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Anonymous's avatar
Anonymous
Not applicable
2 years ago
Solved

dax

p1m = CALCULATE(     DISTINCTCOUNT(PSR[customer_code]),     FILTER(         VALUES(PSR[customer_code]),         [sum of ret] > 0     ) ) this is a measure that calculates distinct customer...
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  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi Anonymous ,

     

    I suggest you to create a "DimDate" table to help calculation.

    DimDate = 
VAR _CALENDAR =
    CALENDARAUTO()
RETURN
    ADDCOLUMNS (
        _CALENDAR,
        "Year", YEAR ( [Date] ),
        "Month", MONTH ( [Date] ),
        "MonthName", FORMAT ( [Date], "MMMM" ),
        "RANKX",
            RANKX ( _CALENDAR, YEAR ( [Date] ) * 100 + MONTH ( [Date] ),, ASC, DENSE )
    )

    Measure:

    Sum 3 month count = 
VAR _SUMMARIZE =
    SUMMARIZE (
        ALL ( DimDate ),
        DimDate[Year],
        DimDate[MonthName],
        DimDate[RANKX],
        "P1M", [p1m]
    )
RETURN
    SUMX (
        FILTER (
            _SUMMARIZE,
            [RANKX]
                > MAX ( DimDate[RANKX] ) - 3
                && [RANKX] <= MAX ( DimDate[RANKX] )
        ),
        [P1M]
    )

    Result is as below.

     

    Best Regards,
    Rico Zhou

     

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

     

Kishore_KVN's avatar
Kishore_KVN
Solution Sage
2 years ago

Hello Anonymous ,

Firstly create a calendar table and create relationship with your fact table. 

Then use below calculation by modifying table names as per your data.

Cal_Measure =
Var CM = CALCULATE(DISTINCTCOUNT(PSR[customer_code]), Filter(Calendar_Table, 'Calendar_Table'[Month] = Month(Today())))
Var PM = CALCULATE(DISTINCTCOUNT(PSR[customer_code]), Filter(Calendar_Table, 'Calendar_Table'[Month] = Month(Today())-1))
Var PPM = CALCULATE(DISTINCTCOUNT(PSR[customer_code]), Filter(Calendar_Table, 'Calendar_Table'[Month] = Month(Today())-2))
Return
CM+PM+PPM

 

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

  • Anonymous's avatar
    Anonymous
    Not applicable
    2 years ago

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