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sborowiec13 Visitor
Visitor

Creating a new table with the first calculated value for each field name

I have a variety of different customer numbers in my data and using fixed calculations, I have been able to get the numbers I need on a category basis (had to do a couple of fixed calculations on age*amount and amount. However, at the very end of this process, I need to sum up those values on a customer basis, not record basis. 

 

Customer 1 50

Customer 1 50

Customer 1 50

Customer 3 30

Customer 3 30

 

Right now I'm getting a sum of 210, when I really need a sum of 80. I've tried EVERYTHING I can think of to get this and I just can't get to it. 

 

Hoping someone can help with a good solution here!

1 ACCEPTED SOLUTION

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v-jiascu-msft Super Contributor
Super Contributor

Re: Creating a new table with the first calculated value for each field name

Hi @sborowiec13,

 

If the sample you posted is the data source, you can try out a measure like this.

Measure =
SUMX (
    SUMMARIZE (
        'Table1',
        'Table1'[Customer],
        Table1[Value],
        "Unique", DISTINCT ( 'Table1'[Value] )
    ),
    [Value]
)

Creating_a_new_table_with_the_first_calculated_value_for_each_field_name

 

Best Regards,

Dale

Community Support Team _ Dale
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

1 REPLY 1
Highlighted
v-jiascu-msft Super Contributor
Super Contributor

Re: Creating a new table with the first calculated value for each field name

Hi @sborowiec13,

 

If the sample you posted is the data source, you can try out a measure like this.

Measure =
SUMX (
    SUMMARIZE (
        'Table1',
        'Table1'[Customer],
        Table1[Value],
        "Unique", DISTINCT ( 'Table1'[Value] )
    ),
    [Value]
)

Creating_a_new_table_with_the_first_calculated_value_for_each_field_name

 

Best Regards,

Dale

Community Support Team _ Dale
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

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