Forum Discussion
Count Unique count for columns in Table Visual in powerBI
I want to display the sum of all the positive values in the Margin column on the Card and the negative values in the Margin Column on another Card. First Card is 9 and the second card is 4 and third card 8
- Anonymous4 years ago
Hi Anonymous ,
If you want to calculate distinct count of Acct of each Product, please try:
Distinct Acct = CALCULATE(DISTINCTCOUNT('Table'[Acct]),ALLEXCEPT('Table','Table'[Product]))Count of Positive Total_Margin = CALCULATE(DISTINCTCOUNT('Table'[Acct]),FILTER('Table',[Product]=MAX('Table'[Product]) && 'Table'[Total_Margin]>0))Count of Negative Total_Margin = CALCULATE(COUNTROWS('Table'),FILTER('Table',[Product]=MAX('Table'[Product]) && 'Table'[Total_Margin]<0))Output:
Best Regards,
Eyelyn Qin
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
3 Replies
- Jbrunson09Frequent Visitor
Something like this should work:
Distinct Count = DISTINCTCOUNT('Table Name'[ACCT])
Distinct Count Negative Values = CALCULATE(DISTINCTCOUNT('Table Name'[ACCT]),FILTER('Table Name, [TOTAL MARGIN] < 0))
Distinct Count Positive Values = CALCULATE(DISTINCTCOUNT('Table Name'[ACCT]),FILTER('Table Name, [TOTAL MARGIN] >= 0))
- AnonymousNot applicable
This works fine but what I failed to put in the question is that the margin in each is an aggregation of transactions that is the group by account number with the same product.
- AnonymousNot applicable
Hi Anonymous ,
If you want to calculate distinct count of Acct of each Product, please try:
Distinct Acct = CALCULATE(DISTINCTCOUNT('Table'[Acct]),ALLEXCEPT('Table','Table'[Product]))Count of Positive Total_Margin = CALCULATE(DISTINCTCOUNT('Table'[Acct]),FILTER('Table',[Product]=MAX('Table'[Product]) && 'Table'[Total_Margin]>0))Count of Negative Total_Margin = CALCULATE(COUNTROWS('Table'),FILTER('Table',[Product]=MAX('Table'[Product]) && 'Table'[Total_Margin]<0))Output:
Best Regards,
Eyelyn Qin
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.