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Anonymous
Not applicable

Getting cumulative total of a grouped value

Hi Everyone,

 

I have following matrix table with grouped total days open and Year/Month

 

Capture2.JPG

I'm trying to get cumulative total and tried following;

 

Cumulative Total = CALCULATE([Count of Issues],FILTER(ALLSELECTED(Query25),Query25[Month_Year]<=MAX(Query25[Month_Year])&&Query25[Total Days Open Excluding Q (groups)]<=MAX(Query25[Total Days Open Excluding Q (groups)])))
 
However, the result is as follows;
 
Capture1.JPG
 

 As it can be seen above, only December column has the correct cummelative total. Cummelative total is working both horizontally and vertically. I just want to get cummelative total for each month.

 

Appreciate any help.

 

Thanks in advance.

 

Best Regards,

Ashan.

 

1 ACCEPTED SOLUTION
Payeras_BI
Super User
Super User

Hi @Anonymous ,

 

Try like this:

 

 

Cumulative Total =
CALCULATE (
    [Count of Issues],
    FILTER (
        ALLSELECTED ( Query25[Total Days Open Excluding Q (groups)] ),
        Query25[Total Days Open Excluding Q (groups)]
            <= MAX ( Query25[Total Days Open Excluding Q (groups)] )
    )
)

 


 Regards,

If this post answered your question, please mark it as a solution to help other users find useful content.
Kudos are another nice way to acknowledge those who tried to help you.

J. Payeras
Mallorca, Spain

View solution in original post

2 REPLIES 2
Anonymous
Not applicable

Hi @Payeras_BI,

 

Thank you so much. It works.

 

Best Regards,

Ashan.

Payeras_BI
Super User
Super User

Hi @Anonymous ,

 

Try like this:

 

 

Cumulative Total =
CALCULATE (
    [Count of Issues],
    FILTER (
        ALLSELECTED ( Query25[Total Days Open Excluding Q (groups)] ),
        Query25[Total Days Open Excluding Q (groups)]
            <= MAX ( Query25[Total Days Open Excluding Q (groups)] )
    )
)

 


 Regards,

If this post answered your question, please mark it as a solution to help other users find useful content.
Kudos are another nice way to acknowledge those who tried to help you.

J. Payeras
Mallorca, Spain

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