如何在Oracle SQL Developer中计算标准偏差？

Question

I have a table employees,

CREATE TABLE employees (
employeeid NUMERIC(9) NOT NULL,
firstname VARCHAR(10),
lastname VARCHAR(20),
deptcode CHAR(5),
salary NUMERIC(9, 2),
  PRIMARY KEY(employeeid)
);

and I want to calculate Standard Deviation for salary.

This is the code I am using:

select avg(salary) as mean, sqrt(sum((salary-avg(salary))*(salary-avg(salary)))/count(employeeid)) as SD 
from employees
group by employeeid;

I am getting this error:

ORA-00979: not a GROUP BY expression
00979. 00000 -  "not a GROUP BY expression"
*Cause:    
*Action:
Error at Line: 260 Column: 12

Line 260 Column 12 is avg(salary)

How can I sort this out?

Answer 1

Oracle has a built-in function to calculate standard deviation: STDDEV .

The usage is as you'd expect for any aggregate function.

select stddev(salary) 
from employees;

Answer 2

I'd just use the stddev function

SELECT avg(salary) as mean, 
       stddev(salary) as sd
  FROM employees

It doesn't make sense to group by employeeid since that is, presumably unique. It doesn't make sense to talk about the average salary by employee, you want the average salary across all employees (or all departments or some other aggregatable unit)

Answer 3

The salary-avg(salary) can't be evaluated; avg(salary) is not available during execution of the query but only after all records are retrieved.

I would suggest to add AVG calculations in a subquery and JOIN it to the main one

select avg(salary) as mean, 
      sqrt(sum((salary-avg_res.avg)*(salary-avg_res.avg))/count(employeeid)) as SD 
from employees JOIN
     (select employeeid,avg(salary) as avg
      from employees 
      group by employeeid) avg_res ON employees.employeeid=avg_res.employeeid
group by employeeid;

Answer 4

I thought you had to include the column in the GROUP BY in the SELECT:

select employeeid, avg(salary) as mean, sqrt(sum((salary-avg(salary))*(salary-avg(salary)))/count(employeeid)) as SD 
from employees
group by employeeid;

But on further reflection the query doesn't make much sense unless it's historical data. An employee id ought to be unique to a single employee. Unless this is an average over time there should be only one salary per employee. Your mean will be the salary and the standard deviation will be zero.

A better query might be average of all salaries. In that case, remove the GROUP BY.

One more nitpick: the formula you're using is more properly called the population standard deviation. The sample deviation divides by (n-1) .