PostgreSQL UNION 不能正确合并行

Question

我在 PostgreSQL 数据库中有 3 个表：

localities (loc, 12561 rows)
plants (pl, 17052 rows)
specimens or samples (esp, 9211 rows)

pl 和 esp 都有一个字段 loc，用于指定标记的植物生活在哪里，或者该样本（通常是带有叶子和花朵的树枝）来自哪里。

我需要一份关于有植物或样本的地方的报告，以及每个地方的植物和样本的数量。到目前为止我做的最好的是两个子查询的联合，它运行得非常快（33 毫秒来获取 69 行）：

(select l.id,l.nome,count(pl.id) pls,null esps
from loc l
left join pl on pl.loc = l.id
where l.id in
    (select distinct pl.loc
    from pl
    where pl.loc > 0)
group by l.id,l.nome
union
select l.id,l.nome,null pls,count(e.id) esps
from loc l
left join esp e on e.loc = l.id
where l.id in
    (select distinct e.loc
    from esp e
    where e.loc > 0)
group by l.id,l.nome)
order by id

重点是，当同一个地方既有植物又有样品，就变成了两条截然不同的线，比如：

11950 | San Martin |    | 5 |
11950 | San Martin | 61 |   |

当然我想要的是：

11950 | San Martin | 61 | 5 |

在此之前，我尝试过在一个查询中完成所有操作：

select l.id,l.nome,count(pl.id),count(e.id) esps
from loc l
left join pl on pl.loc = l.id
left join esp e on e.loc = l.id
where l.id in
    (select distinct pl.loc
    from pl
    where pl.loc > 0)
or l.id in
    (select distinct e.loc
    from esp e
    where e.loc > 0)
group by l.id,l.nome

但它会返回一个奇怪的重复（它将两个结果相乘并显示两次）：

11950 | San Martin | 305 | 305 |

我尝试过不使用子查询，但花了大约 13 秒，这太长了。

Answer 1

我创建了测试布局：

create table localities (id integer, loc_name text);
create table plants (plant_id integer, loc_id integer);
create table samples (sample_id integer, loc_id integer);

insert into localities select x, ('Loc ' || x::text) from generate_series(1, 12561) x ;
insert into plants select x, (random()*12561)::integer from generate_series(1, 17052) x;
insert into samples select x, (random()*12561)::integer from generate_series(1, 9211) x;

诀窍是从植物和样本中创建一个具有相同结构的中间表。如果数据没有意义（植物没有 sample_id），则添加 null：

select loc_id, plant_id, null as sample_id from plants
union all 
select loc_id, null as plant_id, sample_id from samples

此表具有统一的结构，然后您可以在其上进行聚合（我使用WITH 使其更具可读性。）：

with localities_used as (
        select loc_id, plant_id, null as sample_id from plants
        union all 
        select loc_id, null as plant_id, sample_id from samples)
select
    localities_used.loc_id,
    count(localities_used.plant_id) plant_count,
    count(localities_used.sample_id) sample_count
from
    localities_used
group by 
    localities_used.loc_id;

如果您需要来自地区的其他数据，可以将它们加入汇总表中：

with localities_used as (
        select loc_id, plant_id, null as sample_id from plants
        union all 
        select loc_id, null as plant_id, sample_id from samples),
    aggregated as (
        select
            localities_used.loc_id,
            count(localities_used.plant_id) plant_count,
            count(localities_used.sample_id) sample_count
        from
            localities_used
        group by 
            localities_used.loc_id)
select * from aggregated left outer join localities on aggregated.loc_id = localities.id;

这在我的笔记本电脑上总共需要 75 毫秒。

Answer 2

我没有对此进行测试，但我认为可能是这样的：

SELECT 
   l.id,
   l.nome,
   SUM(CASE WHEN pl.id IS NOT NULL THEN 1 ELSE 0 END) as plants_count,
   SUM(CASE WHEN e.id IS NOT NULL THEN 1 ELSE 0 END) as esp_count
FROM loc l
LEFT JOIN pl ON pl.loc = l.id
LEFT JOIN esp e ON e.loc = l.id
GROUP BY l.id,l.nome

重点是统计每种类型的非空id。

Answer 3

这应该很简单

select * from (
  select 
    location.*, 
    (select count(id) from plant where plant.location = location.id) as plants, 
    (select count(id) from sample where sample.location = location.id) as samples 
  from location
) subquery 
where subquery.plants > 0 or subquery.samples > 0;

 id |    name    | plants | samples 
----+------------+--------+---------
  1 | San Martin |      2 |       1
  2 | Rome       |      1 |       2
  3 | Dallas     |      3 |       1
(3 rows)

这是我快速设置的用于试验的数据库：

create table location(id serial primary key, name text);
create table plant(id serial primary key, name text, location integer references location(id));
create table sample(id serial primary key, name text, location integer references location(id));

insert into location (name) values ('San Martin'), ('Rome'), ('Dallas'), ('Ghost Town');
insert into plant (name, location) values ('San Martin Dandelion', 1),('San Martin Camomile', 1), ('Rome Raspberry', 2), ('Dallas Locoweed', 3), ('Dallas Lemongrass', 3), ('Dallas Setaria', 3);
insert into sample (name, location) values ('San Martin Bramble', 1), ('Rome Iris', 2), ('Rome Eucalypt', 2), ('Dallas Dogbane', 3);


tests=# select * from location;
id |    name
----+------------
1 | San Martin
2 | Rome
3 | Dallas
4 | Ghost Town
(4 rows)

tests=# select * from plant;
id |         name         | location
----+----------------------+----------
1 | San Martin Dandelion |        1
2 | San Martin Camomile  |        1
3 | Rome Raspberry       |        2
4 | Dallas Locoweed      |        3
5 | Dallas Lemongrass    |        3
6 | Dallas Setaria       |        3
(6 rows)

tests=# select * from sample;
id |        name        | location
----+--------------------+----------
1 | San Martin Bramble |        1
2 | Rome Iris          |        2
3 | Rome Eucalypt      |        2
4 | Dallas Dogbane     |        3
(4 rows)