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chrisgehm's avatar
chrisgehm
Helper III
8 years ago
Solved

Operate with 2 cards

Hi everyone!   I'm having troubles while operating with 2 cards.   Both of them have filters and I have a numer from each card. Is there any way to make arithmetic operations between both cards?...
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  • TomMartens's avatar
    TomMartens
    8 years ago

    Hey,

     

    try this

    the measure = 
DIVIDE(
  CALCULATE(
    ,COUNTROWS('yourtable')
    ,'yourtable'[colorcolumn] = "red"
  )
  ,CALCULATE(
    ,COUNTROWS('yourtable')
    ,'yourtable'[colorcolumn] = "blue"
  )
)

    Regards

    Tom

TomMartens's avatar
TomMartens
Super User
8 years ago

Hey,

 

unfortunately you cant't reference values from a visual (your card(s)) to calculate a new value.

 

For this reason you have create a measure that rebuilds the filter from the separate card visuals, something like this

3rd measure =
DIVIDE(
  CALCULATE(
    SUM('table'[column])
    ,'table'[filtercolumn2] = "this" &&
    ,,'table'[filtercolumn1] IN {"red", "blue"}
  )
  ,CALCULATE(
    SUM('table'[column])
    ,'table'[filtercolumn2] = "this"
  )
)

Hope this gets you started

 

Regards

Tom

 

  • chrisgehm's avatar
    chrisgehm
    Helper III
    8 years ago

    Hi TomMartens thanks for the answer.

    Actually what I need is to divide the COUNT of 2 things.


    Example:

    I've got 100 values with color red. It does not matter what is the value. It could be A, B, C, anyone.

    And I've got 75 values with clor blue and yellow. Same conditions, it does not matter the value, what I need is to know how many there are.

     

    After that, I need to make the division of that 2 COUNTS.

     

    Is it possible?

     

    Sorry for my bad English.


    Kind regards.

    • TomMartens's avatar
      TomMartens
      Super User
      8 years ago

      Hey,

       

      try this

      the measure = 
DIVIDE(
  CALCULATE(
    ,COUNTROWS('yourtable')
    ,'yourtable'[colorcolumn] = "red"
  )
  ,CALCULATE(
    ,COUNTROWS('yourtable')
    ,'yourtable'[colorcolumn] = "blue"
  )
)

      Regards

      Tom

      • chrisgehm's avatar
        chrisgehm
        Helper III
        8 years ago

        Thanks!