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DKealy
Regular Visitor

Divide sum of one column by sum of another column

A power BI newbie here.

I have a table containing a number of columns, please see below:

Power BI desktop table.png

 

I want to add anew column in Power BI desktop report - table format - which contains the value from "Paid Cases" divided by value from "Collectable Cases" for each row in the table.

Sounds simple.

But I have not been able to get this to work. I thought the formula %Paid = DIVIDE('tablename'[Paid Cases]),'tablename'[Collectable Cases]),0) would work.

It doesn't. This just returned the same values in the "Paid Cases" column.

I have read many answers on forums and the internet, none which have produced a simple answer for what should be a simple action.

I do not want to create a new measure, just a simple formula to divide one column by another to produce a new column in the same table.

 

If it helps, "Paid Cases" is a count created in a table. "Collectable Cases" is a column that was added to the table as a reuslt of a simple calculation - "Cases Received" - ("Closed by Client"+"Closed by ZZPS"+"Accounts on Hold").


Any help and advice is greatly appreciated.

 

 

1 ACCEPTED SOLUTION
DKealy
Regular Visitor

I found the answer to my issue.

More fool me for not wanting to create a new measure, Creating a new measure was indeed the answer.

 

Paid % = DIVIDE(SUM('tablename'[Paid Cases]),SUM('tablename'[Collectable Cases]),0) is the measure that produced the desired results along with changing the formatting to "Percentage".

IT is all so easy once you know how.

Issue resolved.

View solution in original post

2 REPLIES 2
DKealy
Regular Visitor

I found the answer to my issue.

More fool me for not wanting to create a new measure, Creating a new measure was indeed the answer.

 

Paid % = DIVIDE(SUM('tablename'[Paid Cases]),SUM('tablename'[Collectable Cases]),0) is the measure that produced the desired results along with changing the formatting to "Percentage".

IT is all so easy once you know how.

Issue resolved.

Anonymous
Not applicable

@DKealy Your personal problem solving was useful to me. Thank you!

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