I have a database which consists of 5 columns
Each column is an INT
I want to find out which numbers occur most frequently in each column
I also wanted to know which sequence of numbers occurs more frequently
The same thing is true with the least common
I would like to use MYSQL or SQLITE
assuming schema as follows
tbl(..., int_col1, int_col2, int_col3, int_col4, int_col5, ...)
most frequently
do following query for each column
SELECT int_col1, COUNT(int_col1)
FROM tbl
GROUP BY int_col1
ORDER BY COUNT(int_col) DESC LIMIT 1
least frequently
do following query for each column
SELECT int_col1, COUNT(int_col1)
FROM tbl
GROUP BY int_col1
ORDER BY COUNT(int_col) ASC LIMIT 1
sequence of numbers occurs more frequently
SELECT int_col1, int_col2, int_col3, int_col4, int_col5, COUNT(*)
FROM tbl
GROUP BY int_col1, int_col2, int_col3, int_col4, int_col5
ORDER BY COUNT(*) DESC;
assuming column is userid
select UserID, count(UserID)
from myUsers
group by UserID
order by count(UserID) desc
OR
with cte as
(
select user_id,ROW_NUMBER() over (order by UserID) as rn
)
select user_id,count(user_id) as se_count from cte group by user_id
Related
I have a mysql query which will return all the details from table along with i need max_row count i.e total no of rows in a table using COUNT(*) in a single select query without using cross join.
Note: MySQL version is earlier version of 8
Query :
SELECT * FROM tablename ORDER BY column name DESC LIMIT 0,10;
The total count of a table is simple, when you want to add it to every row.
SELECT
*
,(SELECT COUNT(*) FROM tablename ) count1
FROM tablename
ORDER BY column name
DESC LIMIT 0,10;,
I'm a beginner at SQL, how do I get a query which returns the most prevalent column value? Probably there is an answer somewhere but I don't know how to google it.
For example in the user_id column the query should return the value 1 because this is the most prevalent number.
One approach is to do a GROUP BY aggregation and then apply a LIMIT trick:
SELECT user_id, COUNT(*) AS cnt
FROM yourTable
GROUP BY user_id
ORDER BY COUNT(*) DESC
LIMIT 1;
If you want something more complex, then you would be getting into the realm of rank functionality. MySQL (at least as of the current release) does not support built-in rank support, so it can be tricky to perform such queries.
SELECT top 1 user_id, COUNT(*) AS cnt
FROM yourTable
GROUP BY user_id
ORDER BY COUNT(*) DESC
Have a common table expression that counts each user_id. Select user_id where the count is the max count. Will return both user_id's in case of a tie.
with cte as
(
SELECT user_id, COUNT(*) AS cnt
FROM yourTable
GROUP BY user_id
)
select user_id
from cte
where cnt = (select max(cnt) from cte)
I have a below db table structure :
Tablename : agency_mst
Fields:
id
name
is_premium
date_added
I want to sort the data so that the agencies with is_premium=1 comes first and the rest of the data is sorted by date_added desc order.
But also the is_premium=1 records should be sorted in random order. so first set of premium agencies will have random order. How can I do that using MySQL Select query.
I have built this query partially but not sure how to filter specific set of data. Below is that query:
SELECT * FROM agency_mst
ORDER BY is_premium DESC, date_added DESC
How about
SELECT *
FROM agency_mst
ORDER BY IF(is_premium=1, RAND(), -1.0), date_added DESC
That will use random ordering for the matching rows, then put the others last, and order them by date.
