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Seb93's avatar
Seb93
New Member
2 years ago
Solved

Distinctcount and average per month

Hi all,

 

I have a table which contains unique Mac adresses for each user. The table also contains the info which device category and what device type the mac adress is using. Each device gets its own row. The table is updated every month with all active mac adresses and devices. I have also created a date table which is linked to the time stamp of the original data.

 

 

I am trying to calculate the average number of device category and device type per mac adress for each month. I have tried numerous things but I am just not a DAX expert. 

 

E.g. at the moment it would display:

February 2024: 1 accessory as an average because only one mac adress is using it. The correct result should be 0,5 because the total of adresses in 02/2024 is 2.

 

Looking forward to your suggestions and many thanks in advance!

 

 

 

  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi Seb93 ,
    Based on the information you provided, I have created some data hopefully to solve the problem for you. You can follow the steps below:
    1. Add an index column.


    2. Add some Columns.

     

    Access = 
    IF ( 'Table'[Device_category] = "Accessory", 1 )
    
    Mac_Access = 
    VAR _1 = 'Table'[Gateway_mac]
    VAR _SumMac =
        CALCULATE (
            SUM ( 'Table'[Access] ),
            FILTER ( 'Table', 'Table'[Gateway_mac] = _1 )
        )
    VAR _AccessMac =
        IF ( _SumMac >= 1, 1, 0 )
    VAR _rank =
        RANKX (
            FILTER ( 'Table', 'Table'[Gateway_mac] = _1 ),
            'Table'[Index],
            ,
            ASC
        )
    RETURN
        IF ( _AccessMac = _rank, 1, 0 )
    

     


    3. Add some Measure.

     

    CountMac = 
    DISTINCTCOUNT ( 'Table'[Gateway_mac] )
    
    FinalResult = 
    DIVIDE ( SUM ( 'Table'[Mac_Access] ), [CountMac] )
    

     

     

    Final output:

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    If it does not help, please provide more details with your desired out put and pbix file without privacy information.

     

    Best Regards,

    Ada Wang

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

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