Forum Discussion

AZL's avatar
AZL
Frequent Visitor
7 years ago
Solved

Sum the resultant values from Distinctcount function.

Hey All,

 

I deadly need your helps guys.

 

Let's say I have the following table:

 

 

I want firstly to distinct count the values in Col1 based on Col2 to get:

 

A1>>>2

B1>>>1

C1>>>1

 

afterthat I want to count the resultant values based on Col3:

 

DD>>>2

BB>>>2

 

if I will use directly distinct count based on Col3 the result for "BB" will be 1 which is not what I want to get at the end.

 

Thank you all in advance.

 

Regards,

Need Help
  • AlB's avatar
    AlB
    7 years ago

    Hi AZL

     

    How about a DISTINCTCOUNT on the concatenation of the columns you are interested in? For instance, in your first example with Col1 and Col2, you create an additional calculated column

    Table1[NewColum]=Table1[Col] & Table1[Col2]

    and then

    DISTINCTCOUNT(Table1[NewColum])

4 Replies

  • v-juanli-msft's avatar
    v-juanli-msft
    Community Support
    7 years ago

    Hi AZL

    Assume you have the result as below

    Per your requirement, you could create a measure

    Measure = CALCULATE(DISTINCTCOUNT(Sheet3[col1]),ALLEXCEPT(Sheet3,Sheet3[col2]))

     

     

    Best Regards

    Maggie

  • AlB's avatar
    AlB
    Community Champion
    7 years ago

    Hi AZL

     

    How about a DISTINCTCOUNT on the concatenation of the columns you are interested in? For instance, in your first example with Col1 and Col2, you create an additional calculated column

    Table1[NewColum]=Table1[Col] & Table1[Col2]

    and then

    DISTINCTCOUNT(Table1[NewColum])

    • AZL's avatar
      AZL
      Frequent Visitor
      7 years ago

      Hey AlB

       

      That works like heaven to me!!

      Thank you so much for your smart solution.

      • AlB's avatar
        AlB
        Community Champion
        7 years ago

        AZL

        :smileyhappy: you're very welcome. glad it helped.

        Please mark it as solution then, so that it's available for others too.

        Cheers