计算 BigQuery 中数组中的匹配项数

Question

如何计算数组中的匹配项数？例如，对于数组 [1,2,3] 中的数字 [1,3]，将有 2 个匹配项，而对于数组 [1,2]，将有 1 个匹配项。现在我只能检查 [1,3] 是否在数组中。

  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`

我正在尝试得到这样的结果：

Answer 1

考虑以下方法

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

有输出

Answer 2

使用数学，以下似乎是可能的：

如果 arr 和 numbers 的联合与 arr 相同，它将是 numbers is in array
如果arr和numbers的并集大于arr，增加的元素不在arr.
因此，numbers_len - (union_len - arr_len) 将是 check

WITH `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
;

输出：

计算 BigQuery 中数组中的匹配项数

Count the number of matches in an array in BigQuery

sql

google-bigquery