Finding top count of a value in a table using SQL - sql

I'm looking for a way to find the top count value of a column by SQL.
If for example this is my data
id type
----------
1 A
1 B
1 A
2 C
2 D
2 D
I would like the result to be:
1 A
2 D
I'm looking for a way to do it without groping by the column I count (type in the example)
Thanks

Statistically, this is called the "mode". You can calculate it using window functions:
select id, type, cnt
from (select id, type, count(*) as cnt,
row_number() over (partition by id order by count(*) desc) as seqnum
from t
group by id, type
) t
where seqnum = 1;
If there are ties, then an arbitrary value is chosen from among the ties.

You are looking for the statistic mode (the most often ocurring value):
select id, stats_mode(type)
from mytable
group by id
order by id;
Not all DBMS support this however. Check your docs, wheher this function or a similar one is available in your DBMS.

Just GROUP BY id, type and keep the rows with the maximum counter:
select id, type
from tablename
group by id, type
having count(*) = (
select count(*) from tablename group by id, type order by count(*) desc limit 1
)
See the demo
Or
select id, type
from tablename
group by id, type
having count(*) = (
select max(t.counter) from (select count(*) counter from tablename group by id, type) t
)
See the demo

Related

select value based on max of other column

I have a few questions about a table I'm trying to make in Postgres.
The following table is my input:
id
area
count
function
1
100
20
living
1
200
30
industry
2
400
10
living
2
400
10
industry
2
400
20
education
3
150
1
industry
3
150
1
education
I want to group by id and get the dominant function based on max area. With summing up the rows for area and count. When area is equal it should be based on max count, when area and count is equal it should be based on prior function (i still have to decide if education is prior to industry or vice versa). So the result should be:
id
area
count
function
1
300
50
industry
2
1200
40
education
3
300
2
industry
I tried a lot of things and maybe it's easy, but i don't get it. Can someone help to get the right SQL?
One method uses row_number() and conditional aggregation:
select id, sum(area), sum(count),
max(function) over (filter where seqnum = 1) as function
from (select t.*,
row_number() over (partition by id order by area desc) as seqnum
from t
) t
group by id;
Another method uses ``distinct on`:
select id, sum(area) over (partition by id) as area,
sum(count) over (partition by id) as count,
function
from t
order by id, area desc;
Use a scalar sub-query for "function".
select t.id, sum(t.area), sum(t.count),
(
select "function"
from the_table
where id = t.id
order by area desc, count desc, "function" desc
limit 1
) as "function"
from the_table as t
group by t.id order by t.id;
SQL Fiddle
you can use sum as window function:
select distinct on (t.id)
id,
sum(area) over (partition by id) as area,
sum(count) over (partition by id) as count,
( select function from tbl_test where tbl_test.id = t.id order by count desc limit 1 ) as function
from tbl_test t
This is how you get the function for each group based on id:
select id, function
from yourtable yt1
left join yourtable yt2
on yt1.id = yt2.id and yt1.area < yt2.area
where yt2.area.id is null;
(we ensure that no yt2 exists that would be of the same id but of higher areay)
This would work nicely, but you might have several max areas with different values. To cope with this isue, let's ensure that exactly one is chosen:
select id, max(function) as function
from yourtable yt1
left join yourtable yt2
on yt1.id = yt2.id and yt1.area < yt2.area
where yt2.area.id is null
group by id;
Now, let's join this to our main table;
select yourtable.id, sum(yourtable.area), sum(yourtable.count), t.function
from yourtable
join (
select id, max(function) as function
from yourtable yt1
left join yourtable yt2
on yt1.id = yt2.id and yt1.area < yt2.area
where yt2.area.id is null
group by id
) t
on yourtable.id = t.id
group by yourtable.id;

How do I create a new SQL table with custom column names and populate these columns

