How I can count the number of matches in an array? For example, for numbers [1,3] in the array [1,2,3] there will be 2 matches, and for the array [1,2] there will be 1 match. Right now I can only check if [1,3] is in the array or not.
WITH `arrays` AS (
SELECT 1 id, [1,2,3] as arr
UNION ALL
SELECT 2, [1,2]
UNION ALL
SELECT 3, [3]
)
SELECT id, arr, [1,3] as numbers,
CASE
1 IN UNNEST(arr) and
3 IN UNNEST(arr)
WHEN TRUE THEN 'numbers is in array'
ELSE 'numbers is not in array'
END conclusion
FROM `arrays`
I'm trying to get such result:

Consider below approach
with `arrays` as (
select 1 id, [1,2,3] as arr union all
select 2, [1,2] union all
select 3, [3]
)
select *,
( select count(*)
from t.numbers num join t.arr num
using(num)
) check,
( select format('number is %sin array',
if(logical_and(if(num2 is null, false, true)), '', 'not '))
from t.numbers num1 left join t.arr num2
on num1 = num2
) conclusion
from (
select id, arr, [1,3] as numbers
from `arrays`
) t
with output

Using a math, following seems to be possible:
arr and numbers is same as arr, it will be numbers is in arrayarr and numbers is greater than arr, elements as much as the increased number is not in the arr.numbers_len - (union_len - arr_len) will be checkWITH `arrays` AS (
SELECT 1 id, [1,2,3] as arr
UNION ALL
SELECT 2, [1,2]
UNION ALL
SELECT 3, [3]
),
calculated_arrays AS (
SELECT *, [1,3] as numbers,
ARRAY_LENGTH(ARRAY(SELECT DISTINCT * FROM UNNEST(arr || [1, 3]))) AS union_len,
ARRAY_LENGTH(arr) AS arr_len,
ARRAY_LENGTH([1, 3]) AS numbers_len
FROM `arrays`
)
SELECT id, arr, numbers,
numbers_len - union_len + arr_len AS check,
IF (union_len = arr_len, 'numbers is in array', 'numbers is not in array') AS conclusion
FROM calculated_arrays
;
output:

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