如何根据SQL中的条件选择列的70％？

问题描述

这是我现有的表格数据

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。我正在尝试选择70％的C3列，其值为1。总共C3有五个列。所以5的70％是3.5，也就是4。所以我想获得最终数据集，其中包含C3中70％的数据集

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

解决方法

这是答案

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

,

嗯。您似乎不需要随机选择。它们似乎由col1排序。因此，您可以将其计算为：

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，则可以将以上内容简化为：

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