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sandy2405
Frequent Visitor

Ranking on Multiple Measures

Hi All,

 

I am facing issue with calculating rank on mutlple measures. In my case, I have to assign rank to players based on fifties, they have scored.
In my data set, I have tie between Warner and Gambhir (like below screenshot). I am calculating the rank on below formula:

Rank By Most Fifties = RANKX(ALL(Deliveries[batsman]), [Batsman Total Fifties],,DESC,Skip)

1.JPG

 

 

 

 

 

 

 

 

 

 

To deal with tie, I need to re-assign rank based on below three measure:

  1. Total Fifties
  2. Total Runs
  3. Average

I am unable to formulate the DAX for calculating the rank based on these 3 measures. Please provide your input(s) here.

 

Thanks,

Sandy

1 ACCEPTED SOLUTION
v-cherch-msft
Employee
Employee

Hi @sandy2405

 

You may create a measure like below.For further, you may refer to this post.

Measure  = 
RANKX (
    ALL ( 'Table' ),
    RANKX ( ALL ( 'Table' ), [Measure1],, ASC )
        + DIVIDE (
            RANKX ( ALL ( 'Table' ), [Measure2],, ASC ),
            COUNTROWS ( ALL ( 'Table' ) ) + 1
        )
        + DIVIDE (
            DIVIDE (
                RANKX ( ALL ( 'Table' ), [Measure3],, ASC ),
                ( COUNTROWS ( ALL ( 'Table' ) ) + 1 )
            ),
            COUNTROWS ( ALL ( 'Table' ) ) + 1
        ),
    ,
    ASC,
    DENSE
)

Regards,

Cherie

Community Support Team _ Cherie Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

View solution in original post

1 REPLY 1
v-cherch-msft
Employee
Employee

Hi @sandy2405

 

You may create a measure like below.For further, you may refer to this post.

Measure  = 
RANKX (
    ALL ( 'Table' ),
    RANKX ( ALL ( 'Table' ), [Measure1],, ASC )
        + DIVIDE (
            RANKX ( ALL ( 'Table' ), [Measure2],, ASC ),
            COUNTROWS ( ALL ( 'Table' ) ) + 1
        )
        + DIVIDE (
            DIVIDE (
                RANKX ( ALL ( 'Table' ), [Measure3],, ASC ),
                ( COUNTROWS ( ALL ( 'Table' ) ) + 1 )
            ),
            COUNTROWS ( ALL ( 'Table' ) ) + 1
        ),
    ,
    ASC,
    DENSE
)

Regards,

Cherie

Community Support Team _ Cherie Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

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