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Anonymous
Not applicable

Using AND expression in DISTINCT COUNT

Hello, 
 
I'm trying to get a new measure to work, basically I want it to find the percentage of rows that meet two conditions. It says it can't have data from 2 columns. Is there something similar to the measure below that will give me what I need?
 
% C6 NEEDS Deal = CALCULATE(DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]), 'C6 needs TUNED IN'[C6 NEEDS Tuned in]= "Deal" && 'C6 needs TUNED IN'[C6 Needs tuned in response] = "Selected") / DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier])
 
Thanks in advance!
1 ACCEPTED SOLUTION

Hi @Anonymous 

 

try change && by comma... e use DIVIDE... show blank(), when DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]) is zero.

 

% C6 NEEDS Deal = divide(CALCULATE(DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]), 'C6 needs TUNED IN'[C6 NEEDS Tuned in]= "Deal" , 'C6 needs TUNED IN'[C6 Needs tuned in response] = "Selected") , DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]))

 

Regards.

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4 REPLIES 4
Thejeswar
Resident Rockstar
Resident Rockstar

Hi @Anonymous ,

Can you share some sample data ?

 

That would make it easier for us to help..

Anonymous
Not applicable

Hi @Thejeswar , 

 

Unfortunately its client data so im unable to share it. Basically the measure should count the number of rows which meet the conditions in each of those columns (e.g. say 'Deal' and 'Selected'), then divide by the unique number of IDs in the table to get the percentage.

 

Thanks for your help!

Hi @Anonymous 

 

try change && by comma... e use DIVIDE... show blank(), when DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]) is zero.

 

% C6 NEEDS Deal = divide(CALCULATE(DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]), 'C6 needs TUNED IN'[C6 NEEDS Tuned in]= "Deal" , 'C6 needs TUNED IN'[C6 Needs tuned in response] = "Selected") , DISTINCTCOUNT('C6 needs TUNED IN'[uuid: Respondent identifier]))

 

Regards.

Anonymous
Not applicable

Hi @evandrocunico ,

 

I think that works, thanks for the suggestion!

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