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Anonymous's avatar
Anonymous
Not applicable
3 years ago
Solved

GROUP BY expression as a measure

Hello smart people! I have a data like this: And need to build a chart which will show how many books on average each player had daily. So I am trying to get (what should be) a fairly simple...
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  • v-jialluo-msft's avatar
    v-jialluo-msft
    3 years ago

    Hi Anonymous ,

     

    Please follow these steps:

    (1) Create a new measure

    AVG = DISTINCTCOUNT('Table'[book_id])/DISTINCTCOUNT('Table'[user_id])

     

    (2)Final output

     

     

    Best Regards,

    Gallen Luo

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

henkvankester's avatar
henkvankester
Icon for Helper I rankHelper I
3 years ago

or you should use allexcept, something like: 

 

CALCULATE(DISTINCTCOUNT(bookid), ALLEXCEPT(table1, table1[group]))