Forum Discussion
Matrix values doesn't add up
Hello,
I've a matrix
with one column containing Name and one column with values Agile, Bold, Engaged, Not, Smart.
The total of distinct count names is 159, but the number doesn't add.
I have multiple rows per each person.
Smart+ Bold+Engaged+Smart+Not = 159 but I can't manage to find the right one numbers.
- Anonymous2 years ago
Thanks for the reply from Kaviraj11 , please allow me to provide another insight:
Hi Anonymous ,
Are you referring to the incorrect total row calculation? You can consider using the IF() + HASONEVALUE() function to determine if it’s a total. If it is, you can perform calculations based on your requirements, such as average, sum, etc.
Measure = var _table= SUMMARIZE('Table',[Name],"Value",[Count]) return IF( HASONEVALUE('Table'[Name]),[Count],SUMX(_table,[Value]))Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 Replies
- Kaviraj11Solution Sage
Hi,
if you’re trying to get the correct total for distinct counts across multiple rows per person, you might encounter issues because the distinct count in the total row doesn’t simply add up the individual distinct counts from each category. Instead, it calculates the distinct count across the entire dataset.
Create Measure
Total Distinct Names =
SUMX(
VALUES('Table'[Name]),
CALCULATE(
DISTINCTCOUNT('Table'[Name]),
FILTER(
ALL('Table'),
'Table'[Attribute] IN {"Agile", "Bold", "Engaged", "Not", "Smart"}
)
)
)Solved: DISTINCTCOUNT Total not correct - Microsoft Fabric Community
- AnonymousNot applicable
Thanks for the reply from Kaviraj11 , please allow me to provide another insight:
Hi Anonymous ,
Are you referring to the incorrect total row calculation? You can consider using the IF() + HASONEVALUE() function to determine if it’s a total. If it is, you can perform calculations based on your requirements, such as average, sum, etc.
Measure = var _table= SUMMARIZE('Table',[Name],"Value",[Count]) return IF( HASONEVALUE('Table'[Name]),[Count],SUMX(_table,[Value]))Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.