排除匹配的数组元素

Question

如何从另一个数组中排除匹配的元素？

Postgres 代码：

a1 := '{1, 2, 5, 15}'::int[];
a2 := '{1, 2, 3, 6, 7, 9, 15}'::int[];

a3 := a2 ??magic_operator?? a1;

在a3 中，我期望正好是 '{3, 6, 7, 9}'

最终结果

我和 lad2025 的解决方案运行良好。

array_position() 的解决方案需要 PostgreSQL 9.5 及更高版本，执行速度更快。

Answer 1

在数组之间看起来像XOR：

WITH set1 AS
(
 SELECT * FROM unnest('{1, 2, 5, 15}'::int[])
), set2 AS
(
 SELECT * FROM unnest('{1, 2, 3, 6, 7, 9, 15}'::int[])
), xor AS
(
  (SELECT * FROM set1
   UNION 
   SELECT * FROM set2)
  EXCEPT
  (SELECT * FROM set1
   INTERSECT 
   SELECT * FROM set2)
)
SELECT array_agg(unnest ORDER BY unnest)
FROM xor

输出：

"{3,5,6,7,9}"

它是如何工作的：

1. 取消嵌套两个数组
2. 计算总和
3. 计算相交
4. 从 SUM - INTERSECT
5. 组合成数组

或者，您可以使用两个减（除）操作的总和：

(A+B) - (A^B)
<=>
(A-B) + (B-A)

---

利用FULL JOIN:

WITH set1 AS
(
 SELECT *
FROM unnest('{1, 2, 5, 15}'::int[])
), set2 AS
(
 SELECT *
 FROM unnest('{1, 2, 3, 6, 7, 9, 15}'::int[])
)
SELECT array_agg(COALESCE(s1.unnest, s2.unnest) 
                 ORDER BY COALESCE(s1.unnest, s2.unnest))
FROM set1 s1
FULL JOIN set2 s2
  ON s1.unnest = s2.unnest
WHERE s1.unnest IS NULL
  OR s2.unnest IS NULL;

编辑：

如果您只想要第二个数组中不是的元素，请首先使用简单的EXCEPT：

SELECT array_agg(unnest ORDER BY unnest)
FROM (SELECT * FROM unnest('{1, 2, 3, 6, 7, 9, 15}'::int[])
      EXCEPT
      SELECT * FROM unnest('{1, 2, 5, 15}'::int[])) AS sub

输出：

"{3,6,7,9}"

Answer 2

附加模块intarray 为整数数组提供了一个简单快速的subtraction operator -，正是您正在寻找的magic_operator：

test=# SELECT '{1, 2, 3, 6, 7, 9, 15}'::int[] - '{1, 2, 5, 15}'::int[] AS result;
 ?column?
-----------
 {3,6,7,9}

您需要为每个数据库安装一次模块：

CREATE EXTENSION intarray;

它还为索引提供了特殊的运算符类：

• Postgresql intarray error: undefined symbol: pfree

注意only works for:

...无空的整数数组。

Answer 3

找了个小相似的case，修改一下。

那个 SQL 解决了我的问题。

with elements (element) as (
   select unnest(ARRAY[1, 2, 3, 6, 7, 9, 15])
)
select array_agg(element)
from elements
where array_position(ARRAY[1, 2, 5, 15],element) is null

需要 PostgreSQL 9.5 及更高版本。