-Citation Count]
order by a.ID desc
However, I could not get the desired results.
Thank you for your time.
You can simply get all the row where not exist another row with an higher CC
SELECT n.*
FROM tbl_lpNumerical n
WHERE NOT EXISTS ( SELECT 'b'
FROM tbl_lpNumerical n2
WHERE n2.ID = n.ID
AND n2.CC > n.CC
)
In SQL Server, you can use row_number() for this. Based on your sample data`, something like:
select sd.*
from (select sd.*, row_number() over (partition by id order by cc desc) as seqnum
from sampledata sd
) sd
where seqnum = 1;
I have no idea what your query has to do with the sample data. If it generates the data, then you can use a CTE:
with sampledata as (
<some query here>
)
select sd.*
from (select sd.*, row_number() over (partition by id order by cc desc) as seqnum
from sampledata sd
) sd
where seqnum = 1;
The following query will select a single row from each ID partition: the one with the highest CC value:
SELECT *
FROM (SELECT *,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY CC DESC) AS rn
FROM mytable) t
WHERE t.rn = 1
If there can be multiple rows having the same CC max value and you want all of them selected, then you can replace ROW_NUMBER() with RANK().