Forum Discussion

3 Replies

  • amitchandak's avatar
    amitchandak
    Icon for Super User rankSuper User
    6 years ago

    HenryJS , a new column like

    countx(filter(table, [candidate+date] =earlier([candidate+date])),[candidate+date])

  • AllisonKennedy's avatar
    AllisonKennedy
    Icon for Community Champion rankCommunity Champion
    6 years ago
    Sorry, I don't understand what you're asking. A couple of questions:
    1) Why does this need to be a column? What is the end goal and can it be done as measure?
    2) Why is the answer 5? Do you want to count the total rows with the same value in the Candidate-Date column?

    Try new column:
    Rows = COUNTROWS(FILTER(ALL(table), table[Candidate-Date] = EARLIER(table[Candidate-Date]))
  • FrankAT's avatar
    FrankAT
    Icon for Community Champion rankCommunity Champion
    6 years ago

    Hi HenryJS 

    you can do it like this:

     

     

    Count of Candidate + Date = 
CALCULATE(
    COUNT('Table'[Candidate + Date]),
    ALLEXCEPT('Table','Table'[Candidate + Date])
)

    Regards FrankAT