Forum Discussion
YoY% difference
- 1 year ago
Hi Anonymous
Have you tried using Visual Level Calculations:
https://www.wiseowl.co.uk/blog/s3172/visual-calculations-pbi.htm% previous = ([Sum of Price] - PREVIOUS([Sum of Price],1,COLUMNS)) / PREVIOUS([Sum of Price],1,COLUMNS)Otherwise you will need to create a measure like:
% change =var y = selectedValue(table[year column])
ReturnDivide(Sum(table[column]) -Calculate( Sum(table[column]), table[year column] = y -1),Calculate( Sum(table[column]), table[year column] = y -1)) - Anonymous1 year ago
Hi ,
The method SamWiseOwl provided should be helpful. Besides, you can also try the following DAX formula.
Yeartoyear difference = var _year = SELECTEDVALUE('Table'[Year]) VAR _PreviousDate = CALCULATE (MAX('Table'[Year]), 'Table'[Year] < _year) var _cost = SUM('Table'[Cost]) var _previouscost = CALCULATE(SUM('Table'[Cost]), 'Table'[Year] = _PreviousDate) RETURN _cost - _previouscostYou can view the following links to learn more information.
Solved: Calculate difference between columns in matrix vis... - Microsoft Fabric Community
Solved: Calculate % difference between two years - Microsoft Fabric Community
Best Regards,
Wisdom Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi ,
The method SamWiseOwl provided should be helpful. Besides, you can also try the following DAX formula.
Yeartoyear difference =
var _year = SELECTEDVALUE('Table'[Year])
VAR _PreviousDate = CALCULATE (MAX('Table'[Year]), 'Table'[Year] < _year)
var _cost = SUM('Table'[Cost])
var _previouscost = CALCULATE(SUM('Table'[Cost]), 'Table'[Year] = _PreviousDate)
RETURN
_cost - _previouscost
You can view the following links to learn more information.
Solved: Calculate difference between columns in matrix vis... - Microsoft Fabric Community
Solved: Calculate % difference between two years - Microsoft Fabric Community
Best Regards,
Wisdom Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.