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Joop's avatar
Joop
New Member
9 years ago
Solved

cross frequency calculation

Hello community,   I have a question related to the frequency golf players are linked to each other during a competition. Imagine players are grouped together in several flights during one day, an...
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  • Joop's avatar
    Joop
    9 years ago

    Hello Scottsen,

     

    does seem indeed not being trivial !

     

    However, this will definitely work for me. I will build it in my tables using your sample tables and statement.

     

    Thanks very much again!

    Cheers,

    Joop

Joop's avatar
Joop
New Member
9 years ago

Hi Scottsen,

thanks for your reply.

Unfortunately I don't know exactly how to see this duplicate table.

I want to see eventually, as an example, that player BINV1 has played with Bost 6 once, with PEER 3 three times, RABA 3 once and SPIT 1 also once. Also, as a consequence, that BINV 1 never played with BOOR 3, KLOM 1, ROST 5 and TAMI 9.

How would I do this?

 

Thanks again.

Joop

Anonymous's avatar
Anonymous
Not applicable
9 years ago

I literally mean duplicating the table.  So, instead of Players, you would have 2 tables "Home Players" and "Visiting Players" -- or similiar.

 

The easiest way to do that is probably to launch the Query Editor (on Home ribbon, Edit Queries), right click on your table (say Players) and choose "Reference".  That will create another table called Players2, you can rename as appropriate.  Save, done.

  • Joop's avatar
    Joop
    New Member
    9 years ago

    Hello Scottsen,

    yes, I did duplicate the table, but then for the next steps I'm not sure what to do ...

    Could you give me some more hints on how to proceed?

     

    Thanks,

    Joop

    • Anonymous's avatar
      Anonymous
      Not applicable
      9 years ago

      Tried to throw together a sample model here: Golfing.pbix

       

      In fairness, this measure was... non-trivial.  I'm actually wondering if somebody else has a better/easier way.

       

      Together = IF (HASONEVALUE(Away[Person]) && HASONEVALUE(Home[Person]) && VALUES(Home[Person]) <> VALUES(Away[Person]), 
      SUMX(
           SUMMARIZE(Tee, Tee[Date], Tee[Flight], "Together", IF(CONTAINS(Tee, Tee[Player], VALUES(Away[Person])) && CONTAINS(Tee, Tee[Player], VALUES(Home[Person])), 1, 0)),
           [Together]
      )
)
      • Joop's avatar
        Joop
        New Member
        9 years ago

        Hello Scottsen,

         

        does seem indeed not being trivial !

         

        However, this will definitely work for me. I will build it in my tables using your sample tables and statement.

         

        Thanks very much again!

        Cheers,

        Joop