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Thor2022's avatar
Thor2022
Frequent Visitor
3 years ago
Solved

Filter the contents inside MATRIX visual

Dear All  I have a matric that contains the following information:  - Rows: Warehouse names: A, B and C  - Columns: Years  - Values: Profits    and I have a multiselection slicer that contains ...
Tips and Tricks
  • PaulDBrown's avatar
    PaulDBrown
    3 years ago

    Thor2022 

    Actually, thinking about it, the average method could potentially not achieve what you are looking for.

    If you have a set of values for a year like {30, 29, 30, 31}, the average method will hide both 30 values.

    The following  method however will work: compare the minimum value in a set to the maximum value. If they are the same, all values are the same...

    So...

     

    No duplicate profits =
VAR _Min =
    MINX ( ALL ( 'Warehouse table'[Warehouse] ), [Your measure] )
VAR _MAX =
    MAXX ( ALL ( 'Warehouse table'[Warehouse] ), [Your measure] )
RETURN
    IF (
        AND (
            ISINSCOPE ( 'Warehouse table'[Warehouse] ),
            ISINSCOPE ( 'Year Table'[Year] )
        ),
        IF ( _Min = _Max, BLANK (), [Your measure] )
    )

     

     

    PS. I tried deleting the previous message but the forum won't let me for that particular post for some reason....

     

     

pi_eye's avatar
pi_eye
Resolver IV
3 years ago

Hi Thor

 

Can you elaborate on the example? So for example if the column values for year #2 are 5,6 and 7, it shows but if it is 3,3, and 3  it is blank?

 

See example below:

 

Try this expresion in DAX:

Show diff Values = If(Calculate(DISTINCTCOUNT(WareHouse[Value]),ALLSELECTED(WareHouse[WareHouse]))=1,"", Sum(WareHouse[Value]))

 

What it is doing:

Calculate(DISTINCTCOUNT(WareHouse[Value]),ALLSELECTED(WareHouse[WareHouse])) --> Gives you the total of distinct values per warehouse

If(....... =1,"",...) ---> tests for the distinct values =1, thie means they are all the same. Returns blank if that is the case, and the sum if they are different.

 

 

 

Thanks,

 

Pi

Thor2022's avatar
Thor2022
Frequent Visitor
3 years ago

Hi Pi

Thanks a lot, I will try it on Monday. 

Until then, I just need to add that the profit value is a measure, not a column .. so I am not sure if this DAX can be applied? 

Thanks a gain 

Regards 

  • PaulDBrown's avatar
    PaulDBrown
    Community Champion
    3 years ago

    Here is one way. Basically you compare the sum (profit) of each warehouse with the average for all warehouses in a year.

    First the model:

     The calculation (without totals)

     

    No duplicate profits =
VAR _Profit =
    SUM ( fTable[Profit] )
VAR _Average =
    CALCULATE ( AVERAGE ( fTable[Profit] ), ALL ( 'Warehouse table'[Warehouse] ) )
RETURN
    IF (
        AND (
            ISINSCOPE ( 'Warehouse table'[Warehouse] ),
            ISINSCOPE ( 'Year Table'[Year] )
        ),
        IF ( _Profit = _Average, BLANK (), _Profit )
    )

     

    and if you need totals:

     

    No Dups with totals = 
SUMX(SUMMARIZE(fTable, 'Warehouse table'[Warehouse], 'Year Table'[Year]), [No duplicate profits])

     

     

    Sample PBIX attached

    • pi_eye's avatar
      pi_eye
      Resolver IV
      3 years ago

      Hi Thor2022 

       

      This is interesting, because for some reason it's not as simple to aggregate a distinct count on another measure! However it is always possible in DAX 😉

       

      I created a new measure rather than chain it in called "New count" which evaluates to the total number of distinct sums. I am use group by to create an aggregated table which I can pass the summed column through to a  distinct count.

      New Count = calculate(
              countx(
                  distinct(SELECTCOLUMNS( GROUPBY(WareHouse,WareHouse[WareHouse],"SumVal",sumx(CURRENTGROUP(),WareHouse[Value])),[SumVal]) ) ,
                  [SumVal] ),
              ALLSELECTED(WareHouse[WareHouse])
      )
       
      On it's own, it looks like this: (sorry I can't seem to cut and paste images right now)
       12345678 
      WH142123333 
      WH242123333 
      WH342123333 
      WH442123333 
                
       

      Added into the final expression gives you: 

      Show diff Values = If([New Count]=1,"", Sum(WareHouse[Value]))

       

      Let me know if this works,

      Pi

      • Thor2022's avatar
        Thor2022
        Frequent Visitor
        3 years ago

        Hi Pi 

        I am sure its a great solution , but give me some time to digest it , since some of the used functions are new for me 🙂 

        Regards 

    • Thor2022's avatar
      Thor2022
      Frequent Visitor
      3 years ago

      Hi Paul 

      Thank you for your reply .

      I find the logic in your solution excellent but as I mentioned before, the Profit value is a measured value that can't be calculated in the average function !!! 
      looking forward to your reply 

      REgards 

      • PaulDBrown's avatar
        PaulDBrown
        Community Champion
        3 years ago

        My apologies. I did see you mention it and then promptly forgot about it in the solution.

        This measure will work:

        No duplicate profits =
VAR _Profit = [Your measure]
VAR _Average =
    AVERAGEX ( ALL ( 'Warehouse table'[Warehouse] ), [Your measure] )
RETURN
    IF (
        AND (
            ISINSCOPE ( 'Warehouse table'[Warehouse] ),
            ISINSCOPE ( 'Year Table'[Year] )
        ),
        IF ( _Profit = _Average, BLANK (), _Profit )
    )
  • pi_eye's avatar
    pi_eye
    Resolver IV
    3 years ago

    Hi @Thor2022 

     

    This is interesting, because for some reason it's not as simple to aggregate a distinct count on another measure! However it is always possible in DAX ğŸ˜‰

     

     

    I created a new measure rather than chain it in called "New count" which evaluates to the total number of distinct sums. I am use group by to create an aggregated table which I can pass the summed column through to a  distinct count.

    New Count = calculate(
            countx(
                distinct(SELECTCOLUMNS( GROUPBY(WareHouse,WareHouse[WareHouse],"SumVal",sumx(CURRENTGROUP(),WareHouse[Value])),[SumVal]) ) ,
                [SumVal] ),
            ALLSELECTED(WareHouse[WareHouse])
    )
     
    On it's own, it looks like this: (sorry I can't seem to cut and paste images right now)
     12345678 
    WH142123333 
    WH242123333 
    WH342123333 
    WH442123333 
              
     

    Added into the final expression gives you: 

    Show diff Values = If([New Count]=1,"", Sum(WareHouse[Value]))

     

    Let me know if this works,

    Pi