Nothing earth-shattering here, I was just helping out a colleague with this so I thought I’d post up the example I gave him. 70-640 Training .

 

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— sample table:
create table People
(
Person varchar(1) primary key,
City varchar(10),
Age int
)

go

— with some sample data:

insert into People
select ‘A’,’Boston’,23 union all  — odd #
select ‘B’,’Boston’,43 union all
select ‘C’,’Boston’,29 union all

select ‘D’,’Chicago’,15 union all — single #

select ‘E’,’NY’,12 union all  — even #
select ‘F’,’NY’,55 union all
select ‘G’,’NY’,57 union all
select ‘H’,’NY’,61

go

— here’s our query, showing median age per city:

select city,
AVG(age) as MedianAge
from
(
select City, Person, Age,
ROW_NUMBER() over (partition by City order by Age ASC) as AgeRank,
COUNT(*) over (partition by City) as CityCount
from
People
) x
where
x.AgeRank in (x.CityCount/2+1, (x.CityCount+1)/2)
group by
x.City

go

— clean it all up
drop table People

And here’s the result:

city       MedianAge
———- ———–
Boston     29
Chicago    15
NY         56

(3 row(s) affected)

Simply remove “City” from the SELECT clause and the GROUP BY clause to get the median age for all.

There may be more efficient tricks out there, but this is certainly the shortest and simplest technique I am aware of.

 

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