Be careful with performance, though. ORDER BY RAND() in any variant is a notorious performance antipattern in tables with many rows.
select t1.id,t1.name,t1.is_premium,t1.date_added from
(select (ROW_NUMBER() over (order by id))* cast(CRYPT_GEN_RANDOM(10) as int) RND,* from agency_mst) t1
order by t1.RND
SELECT id, name, is_premium , date_added ,if(is_premium ,rand()*-15,0) as first_order
FROM agency_mst
ORDER BY first_order, date_added DESC
Check This Using rand()*-15 you get -velue it show first and remain will be 0 and that will be order by date_added
I have a table that looks like this...
user_id, match_id, points_won
1 14 10
1 8 12
1 12 80
2 8 10
3 14 20
3 2 25
I want to write a MYSQL script that pulls back the most points a user has won in a single match and includes the match_id in the results - in other words...
user_id, match_id, max_points_won
1 12 80
2 8 10
3 2 25
Of course if I didn't need the match_id I could just do...
select user_id, max(points_won)
from table
group by user_id
But as soon as I add match_id to the "select" and "group by" I have a row for every match, and if I only add the match_id to the "select" (and not the "group by") then it won't correctly relate to the points_won.
Ideally I don't want to do the following either because it doesn't feel particularly safe (e.g. if the user has won the same amount of points on multiple matches)...
SELECT t.user_id, max(t.points_won) max_points_won
, (select t2.match_id
from table t2
where t2.user_id = t.user_id
and t2.points_won = max_points_won) as 'match_of_points_maximum'
FROM table t
GROUP BY t.user_id
Are there any more elegant options for this problem?
This is harder than it needs to be in MySQL. One method is a bit of a hack but it works in most circumstances. That is the group_concat()/substring_index() trick:
select user_id, max(points_won),
substring_index(group_concat(match_id order by points_won desc), ',', 1)
from table
group by user_id;
The group_concat() concatenates together all the match_ids, ordered by the points descending. The substring_index() then takes the first one.
Two important caveats:
The resulting expression has a type of string, regardless of the internal type.
The group_concat() uses an internal buffer, whose length -- by default -- is 1,024 characters. This default length can be changed.
You can use the query:
select user_id, max(points_won)
from table
group by user_id
as a derived table. Joining this to the original table gets you what you want:
select t1.user_id, t1.match_id, t2.max_points_won
from table as t1
join (
select user_id, max(points_won) as max_points_won
from table
group by user_id
) as t2 on t1.user_id = t2.user_id and t1.points_won = t2.max_points_won
I think you can optimize your query by add limit 1 in the inner query.
SELECT t.user_id, max(t.points_won) max_points_won
, (select t2.match_id
from table t2
where t2.user_id = t.user_id
and t2.points_won = max_points_won limit 1) as 'match_of_points_maximum'
FROM table t
GROUP BY t.user_id
EDIT : only for postgresql, sql-server, oracle
You could use row_number :
SELECT USER_ID, MATCH_ID, POINTS_WON
FROM
(
SELECT user_id, match_id, points_won, row_number() over (partition by user_id order by points_won desc) rn
from table
) q
where q.rn = 1
For a similar function, have a look at Gordon Linoff's answer or at this article.
In your example, you partition your set of result per user then you order by points_won desc to obtain highest winning point first.
How can I find the most frequent value in a given column in an SQL table?
For example, for this table it should return two since it is the most frequent value:
one
two
two
three
SELECT
<column_name>,
COUNT(<column_name>) AS `value_occurrence`
FROM
<my_table>
GROUP BY
<column_name>
ORDER BY
`value_occurrence` DESC
LIMIT 1;
Replace <column_name> and <my_table>. Increase 1 if you want to see the N most common values of the column.
Try something like:
SELECT `column`
FROM `your_table`
GROUP BY `column`
ORDER BY COUNT(*) DESC
LIMIT 1;
Let us consider table name as tblperson and column name as city. I want to retrieve the most repeated city from the city column:
select city,count(*) as nor from tblperson
group by city
having count(*) =(select max(nor) from
(select city,count(*) as nor from tblperson group by city) tblperson)
Here nor is an alias name.
Below query seems to work good for me in SQL Server database:
select column, COUNT(column) AS MOST_FREQUENT
from TABLE_NAME
GROUP BY column
ORDER BY COUNT(column) DESC
Result:
column MOST_FREQUENT
item1 highest count
item2 second highest
item3 third higest
..
