3

假设我有一个这样的表：

  link_ids  |  length
------------+-----------
 {1,4}      | {1,2}
 {2,5}      | {0,1}

如何找到每个的最小长度link_ids？

所以最终的输出看起来像：

  link_ids  |  length
------------+-----------
 {1,4}      | 1
 {2,5}      | 0
4

5 回答 5

8

假设一个像这样的表：

CREATE TABLE tbl (
  link_ids int[] PRIMARY KEY     -- which is odd for a PK
, length int[]
, CHECK (length <> '{}'::int[])  -- rules out null and empty in length
);

查询 Postgres 9.3 或更高版本：

SELECT link_ids, min(len) AS min_length
FROM   tbl t, unnest(t.length) len  -- implicit LATERAL join
GROUP  BY 1;

或者创建一个小函数（Postgres 8.4+）：

CREATE OR REPLACE FUNCTION arr_min(anyarray)
  RETURNS anyelement LANGUAGE sql IMMUTABLE PARALLEL SAFE AS
'SELECT min(i) FROM unnest($1) i';

仅PARALLEL SAFE在 Postgres 9.6 或更高版本中添加。然后：

SELECT link_ids, arr_min(length) AS min_length FROM t;

该函数可以内联并且速度很快。

或者，对于简单长度integer的数组，使用附加模块及其内置函数（Postgres 8.3+）：intarraysort()

SELECT link_ids, (sort(length))[1] AS min_length FROM t;
于 2015-02-10T00:46:20.417 回答
3

假设表名是t并且每个值link_ids都是唯一的。

select link_ids, min(len)
from (select link_ids, unnest(length) as len from t) as t
group by link_ids;

 link_ids | min
----------+-----
 {2,5}    |   0
 {1,4}    |   1
于 2015-02-10T00:23:41.530 回答
2

Erwin 回答的一个小补充——有时子查询unnest比横向连接更便宜。

我使用了 Erwin 的回答中的表定义并填写了它：

t=# insert into t select '{1}'::int[]||g,'{1}'::int[]||g from generate_series(1,9999,1) g;
INSERT 0 9999
t=# select * from t order by ctid desc limit 1;
 link_ids |  length
----------+----------
 {1,9999} | {1,9999}
(1 row)

然后分析LATERAL JOIN：

t=# explain analyze select link_ids,max(r) from t, unnest(length) r where link_ids = '{1,9999}' group by 1;
                                                      QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------
 GroupAggregate  (cost=0.29..4.81 rows=1 width=33) (actual time=0.030..0.030 rows=1 loops=1)
   ->  Nested Loop  (cost=0.29..4.30 rows=100 width=33) (actual time=0.025..0.027 rows=2 loops=1)
         ->  Index Scan using t_pkey on t  (cost=0.29..2.30 rows=1 width=58) (actual time=0.015..0.016 rows=1 loops=1)
               Index Cond: (link_ids = '{1,9999}'::integer[])
         ->  Function Scan on unnest r  (cost=0.00..1.00 rows=100 width=4) (actual time=0.007..0.007 rows=2 loops=1)
 Total runtime: 0.059 ms
(6 rows)

并尝试子查询：

t=# explain analyze select link_ids, (select max(r) from unnest(length) r) from t where link_ids = '{1,9999}';
                                                      QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------
 Index Scan using t_pkey on t  (cost=0.29..3.56 rows=1 width=58) (actual time=0.030..0.031 rows=1 loops=1)
   Index Cond: (link_ids = '{1,9999}'::integer[])
   SubPlan 1
     ->  Aggregate  (cost=1.25..1.26 rows=1 width=4) (actual time=0.011..0.011 rows=1 loops=1)
           ->  Function Scan on unnest r  (cost=0.00..1.00 rows=100 width=4) (actual time=0.008..0.008 rows=2 loops=1)
 Total runtime: 0.060 ms
(6 rows)

最后确保结果相同：

t=# select link_ids, (select max(r) from unnest(length) r) 
from t 
where link_ids = '{1,9999}';
 link_ids | max
----------+------
 {1,9999} | 9999
(1 row)

t=# select link_ids,max(r) 
from t, unnest(length) r 
where link_ids = '{1,9999}' 
group by 1;
 link_ids | max
----------+------
 {1,9999} | 9999
(1 row)
于 2017-09-15T08:59:18.927 回答
1

对于数组的最小值：

SELECT max(x) from unnest(array_name) as x;
于 2019-03-13T09:02:06.437 回答
0

（我会假设link_ids可以有双打，因为没有 id 列，我们将即兴创作）。

WITH r AS
(SELECT row_number() OVER() as id,
       link_ids,
       length from Table1)
SELECT DISTINCT ON (id) link_ids,
       unnest(length) 
FROM r 
ORDER BY id, length;

小提琴

于 2015-02-10T00:18:07.947 回答