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 JonathanG
		
			JonathanG
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		Hi Guys,
There are such data:
| PLACE | REGION | VALUE1 | VALUE2 | DIFF | 
| 1 | A | 2 | 3 | 1 | 
| 1 | B | 7 | 5 | -2 | 
| 1 | C | 5 | 7 | 2 | 
| 1 | D | 8 | 4 | -4 | 
| 2 | E | 9 | 5 | -4 | 
| 2 | F | 4 | 2 | -2 | 
| 2 | G | 7 | 5 | -2 | 
| 2 | H | 5 | 7 | 2 | 
| 3 | I | 8 | 9 | 1 | 
| 3 | J | 0 | 2 | 2 | 
| 3 | K | 5 | 5 | 0 | 
How to get the highest negative value of "Place"? so the result will be -8 (-4+(-2)+(-2)) (Place "2").
Thankyou
 JonathanG
		
			JonathanG
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		Thank God, I found it!!
The point is I have to aggregate my calculation based on smallest area (region), give condition, sum it, aggregate again based on bigger area (place), and then get the minimum value
so the expression is:
min(aggr(sum(if(aggr(sum(value1)-sum(value2),region)<0,aggr(sum(value1)-sum(value2),region),0)),place))
Thankyou Sunny and Kush!
 sunny_talwar
		
			sunny_talwar
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		May be this
Min(Aggr(
  Sum(DIFF)
, PLACE)) JonathanG
		
			JonathanG
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		Thankyou Sunny, it returns Place "2" but with value = -6 (included positive value). I just need the highest sum of negative value (-8) and ignore the positive ones
 Kushal_Chawda
		
			Kushal_Chawda
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		jus with little bit modification to what @sunny_talwar suggested.
=min(aggr(sum({<DIFF={"<0"}>}DIFF),PLACE))
 JonathanG
		
			JonathanG
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		Sorryy, i forgot to mention
Actually Value 1 and Value 2 are group of value. and DIFF is measure
so i put expression sum(value2)-sum(value1) to get DIFF calculation
when i put it into your expression =min(aggr(sum({<sum(Value2)-sum(Value1)={"<0"}>}sum(value2)-sum(value1)),PLACE)), it returns "-" in my KPI
did i miss something?
 Kushal_Chawda
		
			Kushal_Chawda
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		try below
min(aggr(if(sum(Value2)-sum(Value1)<0,sum(value2)-sum(value1)),PLACE))
 JonathanG
		
			JonathanG
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		Thankyou Kush, but it still returns to -6 😞
I think it still aggregates the every value of "Place", but i have no idea to solve it
Help
 JonathanG
		
			JonathanG
		
		
		
		
		
		
		
		
	
			
		
		
			
					
		Thank God, I found it!!
The point is I have to aggregate my calculation based on smallest area (region), give condition, sum it, aggregate again based on bigger area (place), and then get the minimum value
so the expression is:
min(aggr(sum(if(aggr(sum(value1)-sum(value2),region)<0,aggr(sum(value1)-sum(value2),region),0)),place))
Thankyou Sunny and Kush!
