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Hi Experts, As you can see in first table i have calculated the average by row level and in the second table we have the Average by platform level.
Please help me on this
Thank You
Hi,
Follow these steps:
1-Create a measure as below:
Average By Platform =
VAR _PLATFORM =
SELECTEDVALUE ( PlatformEffTable[Platform] )
RETURN
DIVIDE (
CALCULATE (
SUM ( PlatformEffTable[Efficiency] ),
FILTER ( ALL ( PlatformEffTable ), PlatformEffTable[Platform] = _PLATFORM )
),
CALCULATE (
COUNTROWS ( PlatformEffTable ),
FILTER ( ALL ( PlatformEffTable ), PlatformEffTable[Platform] = _PLATFORM )
)
)
2- Create another measure as below to get what you asked for:
Average of Greater than averages =
VAR _PLATFORM =
SELECTEDVALUE ( PlatformEffTable[Platform] )
RETURN
DIVIDE (
CALCULATE (
SUM ( PlatformEffTable[Efficiency] ),
FILTER (
ALL ( PlatformEffTable ),
PlatformEffTable[Efficiency] > [Average By Platform]
&& PlatformEffTable[Platform] = _PLATFORM
)
),
COUNTROWS (
FILTER (
ALL ( PlatformEffTable ),
PlatformEffTable[Efficiency] > [Average By Platform]
&& PlatformEffTable[Platform] = _PLATFORM
)
)
)
Then the result would be like the attached photo.
If it solves your problem, then please consider Accepting it as the solution
Regards,
Loran
Hi,
Thanks for the reply,
But there is some issue>>
Please look into the below screenshot where we have "Efficiency" which is nothing but Division in (%) of 'Planned Hours' and 'Actual Hours' which should be the average on row level.
And the ask is if we filter on any specific platform, then we will have average "Efficiency", then we need to calculate the "Potential'>>>
For e.g>> We have filter on "Platform" = PDP, and then Average of "Efficiency" is =102.38% which we arrived from 375/366
2. And then we will select all the cells which are greater then 102.38% and get the average of all those greater percents (as compared to 102.38% should be greater) in the another table which will be another percentage.
Please note i have already created
If I understand correctly, you want to make the calculation based on your efficiency measure. You need to change my previous answer a bit as below:
1-
If it solves your problem, then please consider Accepting it as the solution
Regards,
Loran
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