的Oracle SQL：如何使用预定义的贡献获得随机记录各组

Question

这是参照这里所描述的早期问题：Oracle SQL: How to get Random Records by each group

问：

是，能够以比获得随机样本不同的类别。例如：如果我有132个有3个类别（批准，拒绝，取消）的样本的随机记录，我如何按照以下比例获取样本？

total sample = 132 

category  samples % sample Size 
approved  50%  66 
denied  30%  40 
canceled  20%  26 

注：我需要的原始数据，不计

Answer 1

让我们先得到一些样本数据。我使用批准的类别创建了132条记录，以获得具有66行的50％样本。

create table task as 
select 'approved' category, rownum task_id from dual connect by level <= 132 union all 
select 'denied' category, rownum task_id from dual connect by level <= 134 union all 
select 'canceled' category, rownum task_id from dual connect by level <= 130 
; 

的关键步骤是定义方含为每个类别0和1 之间的值。如果你想要的样品的柱RAND_PERC与值小于或等于一个产品类别说50％选择的所有行。 5

通过按随机顺序分配行编号（每个类别都独立）并且除以每个类别中的行数 ，首先计算该列。

select CATEGORY, TASK_ID, 
(row_number() over (partition by task.category order by dbms_random.value))/
(count(*) over (partition by task.category)) as rand_perc 
from task 
order by 1,3; 

CATEGORY TASK_ID RAND_PERC 
-------- ---------- ---------- 
approved   56 ,00757575758 
approved  129 ,0151515152 
approved   61 ,0227272727 

要绘制样本，请根据需要简单定义WHERE条件 - 请参阅下面的示例。

with rnd as (
select CATEGORY, TASK_ID, 
(row_number() over (partition by task.category order by dbms_random.value))/
(count(*) over (partition by task.category)) as rand_perc 
from task 
) 
select CATEGORY, count(*) cnt 
from rnd 
where 
category = 'approved' and rand_perc <= .5 or /* take 50% from active */ 
category = 'denied' and rand_perc <= .3 or 
category = 'canceled' and rand_perc <= .2 
group by CATEGORY 
; 

这给样本大小根据需要

CATEGORY  CNT 
-------- ---------- 
canceled   26 
denied   40 
approved   66