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Add Compute Median Instead of Average as a PostgreSQL TIL

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jbranchaud
2026-02-02 16:55:50 -06:00
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@@ -10,7 +10,7 @@ working across different projects via [VisualMode](https://www.visualmode.dev/).
For a steady stream of TILs, [sign up for my newsletter](https://visualmode.kit.com/newsletter). For a steady stream of TILs, [sign up for my newsletter](https://visualmode.kit.com/newsletter).
_1735 TILs and counting..._ _1736 TILs and counting..._
See some of the other learning resources I work on: See some of the other learning resources I work on:
@@ -863,6 +863,7 @@ If you've learned something here, support my efforts writing daily TILs by
- [Clear The Screen In psql](postgres/clear-the-screen-in-psql.md) - [Clear The Screen In psql](postgres/clear-the-screen-in-psql.md)
- [Clear The Screen In psql (2)](postgres/clear-the-screen-in-psql-2.md) - [Clear The Screen In psql (2)](postgres/clear-the-screen-in-psql-2.md)
- [Compute Hashes With pgcrypto](postgres/compute-hashes-with-pgcrypto.md) - [Compute Hashes With pgcrypto](postgres/compute-hashes-with-pgcrypto.md)
- [Compute Median Instead Of Average](postgres/compute-median-instead-of-average.md)
- [Compute The Levenshtein Distance Of Two Strings](postgres/compute-the-levenshtein-distance-of-two-strings.md) - [Compute The Levenshtein Distance Of Two Strings](postgres/compute-the-levenshtein-distance-of-two-strings.md)
- [Compute The md5 Hash Of A String](postgres/compute-the-md5-hash-of-a-string.md) - [Compute The md5 Hash Of A String](postgres/compute-the-md5-hash-of-a-string.md)
- [Concatenate Strings With A Separator](postgres/concatenate-strings-with-a-separator.md) - [Concatenate Strings With A Separator](postgres/concatenate-strings-with-a-separator.md)

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# Compute Median Instead Of Average
One of the first aggregate functions we might use in PostgreSQL, besides `sum`,
is `avg`.
```sql
select avg(book_count) as average_books_read
from (
select users.id, count(books.id) as book_count
from users
left join books
on books.user_id = users.id
where books.read_in_year = 2025
group by users.id
) as user_book_counts;
```
This computes the average of the set of values which sums them all up
and divides by the count. The average (maybe you've heard this also called the
_mean_) is not always the best way to understand data, especially when there are
outliers.
Instead, we might want to compute the _median_ value of our set of data. There
is no easily identifiable `median` aggregate function. Instead, we can use
`percentile_cont` with a value of `0.5`. This gets us the 50th percentile of our
set of data which is the definition of the _median_.
```sql
select percentile_cont(0.5) within group (
order by book_count
) as median_books_read
from (
select users.id, count(books.id) as book_count
from users
left join books on books.user_id = users.id and books.read_in_year = 2025
group by users.id
) as user_book_counts;
```
The full syntax for `percentile_cont` is `percentile_cong(precision) within
group (order by ...)` because this is an aggregiate that has to work with an
ordered-set of data.
[source](https://www.postgresql.org/docs/current/functions-aggregate.html)