I am practicing SQL in Google Cloud Platform and have a table with the most popular names in USA.
I need to write a query that returns the most popular name in every year.
The Table has 6 columns: id, state, gender, year, name, Number of occurrences of the name
So far I have:
SELECT DISTINCT year
FROM 'table'
But I don't now what next...
What you are looking for is the called the mode in statistics. One way to get this uses aggregation and window fucntions:
select year, name
from (select year, name, count(*) as cnt,
row_number() over (partition by year order by count(*) desc) as seqnum
from t
group by year, name
) yn
where seqnum = 1;
The above version returns one arbitrary name if there are ties for the most frequent. If you do want ties, use rank() instead of row_number().
Related
I'm relatively new to working with SQL and wasn't able to find any past threads to solve my question. I have three columns in a table, columns being name, customer, and location. I'd like to add an additional column determining which location is most frequent, based off name and customer (first two columns).
I have included a photo of an example where name-Jane customer-BEC in my created column would be "Texas" as that has 2 occurrences as opposed to one for California. Would there be anyway to implement this?
If you want 'Texas' on all four rows:
select t.Name, t.Customer, t.Location,
(select t2.location
from table1 t2
where t2.name = t.name
group by name, location
order by count(*) desc
fetch first 1 row only
) as most_frequent_location
from table1 t ;
You can also do this with analytic functions:
select t.Name, t.Customer, t.Location,
max(location) keep (dense_rank first order by location_count desc) over (partition by name) most_frequent_location
from (select t.*,
count(*) over (partition by name, customer, location) as location_count
from table1 t
) t;
Here is a db<>fiddle.
Both of these version put 'Texas' in all four rows. However, each can be tweaks with minimal effort to put 'California' in the row for ARC.
In Oracle, you can use aggregate function stats_mode() to compute the most occuring value in a group.
Unfortunately it is not implemented as a window function. So one option uses an aggregate subquery, and then a join with the original table:
select t.*, s.top_location
from mytable t
inner join (
select name, customer, stats_mode(location) top_location
from mytable
group by name, customer
) s where s.name = t.name and s.customer = t.customer
You could also use a correlated subquery:
select
t.*,
(
select stats_mode(t1.location)
from mytable t1
where t1.name = t.name and t1.customer = t.customer
) top_location
from mytable t
This is more a question about understanding the concepts of a relational database. If you want that information, you would not put that in an additional column. It is calculated data over multiple columns - why would you store that in the table itself ? It is complex to code and it would also be very expensive for the database (imagine all the rows you have to calculate that value for if someone inserted a million rows)
Instead you can do one of the following
Calculate it at runtime, as shown in the other answers
if you want to make it more persisent, you could embed that query above in a view
if you want to physically store the info, you could use a materialized view
Plenty of documentation on those 3 options in the official oracle documentation
Your first step is to construct a query that determines the most frequent location, which is as simple as:
select Name, Customer, Location, count(*)
from table1
group by Name, Customer, Location
This isn't immediately useful, but the logic can be used in row_number(), which gives you a unique id for each row returned. In the query below, I'm ordering by count(*) in descending order so that the most frequent occurrence has the value 1.
Note that row_number() returns '1' to only one row.
So, now we have
select Name, Customer, Location, row_number() over (partition by Name, Customer order by count(*) desc) freq_name_cust
from table1 tb_
group by Name, Customer, Location
The final step puts it all together:
select tab.*, tb_.Location most_freq_location
from table1 tab
inner join
(select Name, Customer, Location, row_number() over (partition by Name, Customer order by count(*) desc) freq_name_cust
from table1
group by Name, Customer, Location) tb_
on tb_.Name = tab.Name
and tb_.Customer = tab.Customer
and freq_name_cust = 1
You can see how it all works in this Fiddle where I deliberately inserted rows with the same frequency for California and Texas for one of the customers for illustration purposes.
I would like to introduce myself as someone who has just recently started to fiddle a bit with SQL. Throughout my learning process I have come across a very specific problem and thus, my question is very specific too. Given the following table:
How should my list of commands look in order to get this following table:
In other words, what should I write to basically show the minimal salary and the id of its owner for each country. I have tried using GROUP BY but all I could get is the minimal salary per country whereas my goal was to show the id that belongs to the minimal salary too.
Hope I got my question clear and I thank everyone for the support.
This is a typical greatest-n-per-group problem.
One cross-database solution is to filter with a subquery:
select t.*
from mytable t
where t.salary = (select min(t1.salary) from mytable t1 where t1.country = t.country)
For performance with this query, you want an index on (country, salary).
You can also use window functions, if your database supports that:
select id, country, salary
from (
select t.*, rank() over(partition by country order by salary) rn
from mytable t
) t
where rn = 1
You can do by
select
id,
country,
salary
from
(
select
id,
country,
salary,
row_number() over (partition by country order by salary) as rnk
from table
)val
where rnk = 1
I need help in writing an efficient query to find a list of toppers (students with maximum total marks in each class) when we are given individual scores for each subject across different classes. We are required to return 3 columns: class, topper_student name and topper_student_total marks.
I have used multiple sub-queries to find a solution. I am sure there would be much better implementations available for this problem (maybe via joins or window functions?).
