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Anonymous
Not applicable

Dinamically calculate % variation in function of selected granularity.

Hello everyone,

how can I calculate a % variation from previous period that is dynamic in function of the selected date granularity?

Many Thanks,
Roberto

 

1 ACCEPTED SOLUTION
ibarrau
Super User
Super User

Hi, your questions is quite ambigous. If you want to work with time intelligence, the first thing you need to know is that you need a Date Table containing one row for each day. Here you can find scripts to build one: https://blog.ladataweb.com.ar/post/185012464051/data-modeling-como-hago-una-tabla-fecha

 

After you relate the data from that table to the date on your table you can build this calculation like this:

Measure = 
CALCULATE (
    Aggregation ( Table[Column] ) 
    , DATEADD ( DateTable[DateColumn] , -N , Period)
)

Where period is year, month or day; -N is the number to go back in period; and "Aggregation" is sum, average, count, etc.

 

Hope this helps,

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


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

Happy to help!

LaDataWeb Blog

View solution in original post

1 REPLY 1
ibarrau
Super User
Super User

Hi, your questions is quite ambigous. If you want to work with time intelligence, the first thing you need to know is that you need a Date Table containing one row for each day. Here you can find scripts to build one: https://blog.ladataweb.com.ar/post/185012464051/data-modeling-como-hago-una-tabla-fecha

 

After you relate the data from that table to the date on your table you can build this calculation like this:

Measure = 
CALCULATE (
    Aggregation ( Table[Column] ) 
    , DATEADD ( DateTable[DateColumn] , -N , Period)
)

Where period is year, month or day; -N is the number to go back in period; and "Aggregation" is sum, average, count, etc.

 

Hope this helps,

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


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

Happy to help!

LaDataWeb Blog

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