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IB
Frequent Visitor

Ranking column base on virtual table column

I need to add a new column  RANKING  "PBND(Hrs)" column in this virtual table

PBN =

SUMMARIZE (

    Table1,

    Table1[Step],

    "Units", DISTINCTCOUNT ( Table1[Job_No] ),

    "AllowTime", MAX ( Table1[Allowable (hrs)] ),

    "StepAccuTime", SUM ( Table1[Bucket-time] ),

    "TimeExpected", ( DISTINCTCOUNT ( Table1[Job_No] ) * MAX ( Table1[Allowable (hrs)] ) ),

    "PBND(Hrs)", SUM ( Table1[Bucket-time] ) - ( DISTINCTCOUNT ( Table1[Job_No] ) * MAX ( Table1[Allowable (hrs)] )),

    "Sort",MAX(Table1[Bucket]))

umn  RANKING  "PBND(Hrs)" column in this virtual table

1 ACCEPTED SOLUTION
Anonymous
Not applicable

Hi @IB ,

 

Please add a column to the PBN table.

RANKING PBND(Hrs) = RANKX(PBN, PBN[PBND(Hrs)], , DESC, Dense)

This formula will rank the values of the PBND(Hrs) column in descending order and assign the same rank to ties.

You can adjust the formula according to your needs, such as changing the order or the tie mode.

 

Please refer to this for more information.

 

Best Regards,

Neeko Tang

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
Anonymous
Not applicable

Hi @IB ,

 

Please add a column to the PBN table.

RANKING PBND(Hrs) = RANKX(PBN, PBN[PBND(Hrs)], , DESC, Dense)

This formula will rank the values of the PBND(Hrs) column in descending order and assign the same rank to ties.

You can adjust the formula according to your needs, such as changing the order or the tie mode.

 

Please refer to this for more information.

 

Best Regards,

Neeko Tang

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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