SQL / Postgresql 用条件计算多列

Question

我有一个简单的表格：

id	gender	a_feature (bool)	b_feature (bool)	...	xyz_feature (bool)

我想根据性别对所有特征列求和。

metric	male	female
a_feature	345	3423
b_feature	65	143
...	...	...
xyz_feature	133	5536

有没有一种简单的方法可以做到这一点，例如使用 information_schema。

我只找到了下面的解决方案，但这很丑：

select
       'a_feature' as feature_name,
       count(case a_feature and gender = 'male') as male,
       count(case a_feature and gender = 'female') as female
from table
union
select
       b_feature as feature_name,
       count(case b_feature and gender = 'male') as male,
       count(case b_feature and gender = 'female') as female
from table
.
.
.
select
       xyz_feature as feature_name,
       count(case xyz_feature and gender = 'male') as male,
       count(case xyz_feature and gender = 'female') as female
from table

Answer 1

您可以取消透视和聚合。一种方法是：

select name,
       sum(case when feature and gender = 'male' then 1 else 0 end) as num_male,
       sum(case when feature and gender = 'female' then 1 else 0 end) as num_female
from ((select 'a_feature' as name, a_feature as feature, gender
       from t
      ) union all
      (select 'b_feature' as name, b_feature, gender
       from t
      ) union all
      . . .
     ) f
group by name;

在 Postgres 中，您可以使用横向连接进行反透视：

select name,
       sum(case when feature and gender = 'male' then 1 else 0 end) as num_male,
       sum(case when feature and gender = 'female' then 1 else 0 end) as num_female
from t cross join lateral
     (values ('a_feature', a_feature),
             ('b_feature', b_feature),
             . . .
     ) v(name, feature)
group by name;

如果您不愿意全部输入，可以使用information_schema.columns 生成values() 的列表。

编辑：

您可以使用以下方式构造values 子句：

select string_agg('(''' || column_name || ''', column_name)', ', ') 
from information_schema.columns
where table_name = ?