我一直被困在这一点上,但我的最终目标是获得总体数据和分组日期(每日)以及类别的负值、正值和中性值的百分比。谢谢。
发布于 2018-07-24 09:12:28
只需使用窗口函数:
select mlsentimentzone,
(count(*) * 1.0 / sum(count(*)) over ()) as ratio
from t
group by mlsentimentzone;
或者,如果您想要此日期,请使用条件聚合:
select date,
avg(case when mlsentimentzone = 'negative' then 1.0 else 0.0 end) as negative,
avg(case when mlsentimentzone = 'neutral' then 1.0 else 0.0 end) as neutral,
avg(case when mlsentimentzone = 'positive' then 1.0 else 0.0 end) as positive
from t
group by date
order by date;
https://stackoverflow.com/questions/51489103
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