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Anonymous
Not applicable

Getting an average from sprints displayed from time slicer

How would I modify this to get an average of story points for the number of Sprints displayed from the time slicer? I can get the total easy enough...

Sprint Done Points = IF('Sprint Burndown'[Date].[Date] = 'Sprint Burndown'[Sprint End Date].[Date] && 'Sprint Burndown'[State] = "Done", 'Sprint Burndown'[Story Points],0)
I know I need to simply divide... but how to capture and calc the number of sprints displayed is what I'm stuck on

singleton2787_0-1664476695581.png

 

 
4 REPLIES 4
Anonymous
Not applicable

I think I can make it simpler... using a slicer to select the Sprint Name(s) add up the number of sprints selected (eg. 4 sprints selected) and divide the sum of the Sprint Points Done by that number. 381/4 in other works

singleton2787_0-1664802664740.png

 

v-henryk-mstf
Community Support
Community Support

Hi @Anonymous ,

 

A separate date table, the purpose of which is to avoid filtering from date fields in the fact table.


If the problem is still not resolved, please provide detailed error information and test data. Looking forward to your reply.


Best Regards,
Henry


If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

Anonymous
Not applicable

where does the date column come from?

v-henryk-mstf
Community Support
Community Support

Hi @Anonymous ,

 

Try formula like below:

Sprint Done Points =
VAR slicer_result =
    SELECTEDVALUE ( date[column] )
RETURN
    IF (
        slicer_result = MAX ( date[column] ),
        IF (
            'Sprint Burndown'[Date].[Date] = 'Sprint Burndown'[Sprint End Date].[Date]
                && 'Sprint Burndown'[State] = "Done",
            'Sprint Burndown'[Story Points],
            0
        )
    )

 

If the problem is still not resolved, please provide detailed error information and test data. Looking forward to your reply.


Best Regards,
Henry


If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

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