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Anonymous's avatar
Anonymous
Not applicable
4 years ago
Solved

Divide Measure by its Max value

I have two tables. They are related to each other on ID.   I want to calculate - Final Speed which is = Speed Measure/max (Speed Measure)   But I am unable to do so. It is not recognising ...
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  • tackytechtom's avatar
    tackytechtom
    4 years ago

    Hi Anonymous ,

     

    How about this:

    ID 11 has the maximum speed, which is why the measure shows 1.

     

    This is the measure I used:

    Speed 1 mes = 
VAR _helpTable  = 
            SUMMARIZE (
                ALLSELECTED ( Table2 ),
                [ID],
                "speed", [speed1]
            )
VAR _maxSpeed = CALCULATE ( MAXX ( _helpTable , [speed] ) )
RETURN
 DIVIDE ( Table2[Speed1], _maxSpeed )

     

    Hope this is it!

    /Tom

    https://www.tackytech.blog

    https://www.instagram.com/tackytechtom

     

     

Anonymous's avatar
Anonymous
Not applicable
4 years ago

tackytechtom Thanks for answering. I am sharing the details.

 

Time = (DATEDIFF('Table1'[start time], 'Table1'[end time],SECOND))
Speed1 = SUM(Table2[Volume])/sum(Table1[Time])
Speed 1 col = DIVIDE(Table2[Volume], SUM(Table1[Time]))
These are two tables linked with each other on ID.If I divide columns as measure, it gives correct answer. But when I divide it as calculated column, answer is not correct.
 
I just wanted to divide the speed with max speed. Thanks in advance!
  • tackytechtom's avatar
    tackytechtom
    Most Valuable Professional
    4 years ago

    Hi Anonymous ,

     

    How about this:

    ID 11 has the maximum speed, which is why the measure shows 1.

     

    This is the measure I used:

    Speed 1 mes = 
VAR _helpTable  = 
            SUMMARIZE (
                ALLSELECTED ( Table2 ),
                [ID],
                "speed", [speed1]
            )
VAR _maxSpeed = CALCULATE ( MAXX ( _helpTable , [speed] ) )
RETURN
 DIVIDE ( Table2[Speed1], _maxSpeed )

     

    Hope this is it!

    /Tom

    https://www.tackytech.blog

    https://www.instagram.com/tackytechtom

     

     

    • Anonymous's avatar
      Anonymous
      Not applicable
      4 years ago

      tackytechtom Hey I use the formula but it is still giving 1. Not sure, what I am doing wrong here.

       

    • Anonymous's avatar
      Anonymous
      Not applicable
      4 years ago

      tackytechtom Thanks for the solution and your help. It worked. Thank you so much!