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Anonymous
Not applicable

Divide filtered value by (non filtered) total

Hello,

 

I need help in something that might be simple.

I have a table with "total transactions per hour" in one column and "50% of the transactions per hour" in another column, something similar to this:

 

HOUR        TOT       50%

3pm          208       104

4pm          212       106 

5pm          218       109

Total         638       319

 

I have to create another column with the percentage of the 50% of transactions per each hour over the total (638). In other words, i need the following:

104 / 638

106 / 638

109 / 638

 

Any help would be greatly appreciated.

 

thanks!

2 ACCEPTED SOLUTIONS
ChrisMendoza
Resident Rockstar
Resident Rockstar

@Anonymous -

You should be apple to add another column as:

% =
DIVIDE ( TableName[50%], SUM ( TableName[TOT] ), 0 )

98.PNG






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View solution in original post

@Anonymous 

 

You may add the measure below.

Measure 2 =
DIVIDE (
    [50%],
    CALCULATE ( SUM ( 'Table'[TOT] ), ALLSELECTED ( 'Table'[HOUR] ) )
)
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.

View solution in original post

4 REPLIES 4
ChrisMendoza
Resident Rockstar
Resident Rockstar

@Anonymous -

You should be apple to add another column as:

% =
DIVIDE ( TableName[50%], SUM ( TableName[TOT] ), 0 )

98.PNG






Did I answer your question? Mark my post as a solution!
Did my answers help arrive at a solution? Give it a kudos by clicking the Thumbs Up!

Proud to be a Super User!



Anonymous
Not applicable

thanks.

but the challenge is that the "50% transactions" is not a straight calculation from "tot transactions", but the result of SELECTEDVALUE from a filter...

@Anonymous 

 

You may add the measure below.

Measure 2 =
DIVIDE (
    [50%],
    CALCULATE ( SUM ( 'Table'[TOT] ), ALLSELECTED ( 'Table'[HOUR] ) )
)
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.
Anonymous
Not applicable

thanks a lot  , this is exactly what i was looking for!

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