with - sql server select last record of each group

Select first row in each GROUP BY group? (8)

As the title suggests, I'd like to select the first row of each set of rows grouped with a GROUP BY.

Specifically, if I've got a purchases table that looks like this:

SELECT * FROM purchases;

My Output:

id | customer | total
 1 | Joe      | 5
 2 | Sally    | 3
 3 | Joe      | 2
 4 | Sally    | 1

I'd like to query for the id of the largest purchase (total) made by each customer. Something like this:

SELECT FIRST(id), customer, FIRST(total)
FROM  purchases
GROUP BY customer

Expected Output:

FIRST(id) | customer | FIRST(total)
        1 | Joe      | 5
        2 | Sally    | 3


Testing the most interesting candidates with Postgres 9.4 and 9.5 with a halfway realistic table of 200k rows in purchases and 10k distinct customer_id (avg. 20 rows per customer).

For Postgres 9.5 I ran a 2nd test with effectively 86446 distinct customers. See below (avg. 2.3 rows per customer).


Main table

CREATE TABLE purchases (
  id          serial
, customer_id int  -- REFERENCES customer
, total       int  -- could be amount of money in Cent
, some_column text -- to make the row bigger, more realistic

I use a serial (PK constraint added below) and an integer customer_id since that's a more typical setup. Also added some_column to make up for typically more columns.

Dummy data, PK, index - a typical table also has some dead tuples:

INSERT INTO purchases (customer_id, total, some_column)    -- insert 200k rows
SELECT (random() * 10000)::int             AS customer_id  -- 10k customers
     , (random() * random() * 100000)::int AS total     
     , 'note: ' || repeat('x', (random()^2 * random() * random() * 500)::int)
FROM   generate_series(1,200000) g;

ALTER TABLE purchases ADD CONSTRAINT purchases_id_pkey PRIMARY KEY (id);

DELETE FROM purchases WHERE random() > 0.9; -- some dead rows

INSERT INTO purchases (customer_id, total, some_column)
SELECT (random() * 10000)::int             AS customer_id  -- 10k customers
     , (random() * random() * 100000)::int AS total     
     , 'note: ' || repeat('x', (random()^2 * random() * random() * 500)::int)
FROM   generate_series(1,20000) g;  -- add 20k to make it ~ 200k

CREATE INDEX purchases_3c_idx ON purchases (customer_id, total DESC, id);


customer table - for superior query

SELECT customer_id, 'customer_' || customer_id AS customer
FROM   purchases

ALTER TABLE customer ADD CONSTRAINT customer_customer_id_pkey PRIMARY KEY (customer_id);


In my second test for 9.5 I used the same setup, but with random() * 100000 to generate customer_id to get only few rows per customer_id.

Object sizes for table purchases

Generated with this query.

               what                | bytes/ct | bytes_pretty | bytes_per_row
 core_relation_size                | 20496384 | 20 MB        |           102
 visibility_map                    |        0 | 0 bytes      |             0
 free_space_map                    |    24576 | 24 kB        |             0
 table_size_incl_toast             | 20529152 | 20 MB        |           102
 indexes_size                      | 10977280 | 10 MB        |            54
 total_size_incl_toast_and_indexes | 31506432 | 30 MB        |           157
 live_rows_in_text_representation  | 13729802 | 13 MB        |            68
 ------------------------------    |          |              |
 row_count                         |   200045 |              |
 live_tuples                       |   200045 |              |
 dead_tuples                       |    19955 |              |


1. row_number() in CTE, (see other answer)

WITH cte AS (
   SELECT id, customer_id, total
        , row_number() OVER(PARTITION BY customer_id ORDER BY total DESC) AS rn
   FROM   purchases
SELECT id, customer_id, total
FROM   cte
WHERE  rn = 1;

2. row_number() in subquery (my optimization)

SELECT id, customer_id, total
FROM   (
   SELECT id, customer_id, total
        , row_number() OVER(PARTITION BY customer_id ORDER BY total DESC) AS rn
   FROM   purchases
   ) sub
WHERE  rn = 1;

3. DISTINCT ON (see other answer)

SELECT DISTINCT ON (customer_id)
       id, customer_id, total
FROM   purchases
ORDER  BY customer_id, total DESC, id;

4. rCTE with LATERAL subquery (see here)

