Forum Discussion

PowerBrowse's avatar
PowerBrowse
Helper I
3 years ago
Solved

Group By Column Type

I have a table with multiple columns. One Column is called Lic Type and another is called User. There are six different Lic Types and multiple users can be in each Lic Type. I want to group my repo...
General Comment
  • bolfri's avatar
    bolfri
    3 years ago

    Something like this?

     

    Number of users = COUNT(SampleData[User])

     

  • Anonymous's avatar
    Anonymous
    3 years ago

    Hi PowerBrowse ,

     

    I think you can try ALLEXCEPT() function to create a measure or a calculated column.

    Total Group by Lic Type =
CALCULATE (
    DISTINCTCOUNT ( 'Table'[User] ),
    ALLEXCEPT ( 'Table', 'Table'[Lic Type] )
)

    Result is as below.

    OR

     

    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.

bolfri's avatar
bolfri
Solution Sage
3 years ago

So.... I still don't understand. Can you post here sample rows with the data you have an what do you expect from that data? It's very hard to understand what do you mean only by description.

PowerBrowse's avatar
PowerBrowse
Helper I
3 years ago
Lic TypeUser
AdminDenise
ReadSimon
WriteCarl
ReadJulia
ReadLaura
AdminPeter
ReadBob

 

I want to Group each Admin, Read, and write group. After each group I want to print the number of each Lic type i.e after Admin I want to print "Total = 2" because there are 2 Admin users. After Read group I want to print "Total = 4" because there are 4 Read Lic Types.

  • bolfri's avatar
    bolfri
    Solution Sage
    3 years ago

    Something like this?

     

    Number of users = COUNT(SampleData[User])

     

  • Anonymous's avatar
    Anonymous
    Not applicable
    3 years ago

    Hi PowerBrowse ,

     

    I think you can try ALLEXCEPT() function to create a measure or a calculated column.

    Total Group by Lic Type =
CALCULATE (
    DISTINCTCOUNT ( 'Table'[User] ),
    ALLEXCEPT ( 'Table', 'Table'[Lic Type] )
)

    Result is as below.

    OR

     

    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.