Orthogonal filtering

[Pavel_Ischenko]

Hello guys!

Is there anybody knows how to implement so called "orthogonal" filtering in PBI. What I mean here.

This is data visual. There are hundreds of products tested in hundreds locations. In the table you can see the results of testing. How easily to filter those locations (TRLID), where only 3 selected products were tested together? If select manually it should be second and fourth stripe from the bottom:

[v-stephen-msft · Accepted solution]

Hi @Pavel_Ischenko ,

Try to create this measure

Measure = 
VAR _count =
    CALCULATE (
        COUNT ( 'Table'[Result] ),
        FILTER ( ALLSELECTED ( 'Table' ), [TRILD] = MAX ( 'Table'[TRILD] ) )
    )
VAR _count2 =
    CALCULATE ( DISTINCTCOUNT ( 'Table'[HGHNM] ), ALLSELECTED ( 'Table' ) )
RETURN
    IF ( ISFILTERED('Table'[HGHNM])&&_count = _count2, 1 )

Then put the measure into Filters and set as follows

When filtering, see the results

Best Regards,

Stephen Tao

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

[Pavel_Ischenko]

Thanks a lot! It works.

[AlexisOlson]

What do your data table(s) look like?

[Pavel_Ischenko]

Hello.

Table is like

[AlexisOlson]

Create a new calculated table to use as your slicer and column values for the matrix.

Slicer  = VALUES ( Table1[HGHNM] )

Then define a measure to check whether the set of selected products is the same as the set of products for the given TRLID.

ResultSliced =
VAR SlicerValues = ALLSELECTED ( Slicer[HGHNM] )
VAR TRLValues = VALUES ( Table1[HGHNM] )
VAR SharedCount = COUNTROWS ( INTERSECT ( SlicerValues, TRLValues ) )
RETURN
    IF (
        COUNTROWS ( SlicerValues ) = SharedCount && COUNTROWS ( TRLValues ) = SharedCount,
        CALCULATE ( SUM ( Table1[Result] ), Table1[HGHNM] IN VALUES ( Slicer[HGHNM] ) )
    )

Note: I check set equality here by taking the intersection of the two sets and checking if that has the same cardinality as each of the sets being intersected. If both A and B only contain values from A ∩ B, then these sets much be equal.