Forum Discussion

adnanarain's avatar
adnanarain
Icon for Helper V rankHelper V
6 years ago
Solved

Rank Measure with ID and Date

Hi All,

Below is the sample data:

IDDateQuantity
101/01/20      2
101/02/20      3
101/03/20      1
201/01/20      3
201/02/20      7
201/03/20      9
301/04/20      3
301/05/20      1
301/06/20      2

 

below is the output I need: Date Slicer dates selected from 1/2/2020 to 1/6/2020

 

IDDateQuantity  Rank
101/02/20      3   1
101/03/20      1   2
201/02/20      7   1
201/03/20      9   2
301/04/20      3   1
301/05/20      1   2
301/06/20      2   3

 

I want to create a measure to display rank within ID category and rank start from the oldest date selected through a slicer.

  • Anonymous's avatar
    Anonymous
    6 years ago

     

    adnanarain 

    Try this one:

    Measure = RANKX(FILTER(ALLSELECTED('Table'),[ID]=MAX('Table'[ID])),CALCULATE(SUM([Date])),,ASC,Dense)

     

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

4 Replies

    • adnanarain's avatar
      adnanarain
      Icon for Helper V rankHelper V

      Thanks, Amit for your reply. This ranking is based on date and quantity so basically I need rank based on ID and date no matter what is the quantity, so for example, if we have ID 1, the date is 1st January and the quantity is 5 and in 2nd row, we have ID 1, date 2nd January and quantity is 8 then the rank will be 1 for 1st January and 2 for 2nd January. No matter what is the quantity.

      I have tried to clear my point here I hope is clear now.

      • Anonymous's avatar
        Anonymous
        Not applicable

         

        adnanarain 

        Try this one:

        Measure = RANKX(FILTER(ALLSELECTED('Table'),[ID]=MAX('Table'[ID])),CALCULATE(SUM([Date])),,ASC,Dense)

         

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