Average Days

[JB17]

Hi everyone,

Can you help me create dax for below? I need to create a measure to calculate the average days of processing date for system ID 0000-0000-0000. So let say for the first two rows, I want to calculate how many days do we get from 02/06/2018 to 08/08/2018? and the average per month? Thanks in advance!

[Anonymous · Accepted solution]

Hi @JB17 ,

According to your description, here are my steps you can follow as a solution.

(1) This is my test data. 

(2) We can create  measures. 

ProcessingDaysDifference = 
VAR CurrentDate = MAX('Table'[Processing Date])
VAR PreviousDate =
    CALCULATE(
        MAX('Table'[Processing Date]),
        FILTER(
            ALL('Table'),
            'Table'[system ID]=MAX('Table'[system ID]) &&
            'Table'[Processing Date] < CurrentDate
        )
    )
RETURN
DATEDIFF(PreviousDate, CurrentDate, DAY)

AverageDaysPerMonth = 
var _min=CALCULATE(MIN('Table'[Processing Date]),FILTER(ALL('Table'),'Table'[system ID]=MAX('Table'[system ID]) && YEAR('Table'[Processing Date])=YEAR(MAX('Table'[Processing Date])) && MONTH('Table'[Processing Date])=MONTH(MAX('Table'[Processing Date]))))
var _max=CALCULATE(MAX('Table'[Processing Date]),FILTER(ALL('Table'),'Table'[system ID]=MAX('Table'[system ID]) && YEAR('Table'[Processing Date])=YEAR(MAX('Table'[Processing Date])) && MONTH('Table'[Processing Date])=MONTH(MAX('Table'[Processing Date]))))
var _day=IF( DATEDIFF(_min, _max, DAY)=0,1,DATEDIFF(_min, _max, DAY))
var _count=COUNTROWS(FILTER(ALL('Table'),'Table'[system ID]=MAX('Table'[system ID]) && YEAR('Table'[Processing Date])=YEAR(MAX('Table'[Processing Date])) && MONTH('Table'[Processing Date])=MONTH(MAX('Table'[Processing Date]))))
return DIVIDE(_day,_count,0)

(3) Then the result is as follows.

If the above one can't help you get the desired result, please provide some sample data in your tables (exclude sensitive data) with Text format and your expected result with backend logic and special examples. It is better if you can share a simplified pbix file. Thank you.

Best Regards,

Neeko Tang

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