   (  -- parentheses required
   SELECT id, customer_id, total
   FROM   purchases
   ORDER  BY customer_id, total DESC
   LIMIT  1
   SELECT u.*
   FROM   cte c
   ,      LATERAL (
      SELECT id, customer_id, total
      FROM   purchases
      WHERE  customer_id > c.customer_id  -- lateral reference
      ORDER  BY customer_id, total DESC
      LIMIT  1
      ) u
SELECT id, customer_id, total
FROM   cte
ORDER  BY customer_id;

5. customer table with LATERAL (see here)

FROM   customer c
,      LATERAL (
   SELECT id, customer_id, total
   FROM   purchases
   WHERE  customer_id = c.customer_id  -- lateral reference
   ORDER  BY total DESC
   LIMIT  1
   ) l;

6. array_agg() with ORDER BY (see other answer)

SELECT (array_agg(id ORDER BY total DESC))[1] AS id
     , customer_id
     , max(total) AS total
FROM   purchases
GROUP  BY customer_id;


Execution time for above queries with EXPLAIN ANALYZE (and all options off), best of 5 runs.

All queries used an Index Only Scan on purchases2_3c_idx (among other steps). Some of them just for the smaller size of the index, others more effectively.

A. Postgres 9.4 with 200k rows and ~ 20 per customer_id

1. 273.274 ms  
2. 194.572 ms  
3. 111.067 ms  
4.  92.922 ms  
5.  37.679 ms  -- winner
6. 189.495 ms

B. The same with Postgres 9.5

1. 288.006 ms
2. 223.032 ms  
3. 107.074 ms  
4.  78.032 ms  
5.  33.944 ms  -- winner
6. 211.540 ms  

C. Same as B., but with ~ 2.3 rows per customer_id

1. 381.573 ms
2. 311.976 ms
3. 124.074 ms  -- winner
4. 710.631 ms
5. 311.976 ms
6. 421.679 ms

Original (outdated) benchmark from 2011

I ran three tests with PostgreSQL 9.1 on a real life table of 65579 rows and single-column btree indexes on each of the three columns involved and took the best execution time of 5 runs.
Comparing @OMGPonies' first query (A) to the above DISTINCT ON solution (B):

  1. Select the whole table, results in 5958 rows in this case.

    A: 567.218 ms
    B: 386.673 ms
  2. Use condition WHERE customer BETWEEN x AND y resulting in 1000 rows.

    A: 249.136 ms
    B:  55.111 ms
  3. Select a single customer with WHERE customer = x.

    A:   0.143 ms
    B:   0.072 ms

Same test repeated with the index described in the other answer

CREATE INDEX purchases_3c_idx ON purchases (customer, total DESC, id);

1A: 277.953 ms  
1B: 193.547 ms

2A: 249.796 ms -- special index not used  
2B:  28.679 ms

3A:   0.120 ms  
3B:   0.048 ms

On Oracle 9.2+ (not 8i+ as originally stated), SQL Server 2005+, PostgreSQL 8.4+, DB2, Firebird 3.0+, Teradata, Sybase, Vertica:

WITH summary AS (
    SELECT p.id, 
           ROW_NUMBER() OVER(PARTITION BY p.customer 
                                 ORDER BY p.total DESC) AS rk
  FROM summary s
 WHERE s.rk = 1

Supported by any database:

But you need to add logic to break ties:

  SELECT MIN(x.id),  -- change to MAX if you want the highest
    JOIN (SELECT p.customer,
                 MAX(total) AS max_total
            FROM PURCHASES p
        GROUP BY p.customer) y ON y.customer = x.customer
                              AND y.max_total = x.total
GROUP BY x.customer, x.total

In PostgreSQL this is typically simpler and faster (more performance optimization below):

       id, customer, total
FROM   purchases
ORDER  BY customer, total DESC, id;

Or shorter (if not as clear) with ordinal numbers of output columns:

       id, customer, total
FROM   purchases
ORDER  BY 2, 3 DESC, 1;

If total can be NULL (won't hurt either way, but you'll want to match existing indexes):

ORDER  BY customer, total DESC NULLS LAST, id;

Major points

  • DISTINCT ON is a PostgreSQL extension of the standard (where only DISTINCT on the whole SELECT list is defined).

  • List any number of expressions in the DISTINCT ON clause, the combined row value defines duplicates. The manual:

    Obviously, two rows are considered distinct if they differ in at least one column value. Null values are considered equal in this comparison.

    Bold emphasis mine.

