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Interact's avatar
Interact
Regular Visitor
2 years ago
Solved

Using a measure in Y axis

Hello everyone, I need some help.

 

I have a database with a hierarchy of Units, Processes, and Risks. Each Risk can have a different Result outcome according to its hierarchy, and I classify it as LOW, NORMAL, and HIGH (measure "AVG Label") based on the filter defined by the user.

 

What I need is a chart (similar to the example in the image), showing distinct count risks and their classification, according to the selected filter. However, I'm not finding a way to create this chart.

 

  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi Interact ,
    Here some steps that I want to share, you can check them if they suitable for your requirement.
    Here is my test data:

    Create a measure

    Average by Risk = 
    VAR _count =
    CALCULATE(
        COUNT('Table'[Risk]),
        ALLEXCEPT(
            'Table',
            'Table'[Risk]
        )
    )
    VAR _sum = 
    CALCULATE(
        SUM('Table'[AVG]),
        ALLEXCEPT(
            'Table',
            'Table'[Risk]
        )
    )
    RETURN
    _sum/_count

    Create two columns

    Average Label risk = SWITCH(TRUE(),
                        [Average by Risk]>0 && [Average by Risk]<=2, "LOW",
                        [Average by Risk]>2 && [Average by Risk]<=4, "NORMAL",
                        [Average by Risk]>4,"HIGH"
    )
    Count = 
    VAR _distinctcount = 
    CALCULATE(
        DISTINCTCOUNT('Table'[Risk]),
        FILTER(
            'Table',
            'Table'[Average Label risk] = EARLIER('Table'[Average Label risk])
        )
    )
    VAR _count = 
    CALCULATE(
        COUNT('Table'[Average Label risk]),
        ALLEXCEPT(
            'Table',
            'Table'[Average Label risk]
        )
    )
    RETURN
    _distinctcount/_count

    Final output


    Because neither the x-axis nor the y-axis can be categorized in power bi using MEASURE as a value, but only COLUMN

    Best regards,

    Albert He

     

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



3 Replies

  • Anonymous's avatar
    Anonymous
    Not applicable

    Create a Count measure - HighLowCount = Count(tablename[columnname]) and place on the same visual as the High Low measure.

    • Interact's avatar
      Interact
      Regular Visitor

      Hello, thanks for your help!

      But there's one more point, in the chart I need to count each Risk only once, and its classification should be the average of its Results.

      In the example, RISK-001 should be 4 = NORMAL ((2 + 4 + 6)/3 = 4).

       

  • Anonymous's avatar
    Anonymous
    Not applicable

    Hi Interact ,
    Here some steps that I want to share, you can check them if they suitable for your requirement.
    Here is my test data:

    Create a measure

    Average by Risk = 
    VAR _count =
    CALCULATE(
        COUNT('Table'[Risk]),
        ALLEXCEPT(
            'Table',
            'Table'[Risk]
        )
    )
    VAR _sum = 
    CALCULATE(
        SUM('Table'[AVG]),
        ALLEXCEPT(
            'Table',
            'Table'[Risk]
        )
    )
    RETURN
    _sum/_count

    Create two columns

    Average Label risk = SWITCH(TRUE(),
                        [Average by Risk]>0 && [Average by Risk]<=2, "LOW",
                        [Average by Risk]>2 && [Average by Risk]<=4, "NORMAL",
                        [Average by Risk]>4,"HIGH"
    )
    Count = 
    VAR _distinctcount = 
    CALCULATE(
        DISTINCTCOUNT('Table'[Risk]),
        FILTER(
            'Table',
            'Table'[Average Label risk] = EARLIER('Table'[Average Label risk])
        )
    )
    VAR _count = 
    CALCULATE(
        COUNT('Table'[Average Label risk]),
        ALLEXCEPT(
            'Table',
            'Table'[Average Label risk]
        )
    )
    RETURN
    _distinctcount/_count

    Final output


    Because neither the x-axis nor the y-axis can be categorized in power bi using MEASURE as a value, but only COLUMN

    Best regards,

    Albert He

     

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