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Anonymous's avatar
Anonymous
Not applicable
7 years ago
Solved

Sum all values dependend on value on another column

Hi all,

 

I have a large database of entered Hours. And I want to calculate the overall average hourRate using an ALL measure so I can show the overal average hourrate in comparison per employee/projectleader etc. 

However my dataset also contains overhead = Hour ID 2 and sales hours Hour ID 3. I want to calculate the average Hour rate only for hour ID 1. And this should not be dependent on filters in the report, therefore I was thinking about an ALL measure. But I cannot seem to use and all measure which does contain a filter in de measure.

 

Does anyone have an idea about how I could do this.  

 

Hour IDAmount hoursSales amount
1440
1660
25100
25120
360
  • Hi Anonymous 

     

    You can create a measure and place it in a card visual for instance.

    If you want the average per line:

     

    Measure =
    AVERAGEX (
        FILTER ( ALL ( Table1 ); Table1[Hour ID] = 1 );
        DIVIDE ( Table1[Sales Amount]; Table1[Amount Hours] )
    )

     

    If you want the weighted average (by number of hours):

     

    Measure =
    DIVIDE (
        CALCULATE ( SUM ( Table1[Sales Amount] ); Table1[HOURS ID] = 1; ALL ( Table1 ) );
        CALCULATE ( SUM ( Table1[Amount Hours] ); Table1[HOURS ID] = 1; ALL ( Table1 ) )
    )

2 Replies

  • AlB's avatar
    AlB
    Icon for Community Champion rankCommunity Champion

    Hi Anonymous 

     

    You can create a measure and place it in a card visual for instance.

    If you want the average per line:

     

    Measure =
    AVERAGEX (
        FILTER ( ALL ( Table1 ); Table1[Hour ID] = 1 );
        DIVIDE ( Table1[Sales Amount]; Table1[Amount Hours] )
    )

     

    If you want the weighted average (by number of hours):

     

    Measure =
    DIVIDE (
        CALCULATE ( SUM ( Table1[Sales Amount] ); Table1[HOURS ID] = 1; ALL ( Table1 ) );
        CALCULATE ( SUM ( Table1[Amount Hours] ); Table1[HOURS ID] = 1; ALL ( Table1 ) )
    )
    • Anonymous's avatar
      Anonymous
      Not applicable

      Thanx for your answer!