Forum Discussion

et123's avatar
et123
New Member
1 year ago
Solved

Totalling a distinct count measure

User Count = DISTINCTCOUNT('View Data'[DisplayName])
 
If I am using his meansure in a matrix I need the total to be a total of the distinct counts.  I had seen a suggesion to use HASONEFILTER but not sure how to apply it to this measure to make it work correctly.
Help
  • timalbers's avatar
    timalbers
    1 year ago

    Hi et123 

     

    try something like this:

    User Count = 
SUMX( 
   VALUES( 'View Data'[DisplayName] ), 
   DISTINCTCOUNT( 'View Data'[DisplayName] )
)
  • mark_endicott's avatar
    mark_endicott
    1 year ago

    et123 - If I interpret your requirement correctly you need every row to show the same total, e.g. each row will show the total of the DISTINCTCOUNT('View Data'[DisplayName]), in which case this should work:

     

    CALCULATE
   DISTINCTCOUNT( 'View Data'[DisplayName] ),
   REMOVEFILTERS( 'View Data' )
)

     

    If this works, please accept as the solution, it helps with visibility for others with the same challenge, if it does not, please provide more information about what you are trying to acheive, and some sample data. 

4 Replies

  • timalbers's avatar
    timalbers
    Icon for Skilled Sharer rankSkilled Sharer
    1 year ago

    Hi et123 

     

    try something like this:

    User Count = 
SUMX( 
   VALUES( 'View Data'[DisplayName] ), 
   DISTINCTCOUNT( 'View Data'[DisplayName] )
)
  • mark_endicott's avatar
    mark_endicott
    Icon for Super User rankSuper User
    1 year ago

    et123 - If I interpret your requirement correctly you need every row to show the same total, e.g. each row will show the total of the DISTINCTCOUNT('View Data'[DisplayName]), in which case this should work:

     

    CALCULATE
   DISTINCTCOUNT( 'View Data'[DisplayName] ),
   REMOVEFILTERS( 'View Data' )
)

     

    If this works, please accept as the solution, it helps with visibility for others with the same challenge, if it does not, please provide more information about what you are trying to acheive, and some sample data. 

  • Anonymous's avatar
    Anonymous
    Not applicable
    1 year ago

    Hi et123 ,

    similar situation

     

    Measure = 
DISTINCTCOUNT('Table'[Column])
    Measure 2 = 
VAR _table = SUMMARIZE('Table',[Type],"Result",[Measure])
RETURN
SUMX(_table,[Result])

     

     

     

    Best Regards

  • mark_endicott's avatar
    mark_endicott
    Icon for Super User rankSuper User
    1 year ago

    et123  - did we resolve your issue? If we did, please select the appropriate solution, it helps with visibility for others and for SuperUsers to keep their status! Thanks!