SUMX Distinct and GroupBy

[Anonymous]

Hi I have a table with Month, Week and number of records. What I would like to do is Sum(Distinct(Records)) groupby MonthNo, WeekNo.

I have a join with another table using MonthNo and WeekNo. So far I used below Dax formula to calculate distinct records

RecordsMeasure = MAXX(DISTINCT(DB[Records]), MAX(DB[Records]))

Next I have another measure to sum the distinct records

SumRecords = SUMX(DISTINCT(DB[WeekNo]), DB[RecordsMeasure])

The issue with the above is that I could either use MonthNo or WeekNo and for each I get different end result

Expected result = 10556

Result when MonthNo used = 9999

Result when MonthNo used = 10060

I would like to use GroupBy so I can use both MonthNo and WeekNo and appreciate if anyone can help with this.

Thanks

[v-jiascu-msft]

Hi @Anonymous ,

Actually, this formula "RecordsMeasure = MAXX(DISTINCT(DB[Records]), MAX(DB[Records]))" equals the one below.

RecordsMeasure = MAX(DB[Records])

What are the values of DB[Records]? Can we calculate a Max value?

Generally, the results base on MonthNo or WeekNo are different. Do you mean you'd like to SUM them base on MonthNo and WeekNo?

Can you share a sample, please? Please mask the sensitive parts first.

Best Regards,

[Anonymous]

@v-jiascu-msft 

Thanks for your reply.

Yes I need to calculate the sum distinct by Monthno and weekno. Please see attached data set. Its basically MonthNo, WeekNo and actual values.

Please see link to data file on OneDrive with DB[Records] values

https://1drv.ms/x/s!AtGoFMN86r67ll-mnGmniDknC2zp

[v-jiascu-msft · Accepted solution]

Hi @Anonymous ,

The correct result should be 10085. There could be two solutions. The basic idea is including the MonthNo and WeekNo at the same time. 

Measure 2 =
SUMX ( ALL ( DB[MonthNo], DB[WeekNo] ), CALCULATE ( MAX ( DB[Records] ) ) )

Measure 3 =
SUMX (
    SUMMARIZE ( DB, DB[MonthNo], DB[WeekNo], "maxRecords", MAX ( DB[Records] ) ),
    [maxRecords]
)

Best Regards,

[Anonymous]

Thanks @v-jiascu-msft  got the desired result now.