您好我正在尝试使用以下hive语句找到高于平均值的计数
Select x, Count(x) as y from data
group by x
Having Count(x) >= (select Avg(z.count1) as aveg
from (select x, Count(x) as count1 from data group by x ) z) ;
我收到错误,因为ParseException第1:87行无法识别'Select''Avg''附近的输入(''在表达式规范中
答案 0 :(得分:0)
select x
,cnt_x
from (select x
,count(x) as cnt_x
,avg (count(x)) over () as avg_cnt_x
from data
group by x
) t
where cnt_x >= avg_cnt_x