Forum Discussion

Anonymous's avatar
Anonymous
Not applicable
4 years ago
Solved

link matrices

Hello,

Is it possible to build relationships between two visual matrices? like have one of them as a percentage from the other one?

 

  • Anonymous's avatar
    Anonymous
    4 years ago

    Hi  Anonymous ,

    You can try to use Measure instead

    I created some data:

    Form a matrix:

    Matrix A: Shows the value of [amount].

    Matrix B: Shows the values of [rand].

    Custom rule to display one column of Group=A&&amount and one column of Group=B&&rand in matrix A

    Here are the steps you can follow:

    1. Create measure.

    A = MAX('Table'[amount])
    Divide =
    var _A=CALCULATE(SUM('Table'[amount]),FILTER(ALL('Table'),'Table'[Group]="A"&&'Table'[Date]=MAX('Table'[Date])))
    var _B=CALCULATE(SUM('Table'[rand]),FILTER(ALL('Table'),'Table'[Group]="B"&&'Table'[Date]=MAX('Table'[Date])))
    return
    DIVIDE(_A,_B)

    2. Result:

     

    Best Regards,

    Liu Yang

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

2 Replies

  • Anonymous ,The information you have provided is not making the problem clear to me. Can you please explain with an example.

    Appreciate your Kudos.

  • Anonymous's avatar
    Anonymous
    Not applicable

    Hi  Anonymous ,

    You can try to use Measure instead

    I created some data:

    Form a matrix:

    Matrix A: Shows the value of [amount].

    Matrix B: Shows the values of [rand].

    Custom rule to display one column of Group=A&&amount and one column of Group=B&&rand in matrix A

    Here are the steps you can follow:

    1. Create measure.

    A = MAX('Table'[amount])
    Divide =
    var _A=CALCULATE(SUM('Table'[amount]),FILTER(ALL('Table'),'Table'[Group]="A"&&'Table'[Date]=MAX('Table'[Date])))
    var _B=CALCULATE(SUM('Table'[rand]),FILTER(ALL('Table'),'Table'[Group]="B"&&'Table'[Date]=MAX('Table'[Date])))
    return
    DIVIDE(_A,_B)

    2. Result:

     

    Best Regards,

    Liu Yang

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