如何比较一行中的一个值，看它是否高于同一列中所有值的 75%？

Question

我有一个 table 看起来像这样：

groups  created_utc            score    count_comments  d_posts ups  downs   ratio
group1  2011-07-11T19:05:19Z    6988    3742             56     8530  1572   .42(8530/20286)
group2  2011-04-23T21:29:12Z    10455   4695             512    11756 1303   .58(11756/20286)

从此查询生成：

SELECT *, ups  / SUM(ups) OVER () AS ratio
FROM table
order by ratio desc;

如何逐行比较 ratio 中的每个值以查看该比率是否大于所有比率的 75% 以创建新的标志列，greater_75p?

新的 table 应该是这样的（无法格式化新的列，但应该是 y/n 作为选项）：

groups  created_utc            score    count_comments  d_posts ups  downs   ratio                greater_75p
                                                                                                   y 
group1  2011-07-11T19:05:19Z    6988    3742             56     8530  1572   .42(8530/20286)
group2  2011-04-23T21:29:12Z    10455   4695             512    11756 1303   .58(11756/20286)

我试过这个查询，但得到错误 Scalar subquery produced more than one element:

SELECT *,ups * 100 / SUM(ups) OVER () AS ratio, 
PERCENT_RANK() OVER(order by (SELECT ups * 100 / SUM(ups) OVER () AS ratio from table )) AS greater_75p
FROM table

不确定我做错了什么以及如何得出 sql 内的百分比比较？

提前致谢。

Answer 1

要获得 percent_rank() 的结果，您可以使用如下常见的 table 表达式：

with cte as
(SELECT *, ups  / SUM(ups) OVER () AS ratio
FROM table) 
select *,(case when percent_rank()over(order by ration) >0.75 then 'yes' else 'no' end) greater_75p from cte

请阐明计算 greater_75p 列的逻辑。

如何比较一行中的一个值，看它是否高于同一列中所有值的 75%？

How to compare one value in a row to see if it is higher than 75% of all values in the same column?

sql

google-bigquery

comparison