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YanaBykovskaya
Helper I
Helper I

Calculate measures on the basis of 2 related tables but ignoring this relationship?

Hi everyone!

 

I have 2 tables: one table is a fact table containing dates field and the number of events, another table is a dimension table containing names and dates of creating these names. These tables are related by the field id_n, but I need calculate measure that will calculate the number of names (from table 2) per each date (from table 1) where created_date (from table 2) is greater than or equal to date (from table 1). 

 

The formula should looks like measure_table_1 = COUNT( IF(table_2[created_date] > = table_1[date], table_2[id_n]) )

 

Screenshot 2021-05-07 110057.png

 

Thanks!

1 ACCEPTED SOLUTION

Hi @YanaBykovskaya ,

You can create a measure as below, please find the details in the attachment.

Count of names = 
CALCULATE (
    DISTINCTCOUNT ( 'Table 2'[name_id] ),
    FILTER (
        ALL('Table 2'),
        'Table 2'[create_date] >= SELECTEDVALUE ( 'Table 1'[date] )
    )
)

yingyinr_1-1620723092415.png

Best Regards

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

View solution in original post

4 REPLIES 4
amitchandak
Super User IV
Super User IV

@YanaBykovskaya , Try like

measure_table_1 = calculate(COUNTX(table_2, IF(table_2[created_date] > = max(table_1[date]), table_2[id_n],blank()) ),values(table_2[id_n]), crossfilter(table_2[id_n],table_1[id_n], none))



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Hi. Thanks for the answer. But is it possible to do calculation without creating a new table?

 

Hi @YanaBykovskaya ,

You can create a measure as below, please find the details in the attachment.

Count of names = 
CALCULATE (
    DISTINCTCOUNT ( 'Table 2'[name_id] ),
    FILTER (
        ALL('Table 2'),
        'Table 2'[create_date] >= SELECTEDVALUE ( 'Table 1'[date] )
    )
)

yingyinr_1-1620723092415.png

Best Regards

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

View solution in original post

Thank you very much!

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