oracle分析函数（3）

1. ntile

--ntile
select e.employee_id,
       e.last_name,
       e.salary,
       e.department_id,
       ntile(10) over(order by e.salary desc) bucket --根据工资排序后，员工分配到10个组里，可以理解为工资相近的会被分配到相同的bucket
  from employees e;

 

2.percentile_cont,percentile_disc

--percentile_cont(P) 如果共有N行数据，及RN = (1 + P * (N - 1))，CRN = CEILING(RN), FRN = FLOOR(RN)
/*If (CRN = FRN = RN) then the result is
  (value of expression from row at RN)
Otherwise the result is
  (CRN - RN) * (value of expression for row at FRN) +
  (RN - FRN) * (value of expression for row at CRN)*/
--percontile_disc(0.5) 如果cume_dist计算的结果中有等于0.5的返回，如果没有取比0.5大的最小的一个值
--作为聚合函数
select e.department_id,
       percentile_cont(0.5) within group(order by e.salary),
       percentile_disc(0.5) within group(order by e.salary)
  from employees e
 group by e.department_id;

select e.employee_id,
       e.last_name,
       e.salary,
       e.department_id,
       percentile_cont(0.5) within group(order by e.salary) over(partition by e.department_id),
       percent_rank() over(partition by e.department_id order by e.salary)
  from employees e;

select e.employee_id,
       e.last_name,
       e.salary,
       e.department_id,
       percentile_disc(0.5) within group(order by e.salary) over(partition by e.department_id),
       cume_dist() over(partition by e.department_id order by e.salary)
  from employees e;

 

3. ratio_to_report

select e.employee_id,
       e.last_name,
       e.department_id,
       e.salary,
       ratio_to_report(e.salary) over(partition by e.department_id) --查询每个人的工资在部门内的占比
  from employees e
 order by e.department_id;