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Artemis1254's avatar
Artemis1254
Frequent Visitor
8 years ago
Solved

Unorthodox Median Calculation

Hi,

 

In Direct Query mode, is there a way to calculate median numbers such that if the number of values selected is odd then the middle one will become the median. If the number of values chosen are odd, then the median would become the value of the "upper" median value instead of being an average. For instance, if the numbers chosen are : 1, 3, 4, 6. The median of this set of numbers is 4.

If the numbers chosen were 1, 3, and 4, then the 3 would be chosen/displayed as the median.

In this case, the number that should be chosen is 2.333. 

Need Help
  • OwenAuger's avatar
    OwenAuger
    8 years ago

     

    Hi again,

    It turns out you can do something like this, assuming you have enabled all functions in DirectQuery.

     

    I have taken this code basically verbatim from

    https://www.daxpatterns.com/statistical-patterns/#median22

     

    Median Unorthodox =
VAR NumValuesHalved = COUNT ( YourTable[Latency] ) / 2
RETURN
    MINX (
        FILTER (
            VALUES ( YourTable[Latency] ),
            CALCULATE (
                COUNT ( YourTable[Latency] ),
                YourTable[Latency] <= EARLIER ( YourTable[Latency] )
            )
                > NumValuesHalved
        ),
        YourTable[Latency]
    )

     The code produces the "upper" median value in the case of an even number of items, otherwise produces the middle value.

     

    Regards,

    Owen :)

2 Replies

  • OwenAuger's avatar
    OwenAuger
    Super User
    8 years ago

     

    Hi again,

    It turns out you can do something like this, assuming you have enabled all functions in DirectQuery.

     

    I have taken this code basically verbatim from

    https://www.daxpatterns.com/statistical-patterns/#median22

     

    Median Unorthodox =
VAR NumValuesHalved = COUNT ( YourTable[Latency] ) / 2
RETURN
    MINX (
        FILTER (
            VALUES ( YourTable[Latency] ),
            CALCULATE (
                COUNT ( YourTable[Latency] ),
                YourTable[Latency] <= EARLIER ( YourTable[Latency] )
            )
                > NumValuesHalved
        ),
        YourTable[Latency]
    )

     The code produces the "upper" median value in the case of an even number of items, otherwise produces the middle value.

     

    Regards,

    Owen :)

    • Artemis1254's avatar
      Artemis1254
      Frequent Visitor
      8 years ago

      Thanks so much for your help. I can't even begin to tell you how much time I spent trying to google the answer and find out. Cheers