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sraj's avatar
sraj
Responsive Resident
3 years ago
Solved

Percent decrease in measure

Hi, Can someone please advise on this?  Probably something very silly, but I really need some input as my value in the visual is showing as '78,860,674,807' instead of what the calculator says '78,8...
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  • DOLEARY85's avatar
    DOLEARY85
    3 years ago

    Hi,

     

    try this measure:

     

    Measure 3 = calculate(sum('Table (3)'[A])*(1-0.0035))
     
    or for 0.35% use a column to transform it into the decimal number 0.0035 and replace that in the measure with the new field
     
    If I answered your question, please mark my post as solution, Appreciate your Kudos 👍

Souvik0812's avatar
Souvik0812
Frequent Visitor
3 years ago

Hi sraj,

Is this value coming from a column?

If this value is from a column, try this:

Measure = SUM('Table'[Column]) - SUM('Table'[Column]) * 0.35
DAX needs aggregation, it can't recongnise column names directly.

If this is a constant value, you can simply do the multiply and then substract:
Measure 2 = 79132267090 * 0.35
 
Feel free to ask if you have any confusion.


  • sraj's avatar
    sraj
    Responsive Resident
    3 years ago

    A is not a column = It is a calculated Value from a total measure i.e.

    (712141429 + 78420125661) = 79,132,267,090

    B 0.35% is a calculated value too and is not going to be constant.

     

    Right now this is what I am doing 

    A = [X] + [Y]

    [X] + [Y] * (1 -([B]))
    • Souvik0812's avatar
      Souvik0812
      Frequent Visitor
      3 years ago

      I am not sure if I can provide a solution to your issue, however, when I back calculate, I see it's actually 0.003432 very near to 0.35%. Is it showing 0.35% sue to 2 decimal places and the actual underlying value is different? This might be the reason for your visual showing a value different than the one in calculator.

      • sraj's avatar
        sraj
        Responsive Resident
        3 years ago

        Yes, that's correct it is showing 0.35% to 2 decimal places.  

    • Souvik0812's avatar
      Souvik0812
      Frequent Visitor
      3 years ago

      Sorry, it's not clear.
      Are you doing(?):

      A = [X] + [Y]

      A * (1 -([B]))
       
      And to achieve this, should you put a parantheses as:

      A = [X] + [Y]

      ([X] + [Y]) * (1 -([B]))