Forum Discussion
Distinctcount and average per month
- Anonymous2 years ago
Hi Seb93 ,
Based on the information you provided, I have created some data hopefully to solve the problem for you. You can follow the steps below:
1. Add an index column.
2. Add some Columns.Access = IF ( 'Table'[Device_category] = "Accessory", 1 )Mac_Access = VAR _1 = 'Table'[Gateway_mac] VAR _SumMac = CALCULATE ( SUM ( 'Table'[Access] ), FILTER ( 'Table', 'Table'[Gateway_mac] = _1 ) ) VAR _AccessMac = IF ( _SumMac >= 1, 1, 0 ) VAR _rank = RANKX ( FILTER ( 'Table', 'Table'[Gateway_mac] = _1 ), 'Table'[Index], , ASC ) RETURN IF ( _AccessMac = _rank, 1, 0 )
3. Add some Measure.CountMac = DISTINCTCOUNT ( 'Table'[Gateway_mac] )FinalResult = DIVIDE ( SUM ( 'Table'[Mac_Access] ), [CountMac] )Final output:
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If it does not help, please provide more details with your desired out put and pbix file without privacy information.
Best Regards,
Ada Wang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi Seb93 ,
Based on the information you provided, I have created some data hopefully to solve the problem for you. You can follow the steps below:
1. Add an index column.
2. Add some Columns.
Access =
IF ( 'Table'[Device_category] = "Accessory", 1 )
Mac_Access =
VAR _1 = 'Table'[Gateway_mac]
VAR _SumMac =
CALCULATE (
SUM ( 'Table'[Access] ),
FILTER ( 'Table', 'Table'[Gateway_mac] = _1 )
)
VAR _AccessMac =
IF ( _SumMac >= 1, 1, 0 )
VAR _rank =
RANKX (
FILTER ( 'Table', 'Table'[Gateway_mac] = _1 ),
'Table'[Index],
,
ASC
)
RETURN
IF ( _AccessMac = _rank, 1, 0 )
3. Add some Measure.
CountMac =
DISTINCTCOUNT ( 'Table'[Gateway_mac] )
FinalResult =
DIVIDE ( SUM ( 'Table'[Mac_Access] ), [CountMac] )
Final output:
How to Get Your Question Answered Quickly - Microsoft Fabric Community
If it does not help, please provide more details with your desired out put and pbix file without privacy information.
Best Regards,
Ada Wang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.