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neonguyen
Frequent Visitor

Sum with group

I have a table as bellow:

 

222.png

And i want to Sum 'Amount' group by: 'Created by' & 'Program'. Then divide them for 10,000,000, and round 0.5 -> 1 to calculate Enrollment. And finally, i will calculate Commission = Enrollment for each user  * 50,000

 

I can do that with Matrix, but in this case, i don't want use it. Anyone have suggestion for me?

 

Regards,

1 ACCEPTED SOLUTION

Hi @neonguyen ,

You can try to use following measure formula if it suitable for your requirement:

Grouped Total =
ROUND (
    DIVIDE (
        CALCULATE (
            SUM ( Table[Amount] ),
            ALLSELECTED ( Table ),
            VALUES ( Table[abc] ),
            VALUES ( Table[Program] )
        ),
        10000000
    ),
    0
) * 50000

Regards,

Xiaoxin Sheng

Community Support Team _ Xiaoxin
If this post helps, please consider accept as solution to help other members find it more quickly.

View solution in original post

3 REPLIES 3
parry2k
Super User
Super User

@neonguyen can you share what you want the expected result to be? 



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@parry2k Please see my matrix for this case:

222.png

Let see user Van Huynh, after group by name, i continue group by Program and calculate Enrollment = 

ROUND(SUMX('table','table'[Amount]) / 10000000,0) ==> meaning Van Huynh's Enrollment = ( sum amount of program 19 / 10000000, round 0.5 to 1) + ( sum amount of program 20 / 10000000, round 0.5 to 1).
 
And now, i want use Rankx for it but still keep formula User's Enrollment = ( sum amoun of program # / 10000000, round 0.5 to 1) + ( sum amoun of program # / 10000000, round 0.5 to 1) + etc

Hi @neonguyen ,

You can try to use following measure formula if it suitable for your requirement:

Grouped Total =
ROUND (
    DIVIDE (
        CALCULATE (
            SUM ( Table[Amount] ),
            ALLSELECTED ( Table ),
            VALUES ( Table[abc] ),
            VALUES ( Table[Program] )
        ),
        10000000
    ),
    0
) * 50000

Regards,

Xiaoxin Sheng

Community Support Team _ Xiaoxin
If this post helps, please consider accept as solution to help other members find it more quickly.

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