How to select the first row in each GROUP BY group? [Answered]

Problem explanation:

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 the table that looks like this:

SELECT * FROM purchases;

My Output:


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:


How to select the first row in each GROUP BY group? Answer #1:

DISTINCT ON is typically simplest and fastest for this in PostgreSQL.
(For performance optimization for certain workloads see 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 in ORDER BY must be in the set of expressions in DISTINCT ON, but you can rearrange order among those freely. Example.
You can add additional expressions to ORDER BY to pick a particular row from each group of peers. Or, as the manual puts it:

The DISTINCT ON expression(s) must match the leftmost ORDER BY expression(s). The ORDER BY clause will normally contain additional expression(s) that determine the desired precedence of rows within each DISTINCT ON group.

I added id as last item to break ties:
“Pick the row with the smallest id from each group sharing the highest total.”

To order results in a way that disagrees with the sort order determining the first per group, you can nest above query in an outer query with another ORDER BYExample.

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

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 – 13. 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. But use it if read performance for the particular query is crucial. If you have DESC NULLS LAST in the query, use the same in the index so that sort order matches and the index is applicable.

Effectiveness / Performance optimization

Weigh cost and benefit before creating tailored indexes for each query. The potential of above index largely depends on data distribution.

The index is used because it delivers pre-sorted data. 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.

For few rows per customer (high cardinality in column customer), this is very efficient. Even more so if you need sorted output anyway. The benefit shrinks with a growing number of rows per customer.
Ideally, you have enough work_mem to process the involved sort step in RAM and not spill to disk. But generally setting work_mem too high can have adverse effects. Consider SET LOCAL for exceptionally big queries. Find how much you need with EXPLAIN ANALYZE. Mention of “Disk:“.

For many rows per customer (low cardinality in column customer), a loose index scan (a.k.a. “skip scan”) would be (much) more efficient, but that’s not implemented up to Postgres 14. (An implementation for index-only scans is in development for Postgres 15. See here and here.)
For now, there are faster query techniques to substitute for this. In particular if you have a separate table holding unique customers, which is the typical use case.

Answer #2:

On databases that support CTE and windowing functions:

WITH summary AS (
           ROW_NUMBER() OVER(PARTITION BY p.customer 
                                 ORDER BY DESC) AS rank
   FROM summary
 WHERE rank = 1

Supported by any database:

But you need to add logic to break ties:

  SELECT MIN(,  -- 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 =
GROUP BY x.customer,

Answer #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).

Added an accented test for Postgres 13 below.


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 a query taken from this related answer:

               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

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;


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

4. rCTE with LATERAL subquery

   (  -- 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 

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 

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. Same as A. 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

Retest with Postgres 13 on 2021-08-11

Simplified test setup: not deleting rows, because VACUUM ANALYZE cleans the table completely for the simple case.

Important changes:

  • General performance improvements.
  • CTEs can be inlined since Postgres 12, so query 1. and 2. now perform mostly identical (same query plan).

D. Like B. ~ 20 rows per customer_id

1. 103 ms
2. 103 ms  
3.  23 ms  -- winner  
4.  71 ms  
5.  22 ms  -- winner
6.  81 ms  

E. Like C. ~ 2.3 rows per customer_id

1. 127 ms
2. 126 ms  
3.  36 ms  -- winner  
4. 620 ms  
5. 145 ms
6. 203 ms  

Accented tests with Postgres 13

1M rows, 10.000 vs. 100 vs. 1.6 rows per customer.

F. with ~ 10.000 rows per customer

1. 526 ms
2. 527 ms  
3. 127 ms
4.   2 ms  -- winner !
5.   1 ms  -- winner !
6. 356 ms  

G. with ~ 100 rows per customer

1. 535 ms
2. 529 ms  
3. 132 ms
4. 108 ms  -- !
5.  71 ms  -- winner
6. 376 ms  

H. with ~ 1.6 rows per customer

1.  691 ms
2.  684 ms  
3.  234 ms  -- winner
4. 4669 ms
5. 1089 ms
6. 1264 ms  


  • DISTINCT ON uses the index effectively and typically performs best for few rows per group. And it performs decently even with many rows per group.
  • For many rows per group, emulating an index skip scan with an rCTE performs best – second only to the query technique with a separate lookup table (if that’s available).
  • Using row_number() (technique of the currently accepted answer) never wins any performance test. Not then, not now. It never comes even close to DISTINCT ON, not even when the data distribution is unfavorable for it. The only good thing about it: it does not scale terribly, just mediocre.

Related benchmarks

Here is a new one by “ogr” testing with 10M rows and 60k unique “customers” on Postgres 11.5 (current as of Sep. 2019).

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 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
  1. Use condition WHERE customer BETWEEN x AND y resulting in 1000 rows.
A: 249.136 ms
B:  55.111 ms
  1. 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

Answer #4:

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.

Answer #5:

The Query:

SELECT purchases.*
FROM purchases
LEFT JOIN purchases as p 
  p.customer = purchases.customer 
  AND <

HOW DOES THAT WORK! (I’ve been there)

We want to make sure that we only have the highest total for each purchase.

Some Theoretical Stuff (skip this part if you only want to understand the query)

Let Total be a function T(customer,id) where it returns a value given the name and id To prove that the given total (T(customer,id)) is the highest we have to prove that We want to prove either

  • ∀x T(customer,id) > T(customer,x) (this total is higher than all other total for that customer)


  • ¬∃x T(customer, id) < T(customer, x) (there exists no higher total for that customer)

The first approach will need us to get all the records for that name which I do not really like.

The second one will need a smart way to say there can be no record higher than this one.

Back to SQL

If we left joins the table on the name and total being less than the joined table:

LEFT JOIN purchases as p 
p.customer = purchases.customer 

we make sure that all records that have another record with the higher total for the same user to be joined:

| |  purchases.customer | | | p.customer | |
|            1 | Tom                 |             200 |    2 | Tom        |     300 |
|            2 | Tom                 |             300 |      |            |         |
|            3 | Bob                 |             400 |    4 | Bob        |     500 |
|            4 | Bob                 |             500 |      |            |         |
|            5 | Alice               |             600 |    6 | Alice      |     700 |
|            6 | Alice               |             700 |      |            |         |

That will help us filter for the highest total for each purchase with no grouping needed:

| | | | | | |
|            2 | Tom            |             300 |      |        |         |
|            4 | Bob            |             500 |      |        |         |
|            6 | Alice          |             700 |      |        |         |

And that’s the answer we need.

Hope you learned something from this post.

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