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Current formula is as follows:
Overall Rank= RANKX(all('EM2016 Participants');[Points])
This ranks all the participants by Sum([Points]), works like a charm..
BUT.. we want the RANK to be affected by a second column/value to determine the leader amongst the people with same score.
As we can see 7 people have an overall rank of 1. But when rank is identical "U poeng" should determine who the leader is. Currently its only sortet alphabetical.
Is it possible to do some sort of nested RANKX that gives rank based on two measures?
Nikil with 3 U points should be the last of Rank=1.
Ideally only one person should be rank 1 here and that is Bjørn since he is the only one with U poeng= 6.
Solved! Go to Solution.
Hi there,
A pattern I have used in this situation is:
Final value to be ranked =
Rank on Primary Measure (ascending)
+ Rank on Secondary Measure (ascending) / (Total Row Count + 1)
The first term is the Primary Measure rank, and the second term is the Secondary Measure rank scaled to be between 0 and 1 so that it can break ties in the Primary Measure rank.
In DAX, assuming you have two measures, [Primary Measure] and [Secondary Measure], to be ranked over all rows of Table:
Final Rank = RANKX ( ALL ( Table ), RANKX ( ALL ( Table ), [Primary Measure],, ASC ) + DIVIDE ( RANKX ( ALL ( Table ), [Secondary Measure],, ASC ), ( COUNTROWS ( ALL ( Table ) ) + 1 ) ) )
Just replace with your table/measure names and it should work.
Let me know how that goes 🙂
Hi there,
A pattern I have used in this situation is:
Final value to be ranked =
Rank on Primary Measure (ascending)
+ Rank on Secondary Measure (ascending) / (Total Row Count + 1)
The first term is the Primary Measure rank, and the second term is the Secondary Measure rank scaled to be between 0 and 1 so that it can break ties in the Primary Measure rank.
In DAX, assuming you have two measures, [Primary Measure] and [Secondary Measure], to be ranked over all rows of Table:
Final Rank = RANKX ( ALL ( Table ), RANKX ( ALL ( Table ), [Primary Measure],, ASC ) + DIVIDE ( RANKX ( ALL ( Table ), [Secondary Measure],, ASC ), ( COUNTROWS ( ALL ( Table ) ) + 1 ) ) )
Just replace with your table/measure names and it should work.
Let me know how that goes 🙂
Hello, can anyone can show how code will change for 5 columns?
Hello. I attempted your solution on my end, but only got the statement "A circular dependency was detected: Query1[Rank_intermediary]."
Here is how I attempted your solution:
Rank_intermediary = RANKX ( ALL ( Query1 ), RANKX ( ALL ( Query1 ), [AvgEmpTimeLength],, True ) + DIVIDE ( RANKX ( ALL ( Query1 ), [EmpFileCount],, False ), ( COUNTROWS ( ALL ( Query1 ) ) + 1 ) ) )
For what it's worth, I can already rank the employees by their first rank criteria, [AvgEmpTimeLength]. Here's how I did that:
Rank_wTies = IF ( HASONEVALUE( Query1[EmployeeName] ), RANKX( ALLSELECTED ( Query1[EmployeeName] ), [AvgEmpTimeLength],, TRUE ) )
Thanks in advance.
Hello,
how to find only top 5 ranking records...
Hi @Anonymous
Use RankX on the measure/numeric columns you want to top N. See below example
Rank_for_TopN = RANKX(
ALL(Datatable[Dimension/attribute]),
[measure/numeric value]
)
Then use Rank_for_TopN in visual level filter, go to advance filtering and use "is less than" and enter 5.
This will limit the values in visual to 5 or number your enter.
This is amazing DAX code! Is it possible to have a dual sort setup by Date then by Value?
My goal is to show daily expenses that are sorted descending by Date first (most recent date at the top) and then by value (higest value on 12/31/2018 for example would be at the top sorting down to the lowest value.
Thanks!
Russ
Thanks, worked great! 🙂
How can i rank as per a dynamic measure, not a fixed column.
I want to create a rank measure, using 2 measures, not 2 columns. Rankx seems to work only for columns in tables.
This works great! I still wish they would add a built-in tiebreaker in RANKX
BTW still haven't had a chance to test the alternate solutions here ( but the + 0 is a nice shortcut )
Anyway as you can see I still do get ties but those really should probably remain tied in my case!
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