Forum Discussion
Filter the contents inside MATRIX visual
- 3 years ago
Actually, thinking about it, the average method could potentially not achieve what you are looking for.
If you have a set of values for a year like {30, 29, 30, 31}, the average method will hide both 30 values.
The following method however will work: compare the minimum value in a set to the maximum value. If they are the same, all values are the same...
So...
No duplicate profits = VAR _Min = MINX ( ALL ( 'Warehouse table'[Warehouse] ), [Your measure] ) VAR _MAX = MAXX ( ALL ( 'Warehouse table'[Warehouse] ), [Your measure] ) RETURN IF ( AND ( ISINSCOPE ( 'Warehouse table'[Warehouse] ), ISINSCOPE ( 'Year Table'[Year] ) ), IF ( _Min = _Max, BLANK (), [Your measure] ) )PS. I tried deleting the previous message but the forum won't let me for that particular post for some reason....
Here is one way. Basically you compare the sum (profit) of each warehouse with the average for all warehouses in a year.
First the model:
The calculation (without totals)
No duplicate profits =
VAR _Profit =
SUM ( fTable[Profit] )
VAR _Average =
CALCULATE ( AVERAGE ( fTable[Profit] ), ALL ( 'Warehouse table'[Warehouse] ) )
RETURN
IF (
AND (
ISINSCOPE ( 'Warehouse table'[Warehouse] ),
ISINSCOPE ( 'Year Table'[Year] )
),
IF ( _Profit = _Average, BLANK (), _Profit )
)
and if you need totals:
No Dups with totals =
SUMX(SUMMARIZE(fTable, 'Warehouse table'[Warehouse], 'Year Table'[Year]), [No duplicate profits])
Sample PBIX attached
Hi Thor2022
This is interesting, because for some reason it's not as simple to aggregate a distinct count on another measure! However it is always possible in DAX 😉
I created a new measure rather than chain it in called "New count" which evaluates to the total number of distinct sums. I am use group by to create an aggregated table which I can pass the summed column through to a distinct count.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||
| WH1 | 4 | 2 | 1 | 2 | 3 | 3 | 3 | 3 | |
| WH2 | 4 | 2 | 1 | 2 | 3 | 3 | 3 | 3 | |
| WH3 | 4 | 2 | 1 | 2 | 3 | 3 | 3 | 3 | |
| WH4 | 4 | 2 | 1 | 2 | 3 | 3 | 3 | 3 | |
Added into the final expression gives you:
Let me know if this works,
Pi
- Thor20223 years agoFrequent Visitor
Hi Pi
I am sure its a great solution , but give me some time to digest it , since some of the used functions are new for me 🙂
Regards