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siuchan Frequent Visitor
Frequent Visitor

Percentage Sum Total

I would like to write a new measure to calculate the % of the total for the period. I can calculate the sum score for the whole table, but not by the period. Thanks,   

 

Company    Period   Score   % total for the period 

A                    Q1       2          16.7%

A                     Q2     3           27.3%

B                    Q1      4           33.3%

B                   Q2       5           45.5%

C                   Q1       6           50%

C                   Q2       3            27.3%

 

 

1 ACCEPTED SOLUTION

Accepted Solutions
Community Support Team
Community Support Team

Re: Percentage Sum Total

@siuchan,

 

You may try using ALLSELECTED Function.

Measure =
DIVIDE (
    SUM ( Table1[Score] ),
    CALCULATE ( SUM ( Table1[Score] ), ALLSELECTED ( Table1[Company] ) )
)
Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
3 REPLIES 3
Super User
Super User

Re: Percentage Sum Total

@siuchan

 

Hi, try with this measure

 

% Total for Period =
VAR TotalPeriod =
    CALCULATE ( SUM ( Table1[ Score ] ), ALLEXCEPT ( Table1, Table1[  Period ] ) )
RETURN
    DIVIDE ( SUM ( Table1[ Score ] ), TotalPeriod )

Regards

 

Victor

 




Lima - Peru
Did I answer your question? Mark my post as a solution!

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siuchan Frequent Visitor
Frequent Visitor

Re: Percentage Sum Total

Thanks for your suggestion. I tried it. But, it gave me the sum of the whole table. only If i specific the time period with the slider, they i can get the total sum % at the specific period. 

 

 

Community Support Team
Community Support Team

Re: Percentage Sum Total

@siuchan,

 

You may try using ALLSELECTED Function.

Measure =
DIVIDE (
    SUM ( Table1[Score] ),
    CALCULATE ( SUM ( Table1[Score] ), ALLSELECTED ( Table1[Company] ) )
)
Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.