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Thad's avatar
Thad
Frequent Visitor
2 years ago

Display movement over time in a matrix

Hello,

 

I am, so far unsuccessfully, trying to display the movement of group membership throughout time in a matrix.

 

My source is a table of 3 columns, "Stichtag" (Date of occurence), "Personennummer" (Customer ID) and "VR-Rating" (Group they belong to at the time), for which I created lookup tables named "Stichtag Aktuell", "Stichtag Vergleich", "Rating Aktuell" and "Rating Vergleich" with their respective distinct in values in them. The aim is to have a matrix that shows how many people moved groups (also from which to which) and how many stayed in their respective groups between two selected dates. Like this:

 

 

When I try it with this measure:

Rating =
VAR ST_Akt = SELECTEDVALUE(Stichtag_Aktuell[Stichtag])
VAR ST_Vgl = SELECTEDVALUE('Stichtag Vergleich'[Stichtag])
VAR VR_Akt = SELECTEDVALUE(Rating_Aktuell[VR-Rating])
VAR VR_Vgl = SELECTEDVALUE(Rating_Vergleich[VR-Rating])
 
RETURN CALCULATE(DISTINCTCOUNT(Rating_Bewegung[Personennummer]), FILTER(Rating_Bewegung, Rating_Bewegung[Stichtag] = ST_Akt && Rating_Bewegung[Stichtag] = ST_Vgl && Rating_Bewegung[VR-Rating] = VR_Akt && Rating_Bewegung[VR-Rating] = VR_Vgl))
it won't show any data because the filters cancel each other out. When I split it into two measures, calculating the values for each date seperately, it displays the values like this:

I know I would still have to calculate the actual movement as right now I just have the amount of members as my output but my first goal would be to get all the data in a matrix that has the same amount of columns as rows. What am I doing wrong? I am sure there is a simple solution to this. Is there another visual I could use? 

 

Thanks in advance

1 Reply

  • Anonymous's avatar
    Anonymous
    Not applicable

    Hi Thad ,

    Sorry, so far, to my knowledge, this may not be achieveable.

     

    Best regards,
    Community Support Team_Binbin Yu
    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.