Compare arrays for equality, ignoring order of elements - arrays

Question

I have a table with 4 array columns.. the results are like:
ids signed_ids new_ids new_ids_signed
{1,2,3} | {2,1,3} | {4,5,6} | {6,5,4}
Anyway to compare ids and signed_ids so that they come out equal, by ignoring the order of the elements?

Answer 1

You can use contained by operator:
(array1 <# array2 and array1 #> array2)

Answer 2

The additional module intarray provides operators for arrays of integer, which are typically (much) faster. Install once per database with (in Postgres 9.1 or later):
CREATE EXTENSION intarray;
Then you can:
SELECT uniq(sort(ids)) = uniq(sort(signed_ids));
Or:
SELECT ids #> signed_ids AND ids <# signed_ids;
Bold emphasis on functions and operators from intarray.
In the second example, operator resolution arrives at the specialized intarray operators if left and right argument are type integer[].
Both expressions will ignore order and duplicity of elements. Further reading in the helpful manual here.
intarray operators only work for arrays of integer (int4), not bigint (int8) or smallint (int2) or any other data type.
Unlike the default generic operators, intarray operators do not accept NULL values in arrays. NULL in any involved array raises an exception. If you need to work with NULL values, you can default to the standard, generic operators by schema-qualifying the operator with the OPERATOR construct:
SELECT ARRAY[1,4,null,3]::int[] OPERATOR(pg_catalog.#>) ARRAY[3,1]::int[]
The generic operators can't use indexes with an intarray operator class and vice versa.
Related:
GIN index on smallint[] column not used or error "operator is not unique"

Answer 3

The simplest thing to do is sort them and compare them sorted. See sorting arrays in PostgreSQL.
Given sample data:
CREATE TABLE aa(ids integer[], signed_ids integer[]);
INSERT INTO aa(ids, signed_ids) VALUES (ARRAY[1,2,3], ARRAY[2,1,3]);
the best thing to do is to if the array entries are always integers is to use the intarray extension, as Erwin explains in his answer. It's a lot faster than any pure-SQL formulation.
Otherwise, for a general version that works for any data type, define an array_sort(anyarray):
CREATE OR REPLACE FUNCTION array_sort(anyarray) RETURNS anyarray AS $$
SELECT array_agg(x order by x) FROM unnest($1) x;
$$ LANGUAGE 'SQL';
and use it sort and compare the sorted arrays:
SELECT array_sort(ids) = array_sort(signed_ids) FROM aa;
There's an important caveat:
SELECT array_sort( ARRAY[1,2,2,4,4] ) = array_sort( ARRAY[1,2,4] );
will be false. This may or may not be what you want, depending on your intentions.
Alternately, define a function array_compare_as_set:
CREATE OR REPLACE FUNCTION array_compare_as_set(anyarray,anyarray) RETURNS boolean AS $$
SELECT CASE
WHEN array_dims($1) <> array_dims($2) THEN
'f'
WHEN array_length($1,1) <> array_length($2,1) THEN
'f'
ELSE
NOT EXISTS (
SELECT 1
FROM unnest($1) a
FULL JOIN unnest($2) b ON (a=b)
WHERE a IS NULL or b IS NULL
)
END
$$ LANGUAGE 'SQL' IMMUTABLE;
and then:
SELECT array_compare_as_set(ids, signed_ids) FROM aa;
This is subtly different from comparing two array_sorted values. array_compare_as_set will eliminate duplicates, making array_compare_as_set(ARRAY[1,2,3,3],ARRAY[1,2,3]) true, whereas array_sort(ARRAY[1,2,3,3]) = array_sort(ARRAY[1,2,3]) will be false.
Both of these approaches will have pretty bad performance. Consider ensuring that you always store your arrays sorted in the first place.

Answer 4

If your arrays have no duplicates and are of the same dimension:
use array contains #>
AND array_length where the length must match the size you want on both sides

Answer 5

select (string_agg(a,',' order by a) = string_agg(b,',' order by b)) from
(select unnest(array[1,2,3,2])::text as a,unnest(array[2,2,3,1])::text as b) A