Input table and my solution can be found at SQL Fiddle link.
http://www.sqlfiddle.com/#!15/2919e/1/0
Input table:
It would be clearer to use temporary tables to store results along the way and make the result traceable, but the solution can be achieved with a single query:
WITH student_marks AS (
SELECT Class_num, Name, SUM(Marks) AS student_total_marks
FROM School
GROUP BY Class_num, Name
)
SELECT Class_num, Name, student_total_marks
FROM (
SELECT Class_num, Name, student_total_marks, ROW_NUMBER() OVER(partition by Class_num order by student_total_marks desc, Class_num) AS beststudentfirst
FROM student_marks
) A
WHERE A.beststudentfirst = 1
The query within WITH statement calculate a sum of marks for every student in a class. At this point, subject is not required anymore. The result is temporarily stored into student_marks.
Next, we need to create a counter (beststudentfirst) using ROW_NUMBER to number the total marks from the highest to the lowest in each class (order by student_total_marks desc, Class_num). The counter should be reinitiated each time the class changes (partition by Class_num order).
From this last result, we only need the counter (beststudentfirst) with the value of one. It is the top student in each class.
Window functions are the most natural way to approach this. If you always want exactly three students, then use row_number():
select Class_num, Name, total_marks
from (select name, class_num, sum(marks) as total_marks,
row_number() over (partition by class_num order by sum(marks) desc) as seqnum
from School
group by Class_num, Name
) s
where seqnum <= 1
order by class_num, total_marks desc;
If you want to take ties into account, then use rank() or dense_rank().
Here is the SQL Fiddle.
select Class_num,[Name],total_marks from
(
select Row_number() over (partition by class_num order by Class_num,SUM(Marks) desc) as
[RN],Class_num,[Name],SUM(Marks) as total_marks
from School
group by Class_num,[Name]
)A
where RN=1
I need to get the Top 10 results for each Region, Market and Name along with those with highest counts (Gaps). There are 4 Regions with 1 to N Markets. I can get the Top 10 but cannot figure out how to do this without using a Union for every Market. Any ideas on how do this?
SELECT DISTINCT TOP 10
Region, Market, Name, Gaps
FROM
TableName
ORDER BY
Region, Market, Gaps DESC
One approach would be to use a CTE (Common Table Expression) if you're on SQL Server 2005 and newer (you aren't specific enough in that regard).
With this CTE, you can partition your data by some criteria - i.e. your Region, Market, Name - and have SQL Server number all your rows starting at 1 for each of those "partitions", ordered by some criteria.
So try something like this:
;WITH RegionsMarkets AS
(
SELECT
Region, Market, Name, Gaps,
RN = ROW_NUMBER() OVER(PARTITION BY Region, Market, Name ORDER BY Gaps DESC)
FROM
dbo.TableName
)
SELECT
Region, Market, Name, Gaps
FROM
RegionsMarkets
WHERE
RN <= 10
Here, I am selecting only the "first" entry for each "partition" (i.e. for each Region, Market, Name tuple) - ordered by Gaps in a descending fashion.
With this, you get the top 10 rows for each (Region, Market, Name) tuple - does that approach what you're looking for??
I think you want row_number():
select t.*
from (select t.*,
row_number() over (partition by region, market order by gaps desc) as seqnum
from tablename t
) t
where seqnum <= 10;
I am not sure if you want name in the partition by clause. If you have more than one name within a market, that may be what you are looking for. (Hint: Sample data and desired results can really help clarify a question.)
I have a table Student in SQL Server with these columns:
[ID], [Age], [Level]
I want the query that returns each age value that appears in Students, and finds the level value that appears most often. For example, if there are more 'a' level students aged 18 than 'b' or 'c' it should print the pair (18, a).
I am new to SQL Server and I want a simple answer with nested query.
You can do this using window functions:
select t.*
from (select age, level, count(*) as cnt,
row_number() over (partition by age order by count(*) desc) as seqnum
from student s
group by age, level
) t
where seqnum = 1;
The inner query aggregates the data to count the number of levels for each age. The row_number() enumerates these for each age (the partition by with the largest first). The where clause then chooses the highest values.
In the case of ties, this returns just one of the values. If you want all of them, use rank() instead of row_number().
One more option with ROW_NUMBER ranking function in the ORDER BY clause. WITH TIES used when you want to return two or more rows that tie for last place in the limited results set.
SELECT TOP 1 WITH TIES age, level
FROM dbo.Student
GROUP BY age, level
ORDER BY ROW_NUMBER() OVER(PARTITION BY age ORDER BY COUNT(*) DESC)
Or the second version of the query using amount each pair of age and level, and max values of count pair age and level per age.
SELECT *
FROM (
SELECT age, level, COUNT(*) AS cnt,
MAX(COUNT(*)) OVER(PARTITION BY age) AS mCnt
FROM dbo.Student
GROUP BY age, level
)x
WHERE x.cnt = x.mCnt
Demo on SQLFiddle
Another option but will require later version of sql-server:
;WITH x AS
(
SELECT age,
level,
occurrences = COUNT(*)
FROM Student
GROUP BY age,
level
)
SELECT *
FROM x x
WHERE EXISTS (
SELECT *
FROM x y
WHERE x.occurrences > y.occurrences
)
I realise it doesn't quite answer the question as it only returns the age/level combinations where there are more than one level for the age.
Maybe someone can help to amend it so it includes the single level ages aswell in the result set: http://sqlfiddle.com/#!3/d597b/9
with combinations as (
select age, level, count(*) occurrences
from Student
group by age, level
)
select age, level
from combinations c
where occurrences = (select max(occurrences)
from combinations
where age = c.age)
This finds every age and level combination in the Students table and counts the number of occurrences of each level.
Then, for each age/level combination, find the one whose occurrences are the highest for that age/level combination. Return the age and level for that row.
This has the advantage of not being tied to SQL Server - it's vanilla SQL. However, a window function like Gordon pointed out may perform better on SQL Server.