所以你想：

select count(*) - count(distinct neighbourhood)
from Neighbourhood;

我可以建议进一步细分：

select cnt, count(*), min(neighbourhood), max(neighbourhood)
from (select neighbourhood, count(*) as cnt
      from neighbourhood
      group by neighbourhood
     ) n
group by cnt
order by cnt;

这显示了复制的频率和示例值 neighbourhood .

使用 count(distinct neighbourhood) 以及 distinct 不是函数

select count(*)- select count(distinct neighbourhood) from Neighbourhood