Advance your Data & AI career with 50 days of live learning, dataviz contests, hands-on challenges, study groups & certifications and more!
Get registeredGet Fabric Certified for FREE during Fabric Data Days. Don't miss your chance! Request now
Goodnight!
I'm trying to create a measure that basically multiplies a column by a rate obtained from a division, but the total return is not the sum of all fields as shown in print:
How can I solve this problem? I'm gonna try to group the data with summarize, but I dont know if it's the right way.
The value that I want is that:
My sample data and pbix (link)
Here how I created the measures:
_Approveds = SUM(DashFraude[approved])
_Bads = CALCULATE(SUM('DashFraude'[fraud]),DashFraude[approved]=1)
_TxFraudApproveds = [_Bads] / [_Approveds]
_Disapprovevds = SUM(DashFraude[disapproved])
_TotalAppDisap = [_Approveds] + [_Disapprovevds]
_InferenceFraudDisapproveds = [_Disapprovevds] * [_TxFraudApproveds]
Thank you for the support.
Solved! Go to Solution.
Hi @fellipea_ao
I hope I understood your requirement to total the [_InferenceFraudDisapproved] measure.
I created a new [_InferenceFraudDisapproveds 2] measure.
Original:
_InferenceFraudDisapproveds = [_Disapprovevds] * [_TxFraudApproveds]
Mine:
_InferenceFraudDisapproveds 2 =
SUMX(
VALUES( DashFraude[fx_score1] ),
[_Disapprovevds] * [_TxFraudApproveds]
)
20240424-dash-sample-test-MINE.pbix
Let me know how that works.
Hi @fellipea_ao
I hope I understood your requirement to total the [_InferenceFraudDisapproved] measure.
I created a new [_InferenceFraudDisapproveds 2] measure.
Original:
_InferenceFraudDisapproveds = [_Disapprovevds] * [_TxFraudApproveds]
Mine:
_InferenceFraudDisapproveds 2 =
SUMX(
VALUES( DashFraude[fx_score1] ),
[_Disapprovevds] * [_TxFraudApproveds]
)
20240424-dash-sample-test-MINE.pbix
Let me know how that works.
Advance your Data & AI career with 50 days of live learning, contests, hands-on challenges, study groups & certifications and more!
Check out the October 2025 Power BI update to learn about new features.