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Hi everyone.
My problem is the present working time of an employee in the company.
In the report, it has a date slicer, a count of working days, and the working hour of the employee in duration time.
Currently, I want to present average of working day and average working hour in duration time from date slicer.
My data below
Table 'working_time' has columns:
emloyee_id, workingdate, worktime_hrs (worktime hour in day), worktime_day (worktime day has default is 1)
Depend on table sum worktime group by employee, the sum worktime is the total value divided by 5 employees.
My expectation is the average of worktime_day should be (3+3+2+2+3) / 5 = 2.6 and the average of worktime_hrs should be (24+24+16+16+24) / 5 = 20.8.
But I don't know how to calculate this average value.
I hope everyone will spare their precious time to help me solve this problem.
Thanks you so much!
Solved! Go to Solution.
Hi @dtdat ,
you can write measures like,
workingtime_avg := var tbl1= summarize (working_table,employee_id, "working_time" ,sum(workingtime_hrs))
return sumx(tbl1,working_time)/distinctcount(employee_id)
and
workingday_avg := var tbl1= summarize (working_table,employee_id, "working_day" ,sum(workingtime_day))
return sumx(tbl1,working_day)/distinctcount(employee_id)
If this post helps, then I would appreciate a thumbs up and mark it as the solution to help the other members find it more quickly.
Hi @Selva-Salimi
Wow, amazing!
I resolved this problem. You help me know more about DAX query processing techniques.
I appreciate your help.
Thank you very much!
Hi @dtdat ,
you can write measures like,
workingtime_avg := var tbl1= summarize (working_table,employee_id, "working_time" ,sum(workingtime_hrs))
return sumx(tbl1,working_time)/distinctcount(employee_id)
and
workingday_avg := var tbl1= summarize (working_table,employee_id, "working_day" ,sum(workingtime_day))
return sumx(tbl1,working_day)/distinctcount(employee_id)
If this post helps, then I would appreciate a thumbs up and mark it as the solution to help the other members find it more quickly.
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