Ranking with FIlters in a Matrix Table

[rennda]

Hi there. I´m a rookie in DAX and Power Bi, so I would be happy if you can help me with my actual issue.  

I build a Matrix table which shows me the Ranking of MTBF by Name. I have tested two different measures (Rank MTBF 1 and RANK MTBF2).

When I drill down the Matrix I can see the "Device Classes" of the single employees with the Ticketcount and the MTBF

Now there is my issue. I have to set a filter on the ticketcount. I only wanna show and rank Device Classes with a Ticketcount >= 10.

If i set the ticketcount filter in the Filter Context from my Visual i´m getting the following result:

Here is the code i use actually for my ranking:

Rank MTBF 1 = RANKX(ALL(Ticket_Kennzahlen_table[Name]); CALCULATE([MTBF Performance]))

RANK MTBF 2 = RANKX(ALL(Ticket_Kennzahlen_table[Name]) ;CALCULATE(Ticket_Kennzahlen_table[MTBF Performance]; ALLEXCEPT(Ticket_Kennzahlen_table; Ticket_Kennzahlen_table[Name]);Ticket_Kennzahlen_table[ticket_gruppe] = 5||6;Ticket_Kennzahlen_table[Name (Gruppen)] = "varV");;DESC)

* In RANK MTBF 2 i set additional Filters on a Group of Names and the Ticketgroup. I don´t know if i really need this.

Thank you very much for your help

[v-xicai]

Hi @rennda ,

You may create the rank measure like DAX below.

Rank MTBF= RANKX(FILTER(ALLSELECTED(Ticket_Kennzahlen_table), Ticket_Kennzahlen_table[Name]=MAX(Ticket_Kennzahlen_table[Name])&&Ticket_Kennzahlen_table[Ticketcount] >10), MAX([MTBF Performance]),,DESC)

Could you please clarify the logical differences between the Rank MTBF 1 and Rank MTBF 2.

I am not sure what desired result would you want, could you please share your sample data and desired output screenshots for further analysis? You can also upload sample pbix to OneDrive and post the link here. Do mask sensitive data before uploading.

Please read this post to get your answer quickly: How to Get Your Question Answered Quickly.

Best Regards,

Amy

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.