Forum Discussion

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!! 🙂

3 years ago

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.

Community Support

3 years ago

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.

Frequent Visitor

3 years ago

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