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Bigrover Frequent Visitor
Frequent Visitor

Percent Column

I am trying to find percent approved, and having trouble. This is in a table by month and year

 

Year      Month       #Received       # Approved           %Approved

2019     Feb              50                     30                        ????

 

Formula?

1 ACCEPTED SOLUTION

Accepted Solutions
Super User
Super User

Re: Percent Column

@Bigrover yes it will, then you have to add a measure

 

% Approved = 
DIVIDE( SUM( Table[Approved] ), SUM( Table[Received] ) )





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5 REPLIES 5
Super User
Super User

Re: Percent Column

@Bigrover is it approved divided by received, add following column and change it to %

 

 

 

% Approved = DIVIDE( Table[Approve], Table[Received])





Did I answer your question? Mark my post as a solution.

Proud to be a Datanaut! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.





Bigrover Frequent Visitor
Frequent Visitor

Re: Percent Column

That doesn't work - it pulls a percent in the thousands, when it should be around 50% (for the real data).

Super User
Super User

Re: Percent Column

@Bigrover doesnt make sense, it should return 0.6 and when you change format to % it would show 60%






Did I answer your question? Mark my post as a solution.

Proud to be a Datanaut! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.





Bigrover Frequent Visitor
Frequent Visitor

Re: Percent Column

the months are summed data. Receieved is a sum of many entries as well as approved. Does this make a difference?

Super User
Super User

Re: Percent Column

@Bigrover yes it will, then you have to add a measure

 

% Approved = 
DIVIDE( SUM( Table[Approved] ), SUM( Table[Received] ) )





Did I answer your question? Mark my post as a solution.

Proud to be a Datanaut! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.





View solution in original post

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