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rfwjr64's avatar
rfwjr64
Frequent Visitor
5 years ago
Solved

Dividing

Hello, I am trying to divide a subset of values in one column by the total of the same column.   What I am trying to determine is how many records have an EAU$ of less than $2000 and divide that number by the total number of EAU$ records.  Because some records are duplicates, I would like to find the distinct count for both the under $2k and the overall.   I have a column labeled 'Total EAU$.  I want divide the distinct count of all EAU$ under $2000 by the distinct count of all of the EAU$.

 

I would like to show the two values in a pie chart to demonstrate how many under $2k records are a part of the total records.

 

I don't have an example of what I am doing, unfortunately. 

 

Any help or guidance would be greatly appreciated. 

  • Hello @rfwjr64 ,

    You can create a measure as follows:

    calcMeasure - DIVIDE(
    CALCULATE(DISTINCTCOUNT(tablename[EAU$]), tablename[EAU$] < 2000),
    DISTINCTCOUNT(table name[EAU$]), 0)

    Use the table name and column name in the previous DAX expression.

    Thank you

    Pragati

4 Replies

  • Hello @rfwjr64 ,

    You can create a measure as follows:

    calcMeasure - DIVIDE(
    CALCULATE(DISTINCTCOUNT(tablename[EAU$]), tablename[EAU$] < 2000),
    DISTINCTCOUNT(table name[EAU$]), 0)

    Use the table name and column name in the previous DAX expression.

    Thank you

    Pragati

    • rfwjr64's avatar
      rfwjr64
      Frequent Visitor

      This does help. Now I can move on. Thank very much.

  • Hi rfwjr64 

     

    The way I would approach this is creating a new column: $ Group = IF([Total EAU$] >2000, "More than $2,000","$2,000 or Less")

     

    Then, put a pie chart in your report, under the Values field, put in the distinct ID you have for each row, e.g. SalesID and change the Count to Count Distinct, and put [$ Group] in the legends.

  • rfwjr64 , Try a new measure like

    divide(calculate(distinctcount(Table[Column), filter(Table, Table[EAU]<2000)),distinctcount(Table[Column))