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TomSinAA's avatar
TomSinAA
Icon for Helper IV rankHelper IV
3 years ago
Solved

dax help

I have data in this format:

CategoryAreaRequirement
Cat1R11
Cat1R1 
Cat1R2 
Cat1R32
Cat1R43
Cat1R5 

 

I need to display the count and % of areas with and without requirements.  So that is looks like this:

AreaCount of Areas with no requirement%Count of Areas with requirement Total%
Cat1240%360%5100

 

Not sure how to approach this. 

  • Anonymous's avatar
    Anonymous
    3 years ago

    Hi TomSinAA 

    You can refer to the following measure

    Count of Areas with no requirement = var a=SELECTCOLUMNS(FILTER('Table',[Requirement]<>BLANK()),"a",[Area])
    return CALCULATE(DISTINCTCOUNT('Table'[Area]),'Table'[Area] in a=FALSE())
    
    Count of Areas with  requirement = DISTINCTCOUNT('Table'[Area])-[Count of Areas with no requirement]
    
    
    Areas with no requiremen t% = DIVIDE([Count of Areas with no requirement],CALCULATE(DISTINCTCOUNT('Table'[Area])))
    
    Areas with  requiremen t% = DIVIDE([Count of Areas with  requirement],CALCULATE(DISTINCTCOUNT('Table'[Area])))

    Output

    Best Regards!

    Yolo Zhu

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

1 Reply

  • Anonymous's avatar
    Anonymous
    Not applicable

    Hi TomSinAA 

    You can refer to the following measure

    Count of Areas with no requirement = var a=SELECTCOLUMNS(FILTER('Table',[Requirement]<>BLANK()),"a",[Area])
    return CALCULATE(DISTINCTCOUNT('Table'[Area]),'Table'[Area] in a=FALSE())
    
    Count of Areas with  requirement = DISTINCTCOUNT('Table'[Area])-[Count of Areas with no requirement]
    
    
    Areas with no requiremen t% = DIVIDE([Count of Areas with no requirement],CALCULATE(DISTINCTCOUNT('Table'[Area])))
    
    Areas with  requiremen t% = DIVIDE([Count of Areas with  requirement],CALCULATE(DISTINCTCOUNT('Table'[Area])))

    Output

    Best Regards!

    Yolo Zhu

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.