Need help with Group functions

I'm a total novice with SQL, so please forgive me if the answer to my question seems to be too obvious
I work with diagrams of the sample (in particular with the employees table):

DESC employees;
result

What I have to do is select all the managers, including the number of subordinates is higher than the average number of subordinates of managers who work in the same Department. What I've done so far is as follows:

SELECT mgr.employee_id manager_id, Director of mgr.last_name, mgr.department_id, COUNT (emp.employee_id)
Employees emp employees JOIN Bishop
ON emp.manager_id = mgr.employee_id
GROUP OF mgr.employee_id, mgr.last_name, mgr.department_id
ORDER BY mgr.department_id;
result

As you can see, I'm almost done. Now, I need only to calculate the average of the result of the COUNT function for each Department. But I'm totally stuck at this point.
All advice?

Hello

Welcome to the forum!

user12107811 wrote:
I'm a total novice with SQL, so please forgive me if the answer to my question seems to be too obvious

Just the opposite! Looks like a very difficult mission.

I work with diagrams of the sample (in particular with the employees table):

DESC employees;
result

What I have to do is select all the managers, including the number of subordinates is higher than the average number of subordinates of managers who work in the same Department. What I've done so far is as follows:

SELECT mgr.employee_id manager_id, Director of mgr.last_name, mgr.department_id, COUNT (emp.employee_id)
Employees emp employees JOIN Bishop
ON emp.manager_id = mgr.employee_id
GROUP OF mgr.employee_id, mgr.last_name, mgr.department_id
ORDER BY mgr.department_id;
result

As you can see, I'm almost done. Now, I need only to calculate the average of the result of the COUNT function for each Department. But I'm totally stuck at this point.
All advice?

Yes, you're almost done. You just need to add one more condition. You have to calculate the average value of total_cnt (the COUNT (*) you already do) of a Department and compare that to total_cnt.

There are several ways to do this, including
a scalar subquery (in a HAVING clause)
(b) make a result set with one line per Department, containing the average_cnt and reach than your current result set
(c) analytical functions. Analytical functions are calculated after the GROUP BY clause is applied and aggregate functions are calculated, it is legitimate to say "AVG (COUNT (*)) MORE (...)").

If thinking (c) is the simplest. It involves the use of a query of Tahina, but (a) and (b) also require subqueries.

This sounds like homework, so I'll do it for you.
Instead, here is a very similar problem with the hr.employees table.
Let's say that we are interested in total wages given each type of work in each Department.

SELECT        department_id
,        job_id
,        SUM (salary)     AS sum_sal
FROM        hr.employees
GROUP BY   department_id
,             job_id
ORDER BY   department_id
,             job_id
;

Results:

DEPARTMENT_ID JOB_ID        SUM_SAL
------------- ---------- ----------
           10 AD_ASST          4400
           20 MK_MAN          13000
           20 MK_REP           6000
           30 PU_CLERK        13900
           30 PU_MAN          11000
           40 HR_REP           6500
           50 SH_CLERK        64300
           50 ST_CLERK        55700
           50 ST_MAN          36400
           60 IT_PROG         28800
           70 PR_REP          10000
           80 SA_MAN          61000
           80 SA_REP         243500
           90 AD_PRES         24000
           90 AD_VP           34000
          100 FI_ACCOUNT      39600
          100 FI_MGR          12000
          110 AC_ACCOUNT       8300
          110 AC_MGR          12000
              SA_REP           7000

Now suppose we want to find out which of these sum_sals is higher than the average sum_sal of his Department.
For example, in detriment 110 (near the end OIF the list) there two types of work (AC_ACCOUND and AC_MGR) that have sum_sals of 8300 and 12000. The average of these two numbers is 10150, so we selected AC_MGR (because its sum_sal, 12000, is superior to 10150, and we do not want to include AC_ACCOUNT, because its sum_sal, 8300, is less than or equal to the average of the Department.
In departments where there is only one job type (for example, Department 70, or null "Department" at the end of the list above) the only sum_sal will be the average; and because the sum_sal is not greater than the average, we want to exclude this line.

Let's start with the calculation of the avg_sum_sal using the analytical function AVG:

SELECT        department_id
,        job_id
,        SUM (salary)                              AS sum_sal
,        AVG (SUM (salary)) OVER (PARTITION BY department_id)     AS avg_sum_sal
FROM        hr.employees
GROUP BY   department_id
,             job_id
ORDER BY   department_id
,             job_id
;

Output:

DEPARTMENT_ID JOB_ID        SUM_SAL AVG_SUM_SAL
------------- ---------- ---------- -----------
           10 AD_ASST          4400        4400
           20 MK_MAN          13000        9500
           20 MK_REP           6000        9500
           30 PU_CLERK        13900       12450
           30 PU_MAN          11000       12450
           40 HR_REP           6500        6500
           50 SH_CLERK        64300  52133.3333
           50 ST_CLERK        55700  52133.3333
           50 ST_MAN          36400  52133.3333
           60 IT_PROG         28800       28800
           70 PR_REP          10000       10000
           80 SA_MAN          61000      152250
           80 SA_REP         243500      152250
           90 AD_PRES         24000       29000
           90 AD_VP           34000       29000
          100 FI_ACCOUNT      39600       25800
          100 FI_MGR          12000       25800
          110 AC_ACCOUNT       8300       10150
          110 AC_MGR          12000       10150
              SA_REP           7000        7000

Now all we have to do is to compare the sum_sal and avg_sum_sal columns.
Given that the analytic functions are calculated after the WHERE clause is applied, we cannot use avg_sum_sal in the WHERE clause of the query, even where it has been calculated. But we can do that in a subquery; Then, we can use avg_sum_sal in any way that we love in the Super-requete:

WITH     got_avg_sum_sal       AS
(
     SELECT        department_id
     ,        job_id
     ,        SUM (salary)                              AS sum_sal
     ,        AVG (SUM (salary)) OVER (PARTITION BY department_id)     AS avg_sum_sal
     FROM        hr.employees
     GROUP BY   department_id
     ,             job_id
)
SELECT        department_id
,        job_id
,        sum_sal
FROM        got_avg_sum_sal
WHERE        sum_sal     > avg_sum_sal
ORDER BY   department_id
,             job_id
;

Results:

DEPARTMENT_ID JOB_ID        SUM_SAL
------------- ---------- ----------
           20 MK_MAN          13000
           30 PU_CLERK        13900
           50 SH_CLERK        64300
           50 ST_CLERK        55700
           80 SA_REP         243500
           90 AD_VP           34000
          100 FI_ACCOUNT      39600
          110 AC_MGR          12000

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