这不是关系数据库的工作方式，但有时您必须做您必须做的事情。
您可以：

select a.name, b.age, c.city
from (select distinct name, row_number() over() as rn from t) a
full join (select distinct age, row_number() over() as rn from t) b on b.rn = a.rn
full join (select distinct city, row_number() over() as rn from t) c
  on c.rn = coalesce(a.rn, b.rn)

一个选项是聚合到数组中，然后取消嵌套这些数组：

select x.*
from (
  select array_agg(distinct name) as names, 
         array_agg(distinct age) as ages,
         array_agg(distinct city) as cities
  from the_table
) d
  cross join lateral unnest(d.names, d.ages, d.cities) with ordinality as x(name, age, city);

如果确实有许多不同的值（“millows”），我认为这会相当慢，但如果只期望很少的不同值（“bunders”或“thousand”），那么这可能是可以的。