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borniex's avatar
borniex
Frequent Visitor
3 years ago
Solved

aggregate series

Hello,
I want to aggregate series, I am attaching a picture of my idea, does anyone have an existing solution?
Thanks for your help!! 🙂

  • v-zhangti's avatar
    v-zhangti
    3 years ago

    Hi, borniex 

     

    You can try the following methods.

    Column:

    previous part = CALCULATE(MAX('Table'[part]),FILTER('Table',[id]=EARLIER('Table'[id])-1))

    judgement = IF([part]=[previous part],1,0)//Determine if the part is the same as the previous id
    First part = IF([judgement]=0,1,0)
    Max ID = CALCULATE(Max('Table'[id]),FILTER('Table',[id]<=EARLIER('Table'[id])&&[First part]=1))
    Sum value = CALCULATE(SUM('Table'[value]),ALLEXCEPT('Table','Table'[Max ID]))

    Is this the result you expect?

     

    Best Regards,

    Community Support Team _Charlotte

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

     

     

3 Replies

  • v-zhangti's avatar
    v-zhangti
    Community Support
    3 years ago

    Hi, borniex 

     

    You can try the following methods.

    Column:

    previous part = CALCULATE(MAX('Table'[part]),FILTER('Table',[id]=EARLIER('Table'[id])-1))

    judgement = IF([part]=[previous part],1,0)//Determine if the part is the same as the previous id
    First part = IF([judgement]=0,1,0)
    Max ID = CALCULATE(Max('Table'[id]),FILTER('Table',[id]<=EARLIER('Table'[id])&&[First part]=1))
    Sum value = CALCULATE(SUM('Table'[value]),ALLEXCEPT('Table','Table'[Max ID]))

    Is this the result you expect?

     

    Best Regards,

    Community Support Team _Charlotte

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

     

     

    • borniex's avatar
      borniex
      Frequent Visitor
      3 years ago

      Thank you very much, I was interested in this solution! Elegant, I thought more complicated 😄