39

I'm looking for a first() aggregate function.
Here I found something that almost works:

CREATE OR REPLACE FUNCTION public.first_agg (anyelement, anyelement)
RETURNS anyelement LANGUAGE sql IMMUTABLE STRICT AS $$
    SELECT $1;
$$;

-- And then wrap an aggregate around it
CREATE AGGREGATE public.first (
    sfunc    = public.first_agg,
    basetype = anyelement,
    stype    = anyelement
);

The problem is that when a varchar(n) column passes through the first() function, it's converted into simple varchar (without size modifier). Trying to return the query in a function as RETURNS SETOF anyelement, I get the following error:

ERROR: structure of query does not match function result type Estado de
SQL:42804
Detalhe:Returned type character varying does not match expected type character varying(40) in column 2.
Contexto:PL/pgSQL function vsr_table_at_time(anyelement,timestamp without time zone) line 31 at RETURN QUERY

In the same wiki page there is a link to a C Version of the function that would replace the above. I don't know how to install it, but I wonder if this version could solve my problem.

Meanwhile, is there a way I can change the above function so it returns the exact same type of the input column?

Erwin Brandstetter
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Alexandre Neto
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5 Answers5

45

DISTINCT ON()

Just as a side note, this is precisely what DISTINCT ON() does (not to be confused with DISTINCT)

SELECT DISTINCT ON ( expression [, ...] ) keeps only the first row of each set of rows where the given expressions evaluate to equal. The DISTINCT ON expressions are interpreted using the same rules as for ORDER BY (see above). Note that the "first row" of each set is unpredictable unless ORDER BY is used to ensure that the desired row appears first. For example

So if you were to write,

SELECT myFirstAgg(z)
FROM foo
GROUP BY x,y;

It's effectively 

SELECT DISTINCT ON(x,y) z
FROM foo;
-- ORDER BY z;

In that it takes the first z. There are two important differences,

  1. You can also select other columns at no cost of further aggregation..

    SELECT DISTINCT ON(x,y) z, k, r, t, v
FROM foo;
-- ORDER BY z, k, r, t, v;

  2. Because there is no GROUP BY you can not use (real) aggregates with it.

    CREATE TABLE foo AS
SELECT * FROM ( VALUES
  (1,2,3),
  (1,2,4),
  (1,2,5)
) AS t(x,y,z);

SELECT DISTINCT ON (x,y) z, sum(z)
FROM foo;

-- fails, as you should expect.
SELECT DISTINCT ON (x,y) z, sum(z)
FROM foo;

-- would not otherwise fail.
SELECT myFirstAgg(z), sum(z)
FROM foo
GROUP BY x,y;

Don't forget ORDER BY

Also, while I didn't bold it then I will now

Note that the "first row" of each set is unpredictable unless ORDER BY is used to ensure that the desired row appears first. For example

Always use an ORDER BY with DISTINCT ON

Using an Ordered-Set Aggregate Function

I imagine a lot of people are looking for first_value, Ordered-Set Aggregate Functions. Just wanted to throw that out there. It would look like this, if the function existed:

SELECT a, b, first_value() WITHIN GROUP (ORDER BY z)    
FROM foo
GROUP BY a,b;

But, alas you can do this.

SELECT a, b, percentile_disc(0) WITHIN GROUP (ORDER BY z)   
FROM foo
GROUP BY a,b;
Evan Carroll
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17

Yay, I've found out an easy way with your case by using some features in PostgreSQL 9.4+

Let's see this example:

select  (array_agg(val ORDER BY i))[1] as first_value_orderby_i,
    (array_agg(val ORDER BY i DESC))[1] as last_value_orderby_i,
    (array_agg(val))[1] as last_value_all,
    (array_agg(val))[array_length(array_agg(val),1)] as last_value_all
   FROM (
        SELECT i, random() as val
        FROM generate_series(1,100) s(i)
        ORDER BY random()
    ) tmp_tbl

I hope it will help you at your case.

Michael Green
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Mabu Kloesen
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8

Not a direct answer to your question but you should try the first_value window function. It works like this:

CREATE TABLE test (
    id SERIAL NOT NULL PRIMARY KEY,
    cat TEXT,
    value VARCHAR(2)
    date TIMESTAMP WITH TIME ZONE

);

Then, if you want the first item in each cat (category) you will query like that:

SELECT
    cat,
    first_value(date) OVER (PARTITION BY cat ORDER BY date)
FROM
    test;

or:

SELECT
    cat,
    first_value(date) OVER w
FROM
    test
WINDOW w AS (PARTITION BY cat ORDER BY date);
Ghislain Leveque
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1

Get the first value of a column and order by another in a group by

To get the first value of a column and order by another, you can:

The column to order by should be the first field in the type definition.

Note that, if the ordering column and the value column have the same type you can skip the type creation and use an ARRAY instead.

Example:

For example, if we want to find, for each group id, the intial value of column value based on timestamp time. We can define the type

CREATE TYPE time_value AS (
  time TIMESTAMP,  -- ordering column
  value NUMERIC
)

and then aggregate as follows:

SELECT
  id,
  (
    PERCENTILE_DISC(0) WITHIN GROUP (ORDER BY (time, value)::time_value)
  ).value AS first_value
FROM my_table
GROUP BY id

With a value column time_value that has the same type as the ordering column time, we can skip the type creation. The aggregation is then as follows:

SELECT
  id,
  (
    PERCENTILE_DISC(0) WITHIN GROUP (ORDER BY ARRAY[time, time_value])
  )[2] AS first_time_value
FROM my_table
GROUP BY id

Test data:

CREATE TABLE my_table AS SELECT
    FLOOR(2 * RANDOM()) AS id,
    TIMESTAMP '2023-01-01' + time * INTERVAL '1 hour' AS time,
    FLOOR(10 * RANDOM()) AS value,
    TIMESTAMP '2023-01-01' + FLOOR(10 * RANDOM()) * INTERVAL '1 hour' AS time_value
FROM GENERATE_SERIES(1, 10) AS time
MaFF
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0

I'm sure the question is no more actual for the topic starter, but for those who looking solution in 2021 there is a simple trick - array_agg

e.g. :

SELECT (array_agg(t.f0) FILTER (WHERE t.f0 IS NOT NULL))[1] agg FROM t GROUP BY t.f1

foxen
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