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Nagin's avatar
Nagin
Frequent Visitor
4 years ago
Solved

Countrows ALL

Hi All,

 

I am trying to create a measure to countrows and ignore all slicer filters, to enable me to calculate a % rate, so for example in the image below, the % calculation is working fine as no provider is selected.

 

However once I select a provider the 'STC All incidents' column gets filtered to that particular provider and I get a 100% rate

Is there a way that I can count all incidents when a provider is selected, e.g for the selected provider for June 22 should be 3111/14072 to get the provider rate.

 

I have used the following measure to calculate all incidents

 

STC ALL Incidents = CALCULATE(
COUNTROWS(EMAS),EMAS[EMAS Call Connect Date],
ALL(EMAS[Provider Code]))

 

Thanks

 

Nagin

  • Hi Nagin ,

     

    Based on your description, I have created a simple sample:

    Please try:

    ALL = COUNTROWS(FILTER(ALL('Table'),'Table'[MonthAndYear]=MAX('Table'[MonthAndYear])))
    
    Percentage = DIVIDE(COUNTROWS('Table'),[ALL])

    Final output:

    Best Regards,

    Jianbo Li

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

     

2 Replies

  • ValtteriN's avatar
    ValtteriN
    Community Champion

    Hi,

    I am not sure I follow but here are a few examples:

    Data:

     


    ALL rows of a table:

    Measure 19 = COUNTROWS(ALL(Category))

    rows of all selected:

    Measure 20 = COUNTROWS(Category)

    Percentage is just DIVIDE([m 20], [m 19])

    End result:

    I hope this post helps to solve your issue and if it does consider accepting it as a solution and giving the post a thumbs up!

    My LinkedIn: https://www.linkedin.com/in/n%C3%A4ttiahov-00001/



  • Hi Nagin ,

     

    Based on your description, I have created a simple sample:

    Please try:

    ALL = COUNTROWS(FILTER(ALL('Table'),'Table'[MonthAndYear]=MAX('Table'[MonthAndYear])))
    
    Percentage = DIVIDE(COUNTROWS('Table'),[ALL])

    Final output:

    Best Regards,

    Jianbo Li

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.