如何根据 SQL 中的条件 select 70% 的列？

How to select 70 percent of a column based on condition in SQL?

这是我现有的 table 数据

C1 C2 C3
1  A  1                                                                                                  
2  B  1 
3  C  0 
4  D  0
5  E  0
6  F  0
7  G  1
8  H  1
9  I  1
10 J  0

我想要这个。我正在尝试的是我想要 select 70% C3 列，值为 1。C3 总共有五个。所以 5 的 70% 是 3.5，也就是 4。所以我想得到我的最终数据集，其中 70% 在 C3

C1 C2 C3
1  A  1  
2  B  1 
3  C  0 
4  D  0
5  E  0
7  G  1
8  H  1

嗯。您似乎不需要 random 选择。它们似乎是按 col1 排序的。所以，你可以这样计算：

select t.*
from (select t.*,
             sum(case when col3 = 1 then 1 else 0 end) over (order by col1) as running_col3,
             sum(case when col3 = 1 then 1 else 0 end) over () as total_col3
      from t
     ) t
where running_col3 >= 0.8 * total_col3 and
      (running_col3 - col3) < 0.8 * total_col3;

注意：如果col3只有0和1，可以将上面的简化为：

select t.*
from (select t.*,
             sum(col3) over (order by col1) as running_col3,
             sum(col3) over () as total_col3
      from t
     ) t
where running_col3 >= 0.8 * total_col3 and
      (running_col3 - col3) < 0.8 * total_col3

答案在这里

select * 
from 
    (SELECT *,
        (SELECT SUM(C3) FROM table_name t1 WHERE t1.C1 <= t.C1) AS cumulative_sum,
        (select sum(C3) from table_name) as total_sum
    FROM table_name t) t
where (cumulative_sum - C3) < 0.8 * total_sum