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Anubis0815
Frequent Visitor

Column caluclation with multiple cross table filteres

Hi Community

 

I have a problem to calculate the following. In one data source I have all my revenues splitted by quarter. In another table I have a list with margin per quarter. Now I would like to calculate in my revenue table the effictive margin related to the margin for the quarter.

 

For example:

Table 1   
Product QuarterRevenue Margin(Calc)
AQ2500?
BQ3300?
CQ4600?

 

Table 2  
   
ProductMarginQuartal
A0.2Q2
B0.3Q2
C0.4Q2
A0.2Q3
B0.3Q3
C0.4Q3
A0.2Q4
B0.3Q4
C0.4Q4

 

Thank you very much for your help!

1 ACCEPTED SOLUTION
v-xicai
Community Support
Community Support

Hi @Anubis0815 ,

 

You may create column or measure in Table1 like DAX below.

 

Column: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = Table1[Product]&& Table2[Quartal] = Table1[Quarter]))  
 
Measure: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = MAX(Table1[Product])&& Table2[Quartal] = MAX(Table1[Quarter])))

Or you can also try to merge the two tables on the columns Product and Quarter in Query Editor.

 

Best Regards,

Amy

 

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-xicai
Community Support
Community Support

Hi @Anubis0815 ,

 

You may create column or measure in Table1 like DAX below.

 

Column: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = Table1[Product]&& Table2[Quartal] = Table1[Quarter]))  
 
Measure: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = MAX(Table1[Product])&& Table2[Quartal] = MAX(Table1[Quarter])))

Or you can also try to merge the two tables on the columns Product and Quarter in Query Editor.

 

Best Regards,

Amy

 

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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