So I currently have an SQL statement that generates a table with the most frequent occurring value as well as the least frequent occurring value in a table. However this table has 2 rows with the row values as well as the fields. I need to create a custom table with 2 columns with min and max. Then have one row with one value for each. The value for these columns needs to be from the same row.
(SELECT name, COUNT(name) AS frequency
FROM firefighter_certifications
GROUP BY name
ORDER BY frequency DESC limit 1)
UNION
(SELECT name, COUNT(name) AS frequency
FROM firefighter_certifications
GROUP BY name
ORDER BY frequency ASC limit 1);
So for the above query I would need the names of the min and max values in one row. I need to be able to define the name of new columns for the generated SQL query as well.
Min_Name | Max_Name
Certif_1 | Certif_2
I think this query should give you the results you want. It ranks each name according to the number of times it appears in the table, then uses conditional aggregation to select the min and max frequency names in one row:
with cte as (
select name,
row_number() over (order by count(*) desc) as maxr,
row_number() over (order by count(*)) as minr
from firefighter_certifications
group by name
)
select max(case when minr = 1 then name end) as Min_Name,
max(case when maxr = 1 then name end) as Max_Name
from cte
Postgres doesn't offer "first" and "last" aggregation functions. But there are other, similar methods:
select distinct first_value(name) over (order by cnt desc, name) as name_at_max,
first_value(name) over (order by cnt asc, name) as name_at_min
from (select name, count(*) as cnt
from firefighter_certifications
group by name
) n;
Or without any subquery at all:
select first_value(name) over (order by count(*) desc, name) as name_at_max,
first_value(name) over (order by count(*) asc, name) as name_at_min
from firefighter_certifications
group by name
limit 1;
Here is a db<>fiddle

DB ORACLE QUERY

I have a table where storing details
ID NAME
1 A
2 A
1 A
I need the output like
ID Name Count
1,2 A 3
Please help to get the output like that in oracle select query
In Oracle, you can use listagg(), but it has no distinct option. So, use a subquery and two levels of aggregation:
select listagg(id, ',') within group (order by id) as id, name, sum(cnt)
from (select id, name, count(*) as cnt
from t
group by id, name
) x
group by name;

How to get max value and using group by clause

I have a query like this:
select transactions_id,
time_stamp,
clock
from times
group by transactions_id
having sum(distinct type) = 1
now, I would like to get max value depending on id.
I used below queries but not worked:
select max(id),
transactions_id,
time_stamp,
clock
from times
group by transactions_id
having sum(distinct type) = 1
or
select transactions_id,
time_stamp,
clock
from times
group by transactions_id
having sum(distinct type) = 1
and max(id)
for example:
I have three conditions:
type must be 1
group by transactions_id
max id
You can find aggregates in one query and join its result with the table to get the relevant rows.
select *
from times t1
join (
select transactions_id,
max(id) as id
from times
where type = 1
group by transactions_id
) t2 using (transactions_id, id);
If I understand correctly, you can use the ANSI standard row_number() function:
select t.*
from (select t.*,
row_number() over (partition by transactions_id order by id desc) as seqnum
from times t
) t
where seqnum = 1;
I am not sure what having sum(distinct type) = 1. That condition is not explained in the question.

Aggregate function like MAX for most common cell in column?

Group by the highest Number in a column worked great with MAX(), but what if I would like to get the cell that is at most common.
As example:
ID
100
250
250
300
200
250
So I would like to group by ID and instead of get the lowest (MIN) or highest (MAX) number, I would like to get the most common one (that would be 250, because there 3x).
Is there an easy way in SQL Server 2012 or am I forced to add a second SELECT where I COUNT(DISTINCT ID) and add that somehow to my first SELECT statement?
You can use dense_rank to return all the id's with the highest counts. This would handle cases when there are ties for the highest counts as well.
select id from
(select id, dense_rank() over(order by count(*) desc) as rnk from tablename group by id) t
where rnk = 1
A simple way to do what you want uses top and order by:
SELECT top 1 id
FROM t
GROUP BY id
ORDER BY COUNT(*) DESC;
This is a statistic called the mode. Getting the mode and max is a bit challenging in SQL Server. I would approach it as:
WITH cte AS (
SELECT t.id, COUNT(*) AS cnt,
row_number() OVER (ORDER BY COUNT(*) DESC) AS seqnum
FROM t
GROUP BY id
)
SELECT MAX(id) AS themax, MAX(CASE WHEN seqnum = 1 THEN id END) AS MODE
FROM cte;