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Afkkek
Frequent Visitor

Calculating an Average of previous rows in the Total

Hello, was having trouble with a calculation so thought I'd drop by here and see if anyone had any ideas to solve the issue.

 

In the below screenshot of an excel mockup, I have Column A which is a column from one table, and Column B which is a column from a different table, and then A/C which is just a measure which divides Column A by Column B.  This is repeated for each year.

 

Afkkek_0-1618862623154.png

 

The Grand Total would end up summing all of Column A and Column B, then repeating the measure's instructions of dividing A/C which I understand is what it's supposed to be doing, but any suggestions on how I would I go about it if I wanted to take a simple average of all the results of Column A/C?  The result of dividing the totals would be 0.05 in this case, but the average of (0.1, 0.025, 0.0833 and -0.05) would be 0.039583 in this example.

 

Thanks in advance!

1 ACCEPTED SOLUTION
FrankAT
Community Champion
Community Champion

Hi @Afkkek ,

here is my suggestion:

Sum of A = SUM('Table'[A])

Sum of B = SUM('Table'[B])

A/B = 
IF (
    HASONEVALUE ( 'Table'[Year] ),
    DIVIDE ( [Sum of A], [Sum of B] ),
    AVERAGEX ( 'Table', DIVIDE ( [Sum of A], [Sum of B] ) )
)

 

19-04-_2021_23-04-43.png

With kind regards from the town where the legend of the 'Pied Piper of Hamelin' is at home
FrankAT (Proud to be a Datanaut)

View solution in original post

2 REPLIES 2
FrankAT
Community Champion
Community Champion

Hi @Afkkek ,

here is my suggestion:

Sum of A = SUM('Table'[A])

Sum of B = SUM('Table'[B])

A/B = 
IF (
    HASONEVALUE ( 'Table'[Year] ),
    DIVIDE ( [Sum of A], [Sum of B] ),
    AVERAGEX ( 'Table', DIVIDE ( [Sum of A], [Sum of B] ) )
)

 

19-04-_2021_23-04-43.png

With kind regards from the town where the legend of the 'Pied Piper of Hamelin' is at home
FrankAT (Proud to be a Datanaut)

Thanks FrankAT! I ended up making it work on the side with a needlessly complicated approach, but your solution is much more intuitive.

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