Forum Discussion
Customize the matrix component totals.
- Anonymous2 years ago
Hi railtons ,
You cannot have the Total in the matrix calculate the sum of different types separately and present it in a hierarchy. There is no setting that can achieve this effect.
And according to your screenshot, there should be fields placed in the rows, columns, and values of your matrix, which means that you cannot add additional measures or columns to the existing matrix to achieve your effect unless you rebuild the entire matrix.
I suggest you create an additional matrix to show the two sums.
Here is my sample data:Use this DAX to create a measure:
Total = VAR _Type1 = MAX('Table'[Type1]) VAR _Type2 = MAX('Table'[Type2]) RETURN CALCULATE( SUM('Table'[Value]), ALLEXCEPT('Table', 'Table'[Group]), 'Table'[Type1] = _Type1 && 'Table'[Type2] = _Type2 )Then create the matrix:
Best Regards,
Dino Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Thanks a lot for you feedback.
Look the option that is on in my case:
I need to have a "+" before the total line to see the total by group.
Hi railtons ,
You cannot have the Total in the matrix calculate the sum of different types separately and present it in a hierarchy. There is no setting that can achieve this effect.
And according to your screenshot, there should be fields placed in the rows, columns, and values of your matrix, which means that you cannot add additional measures or columns to the existing matrix to achieve your effect unless you rebuild the entire matrix.
I suggest you create an additional matrix to show the two sums.
Here is my sample data:
Use this DAX to create a measure:
Total =
VAR _Type1 = MAX('Table'[Type1])
VAR _Type2 = MAX('Table'[Type2])
RETURN
CALCULATE(
SUM('Table'[Value]),
ALLEXCEPT('Table', 'Table'[Group]),
'Table'[Type1] = _Type1 && 'Table'[Type2] = _Type2
)
Then create the matrix:
Best Regards,
Dino Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
- railtons2 years agoHelper I
Thnaks a lot!!!
It works as you suggested