..
For use with SQL Server.
As there is no limit command support in that.
Yo can use the top 1 command to find the maximum occurring value in the particular column in this case (value)
SELECT top1
`value`,
COUNT(`value`) AS `value_occurrence`
FROM
`my_table`
GROUP BY
`value`
ORDER BY
`value_occurrence` DESC;
Assuming Table is 'SalesLT.Customer' and the Column you are trying to figure out is 'CompanyName' and AggCompanyName is an Alias.
Select CompanyName, Count(CompanyName) as AggCompanyName from SalesLT.Customer
group by CompanyName
Order By Count(CompanyName) Desc;
If you can't use LIMIT or LIMIT is not an option for your query tool. You can use "ROWNUM" instead, but you will need a sub query:
SELECT FIELD_1, ALIAS1
FROM(SELECT FIELD_1, COUNT(FIELD_1) ALIAS1
FROM TABLENAME
GROUP BY FIELD_1
ORDER BY COUNT(FIELD_1) DESC)
WHERE ROWNUM = 1
If you have an ID column and you want to find most repetitive category from another column for each ID then you can use below query,
Table:
Query:
SELECT ID, CATEGORY, COUNT(*) AS FREQ
FROM TABLE
GROUP BY 1,2
QUALIFY ROW_NUMBER() OVER(PARTITION BY ID ORDER BY FREQ DESC) = 1;
Result:
Return all most frequent rows in case of tie
Find the most frequent value in mysql,display all in case of a tie gives two possible approaches:
Scalar subquery:
SELECT
"country",
COUNT(country) AS "cnt"
FROM "Sales"
GROUP BY "country"
HAVING
COUNT("country") = (
SELECT COUNT("country") AS "cnt"
FROM "Sales"
GROUP BY "country"
ORDER BY "cnt" DESC,
LIMIT 1
)
ORDER BY "country" ASC
With the RANK window function, available since MySQL 8+:
SELECT "country", "cnt"
FROM (
SELECT
"country",
COUNT("country") AS "cnt",
RANK() OVER (ORDER BY COUNT(*) DESC) "rnk"
FROM "Sales"
GROUP BY "country"
) AS "sub"
WHERE "rnk" = 1
ORDER BY "country" ASC
This method might save a second recount compared to the first one.
RANK works by ranking all rows, such that if two rows are at the top, both get rank 1. So it basically directly solves this type of use case.
RANK is also available on SQLite and PostgreSQL, I think it might be SQL standard, not sure.
In the above queries I also sorted by country to have more deterministic results.
Tested on SQLite 3.34.0, PostgreSQL 14.3, GitHub upstream.
Most frequent for each GROUP BY group
MySQL: MySQL SELECT most frequent by group
PostgreSQL:
Get most common value for each value of another column in SQL
https://dba.stackexchange.com/questions/193307/find-most-frequent-values-for-a-given-column
SQLite: SQL query for finding the most frequent value of a grouped by value
SELECT TOP 20 WITH TIES COUNT(Counted_Column) AS Count, OtherColumn1,
OtherColumn2, OtherColumn3, OtherColumn4
FROM Table_or_View_Name
WHERE
(Date_Column >= '01/01/2023') AND
(Date_Column <= '03/01/2023') AND
(Counted_Column = 'Desired_Text')
GROUP BY OtherColumn1, OtherColumn2, OtherColumn3, OtherColumn4
ORDER BY COUNT(Counted_Column) DESC
20 can be changed to any desired number
WITH TIES allows all ties in the count to be displayed
Date range used if date/time column exists and can be modified to search a date range as desired
Counted_Column 'Desired_Text' can be modified to only count certain entries in that column
Works in INSQL for my instance
One way I like to use is:
select *<given_column>*,COUNT(*<given_column>*)as VAR1 from Table_Name
group by *<given_column>*
order by VAR1 desc
limit 1