  • DISTINCT ON can be combined with ORDER BY. Leading expressions have to match leading DISTINCT ON expressions in the same order. You can add additional expressions to ORDER BY to pick a particular row from each group of peers. I added id as last item to break ties:

    "Pick the row with the smallest id from each group sharing the highest total."

    If total can be NULL, you most probably want the row with the greatest non-null value. Add NULLS LAST like demonstrated. Details:

  • The SELECT list is not constrained by expressions in DISTINCT ON or ORDER BY in any way. (Not needed in the simple case above):

    • You don't have to include any of the expressions in DISTINCT ON or ORDER BY.

    • You can include any other expression in the SELECT list. This is instrumental for replacing much more complex queries with subqueries and aggregate / window functions.

  • I tested with Postgres versions 8.3 – 10. But the feature has been there at least since version 7.1, so basically always.


The perfect index for the above query would be a multi-column index spanning all three columns in matching sequence and with matching sort order:

CREATE INDEX purchases_3c_idx ON purchases (customer, total DESC, id);

May be too specialized for real world applications. But use it if read performance is crucial. If you have DESC NULLS LAST in the query, use the same in the index so Postgres knows sort order matches.

Effectiveness / Performance optimization

You have to weigh cost and benefit before you create a tailored index for every query. The potential of above index largely depends on data distribution.

The index is used because it delivers pre-sorted data, and in Postgres 9.2 or later the query can also benefit from an index only scan if the index is smaller than the underlying table. The index has to be scanned in its entirety, though.


I had a simple benchmark here which is outdated by now. I replaced it with a detailed benchmark in this separate answer.

In Postgres you can use array_agg like this:

SELECT  customer,
        (array_agg(id ORDER BY total DESC))[1],
FROM purchases
GROUP BY customer

This will give you the id of each customer's largest purchase.

Some things to note:

  • array_agg is an aggregate function, so it works with GROUP BY.
  • array_agg lets you specify an ordering scoped to just itself, so it doesn't constrain the structure of the whole query. There is also syntax for how you sort NULLs, if you need to do something different from the default.
  • Once we build the array, we take the first element. (Postgres arrays are 1-indexed, not 0-indexed).
  • You could use array_agg in a similar way for your third output column, but max(total) is simpler.
  • Unlike DISTINCT ON, using array_agg lets you keep your GROUP BY, in case you want that for other reasons.

The accepted OMG Ponies' "Supported by any database" solution has good speed from my test.

Here I provide a same-approach, but more complete and clean any-database solution. Ties are considered (assume desire to get only one row for each customer, even multiple records for max total per customer), and other purchase fields (e.g. purchase_payment_id) will be selected for the real matching rows in the purchase table.

Supported by any database:

select * from purchase
join (
    select min(id) as id from purchase
    join (
        select customer, max(total) as total from purchase
        group by customer
    ) t1 using (customer, total)
    group by customer
) t2 using (id)
order by customer

This query is reasonably fast especially when there is a composite index like (customer, total) on the purchase table.


  1. t1, t2 are subquery alias which could be removed depending on database.

  2. Caveat: the using (...) clause is currently not supported in MS-SQL and Oracle db as of this edit on Jan 2017. You have to expand it yourself to e.g. on t2.id = purchase.id etc. The USING syntax works in SQLite, MySQL and PostgreSQL.

The solution is not very efficient as pointed by Erwin, because of presence of SubQs

select * from purchases p1 where total in
(select max(total) from purchases where p1.customer=customer) order by total desc;

Very fast solution

    purchases a 
    JOIN ( 
        SELECT customer, min( id ) as id 
        FROM purchases 
        GROUP BY customer 
    ) b USING ( id );

and really very fast if table is indexed by id:

create index purchases_id on purchases (id);

  • If you want to select any (by your some specific condition) row from the set of aggregated rows.

  • If you want to use another (sum/avg) aggregation function in addition to max/min. Thus you can not use clue with DISTINCT ON

You can use next subquery:

       SELECT **id** FROM t2   
       WHERE id = ANY ( ARRAY_AGG( tf.id ) ) AND amount = MAX( tf.amount )   
    ) id,  
    MAX(amount) ma,  
    SUM( ratio )  
FROM t2  tf  

You can replace amount = MAX( tf.amount ) with any condition you want with one restriction: This subquery must not return more than one row

But if you wanna to do such things you probably looking for window functions