Forum Discussion
Count Consecutive Increasing Amounts
- Anonymous2 years ago
You can try the following modified calculated columns
Rank = RANKX ( FILTER ( 'Table', [Item] = EARLIER ( 'Table'[Item] ) ), [LotNumber], , ASC )Flag = VAR a = CALCULATE ( SUM ( 'Table'[Amount] ), ALLEXCEPT ( 'Table', 'Table'[Item] ), 'Table'[Rank] = EARLIER ( 'Table'[Rank] ) - 1 ) RETURN IF ( [Amount] > a, 1, 0 )ConsecutiveCount = VAR a = MAXX ( FILTER ( 'Table', [Flag] = 0 && [Item] = EARLIER ( 'Table'[Item] ) && [Rank] <= EARLIER ( 'Table'[Rank] ) ), [Rank] ) RETURN SWITCH ( TRUE (), a = BLANK (), [Rank], [Rank] = a, 1, [Rank] - a + 1 )Output
Best Regards!
Yolo Zhu
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- PowerBITestingG2 years agoResolver I
Thank you for that. Would you know why when you add another column everything turns to 1?
- Anonymous2 years agoNot applicable
Hi,
Thanks for the solution ThxAlot provided, it is excellent, and i want to offer some more information for user to refer to.
hello PowerBITestingG , based on your descriotion, you can create the following calcualted columns.
Rank = RANKX('Table',[LotNumber],,ASC)Flag = VAR a = LOOKUPVALUE ( 'Table'[Amount], [Rank], [Rank] - 1 ) RETURN IF ( [Amount] > a, 1, 0 )ConsecutiveCount = VAR a = MAXX ( FILTER ( 'Table', [Flag] = 0 && [Rank] <= EARLIER ( 'Table'[Rank] ) ), [Rank] ) RETURN SWITCH ( TRUE (), a = BLANK (), [Rank], [Rank] = a, 1, [Rank] - a + 1 )Output
Best Regards!
Yolo Zhu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
- PowerBITestingG2 years agoResolver I
The issue now is that if you use the entire dummy data I shared (with Item Chair), the code stops working