Get values from first and last row per group - sql

Question

I'm new to Postgres, coming from MySQL and hoping that one of y'all would be able to help me out.
I have a table with three columns: name, week, and value. This table has a record of the names, the week at which they recorded the height, and the value of their height.
Something like this:
Name | Week | Value
------+--------+-------
John | 1 | 9
Cassie| 2 | 5
Luke | 6 | 3
John | 8 | 14
Cassie| 5 | 7
Luke | 9 | 5
John | 2 | 10
Cassie| 4 | 4
Luke | 7 | 4
What I want is a list per user of the value at the minimum week and the max week. Something like this:
Name |minWeek | Value |maxWeek | value
------+--------+-------+--------+-------
John | 1 | 9 | 8 | 14
Cassie| 2 | 5 | 5 | 7
Luke | 6 | 3 | 9 | 5
In Postgres, I use this query:
select name, week, value
from table t
inner join(
select name, min(week) as minweek
from table
group by name)
ss on t.name = ss.name and t.week = ss.minweek
group by t.name
;
However, I receive an error:
column "w.week" must appear in the GROUP BY clause or be used in an aggregate function
Position: 20
This worked fine for me in MySQL so I'm wondering what I'm doing wrong here?

Answer 1

There are various simpler and faster ways.
2x DISTINCT ON
SELECT *
FROM (
SELECT DISTINCT ON (name)
name, week AS first_week, value AS first_val
FROM tbl
ORDER BY name, week
) f
JOIN (
SELECT DISTINCT ON (name)
name, week AS last_week, value AS last_val
FROM tbl
ORDER BY name, week DESC
) l USING (name);
Or shorter:
SELECT *
FROM (SELECT DISTINCT ON (1) name, week AS first_week, value AS first_val FROM tbl ORDER BY 1,2) f
JOIN (SELECT DISTINCT ON (1) name, week AS last_week , value AS last_val FROM tbl ORDER BY 1,2 DESC) l USING (name);
Simple and easy to understand. Also fastest in my old tests. Detailed explanation for DISTINCT ON:
Select first row in each GROUP BY group?
2x window function, 1x DISTINCT ON
SELECT DISTINCT ON (name)
name, week AS first_week, value AS first_val
, first_value(week) OVER w AS last_week
, first_value(value) OVER w AS last_value
FROM tbl t
WINDOW w AS (PARTITION BY name ORDER BY week DESC)
ORDER BY name, week;
The explicit WINDOW clause only shortens the code, no effect on performance.
first_value() of composite type
The aggregate functions min() or max() do not accept composite types as input. You would have to create custom aggregate functions (which is not that hard).
But the window functions first_value() and last_value() do. Building on that we can devise simple solutions:
Simple query
SELECT DISTINCT ON (name)
name, week AS first_week, value AS first_value
,(first_value((week, value)) OVER (PARTITION BY name ORDER BY week DESC))::text AS l
FROM tbl t
ORDER BY name, week;
The output has all data, but the values for the last week are stuffed into an anonymous record (optionally cast to text). You may need decomposed values.
Decomposed result with opportunistic use of table type
For that we need a well-known composite type. An adapted table definition would allow for the opportunistic use of the table type itself directly:
CREATE TABLE tbl (week int, value int, name text); -- optimized column order
week and value come first, so now we can sort by the table type itself:
SELECT (l).name, first_week, first_val
, (l).week AS last_week, (l).value AS last_val
FROM (
SELECT DISTINCT ON (name)
week AS first_week, value AS first_val
, first_value(t) OVER (PARTITION BY name ORDER BY week DESC) AS l
FROM tbl t
ORDER BY name, week
) sub;
Decomposed result from user-defined row type
That's probably not possible in most cases. Register a composite type with CREATE TYPE (permanent) or with CREATE TEMP TABLE (for the duration of the session):
CREATE TEMP TABLE nv(last_week int, last_val int); -- register composite type
SELECT name, first_week, first_val, (l).last_week, (l).last_val
FROM (
SELECT DISTINCT ON (name)
name, week AS first_week, value AS first_val
, first_value((week, value)::nv) OVER (PARTITION BY name ORDER BY week DESC) AS l
FROM tbl t
ORDER BY name, week
) sub;
Custom aggregate functions first() & last()
Create functions and aggregates once per database:
CREATE OR REPLACE FUNCTION public.first_agg (anyelement, anyelement)
RETURNS anyelement
LANGUAGE sql IMMUTABLE STRICT PARALLEL SAFE AS
'SELECT $1';
CREATE AGGREGATE public.first(anyelement) (
SFUNC = public.first_agg
, STYPE = anyelement
, PARALLEL = safe
);
CREATE OR REPLACE FUNCTION public.last_agg (anyelement, anyelement)
RETURNS anyelement
LANGUAGE sql IMMUTABLE STRICT PARALLEL SAFE AS
'SELECT $2';
CREATE AGGREGATE public.last(anyelement) (
SFUNC = public.last_agg
, STYPE = anyelement
, PARALLEL = safe
);
Then:
SELECT name
, first(week) AS first_week, first(value) AS first_val
, last(week) AS last_week , last(value) AS last_val
FROM (SELECT * FROM tbl ORDER BY name, week) t
GROUP BY name;
Probably the most elegant solution. Faster with the additional module first_last_agg providing a C implementation.
Compare instructions in the Postgres Wiki.
Related:
Calculating follower growth over time for each influencer
db<>fiddle here (showing all)
Old sqlfiddle
Each of these queries was substantially faster than the currently accepted answer in a quick test on a table with 50k rows with EXPLAIN ANALYZE.
There are more ways. Depending on data distribution, different query styles may be (much) faster, yet. See:
Optimize GROUP BY query to retrieve latest row per user

Answer 2

This is a bit of a pain, because Postgres has the nice window functions first_value() and last_value(), but these are not aggregation functions. So, here is one way:
select t.name, min(t.week) as minWeek, max(firstvalue) as firstvalue,
max(t.week) as maxWeek, max(lastvalue) as lastValue
from (select t.*, first_value(value) over (partition by name order by week) as firstvalue,
last_value(value) over (partition by name order by week) as lastvalue
from table t
) t
group by t.name;