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johndolan2
Resolver I
Resolver I

Distnict count multiple columns

Hi -  am trying to come up with measures to show a distinct total and a percentage of total for the table below labeled "Here is what I have". If you could help that would be great. Thanks, John

 

Here is what I am looking for

ProductCount of Status OnCounr of Status OffStatus On %Status Off %
A1225%75%
B20100%0%
CNANANANA

 

Here is what I have

ProductSevice IDStatus
A14ON
A14ON
A21OFF
A21OFF
A22OFF
B8ON
B9ON
C11Not Applicable
3 ACCEPTED SOLUTIONS
Ashish_Mathur
Super User
Super User

Hi,

For ProductA, the Status on % should be 1/3 i.e. 33.33%.  Am i correct?


Regards,
Ashish Mathur
http://www.ashishmathur.com
https://www.linkedin.com/in/excelenthusiasts/

View solution in original post

amitchandak
Super User
Super User

@johndolan2 , % for A does not seems correct, Try like

 

Count of Status On = calculate(distinctcount(Table[Sevice ID]) , filter(Table, Table[Status] = "ON"))
Count of Status Off = calculate(distinctcount(Table[Sevice ID]) , filter(Table, Table[Status] = "OFF"))

Off Status On % = divide([Count of Status On], [Count of Status On] +[Count of Status Off])
Status Off % =divide([Count of Status Off], [Count of Status On] +[Count of Status Off])

or

Off Status On % = divide(calculate(count(Table[Sevice ID]) , filter(Table, Table[Status] = "ON")), count(Table[Sevice ID]))
Status Off % =divide(calculate(distinctcount(Table[Sevice ID]) , filter(Table, Table[Status] = "OFF")) , count(Table[Sevice ID]))

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View solution in original post

Hi Amitchandak - Thank you very much!!!

John

View solution in original post

7 REPLIES 7
amitchandak
Super User
Super User

@johndolan2 , % for A does not seems correct, Try like

 

Count of Status On = calculate(distinctcount(Table[Sevice ID]) , filter(Table, Table[Status] = "ON"))
Count of Status Off = calculate(distinctcount(Table[Sevice ID]) , filter(Table, Table[Status] = "OFF"))

Off Status On % = divide([Count of Status On], [Count of Status On] +[Count of Status Off])
Status Off % =divide([Count of Status Off], [Count of Status On] +[Count of Status Off])

or

Off Status On % = divide(calculate(count(Table[Sevice ID]) , filter(Table, Table[Status] = "ON")), count(Table[Sevice ID]))
Status Off % =divide(calculate(distinctcount(Table[Sevice ID]) , filter(Table, Table[Status] = "OFF")) , count(Table[Sevice ID]))

Share with Power BI Enthusiasts: Full Power BI Video (20 Hours) YouTube
Microsoft Fabric Series 60+ Videos YouTube
Microsoft Fabric Hindi End to End YouTube

Hi Amitchandak - Thank you very much!!!

John

Ashish_Mathur
Super User
Super User

Hi,

For ProductA, the Status on % should be 1/3 i.e. 33.33%.  Am i correct?


Regards,
Ashish Mathur
http://www.ashishmathur.com
https://www.linkedin.com/in/excelenthusiasts/

Hi Ashish and Ebeery - to get the % totals I did this: I created a measure to sum the two measures of "distinct count on" with the "distinct count off". I then divided that number against the "distinct count on" and then against the "distinct count off". The %s look good to me. I just want to say thank you! thank you! and thank you! I had been fumbling for hours. 

Hi Ashish Yes, product A should be 1/3 i.e. 33.33%.  .... I was just in the process of fumbling around trying to figure out how to get that number in a measure.  You have been great ... How do I now get that 33.33% in a measure. Thanks, John

I was thinking there was an error in OP's original table - definitely seems like it shouldn't be 25%, but to me it looks like it should be 40% (2/5).

ebeery
Memorable Member
Memorable Member

Hi @johndolan2 

What you're looking for is easily achievable with a Matrix visual.

The measures would look something like this:

DistinctCountOn =
CALCULATE ( DISTINCTCOUNT ( 'Table'[ID] ), 'Table'[Status] = "On" )

DistinctCountOff =
CALCULATE ( DISTINCTCOUNT ( 'Table'[ID] ), 'Table'[Status] = "Off" )


% On =
DIVIDE (
CALCULATE ( COUNT ( 'Table'[ID] ), 'Table'[Status] = "On" ),
COUNT ( 'Table'[ID] )
)

% Off =
DIVIDE(
CALCULATE(
COUNT('Table'[ID]),
'Table'[Status] = "Off"
),
COUNT('Table'[ID])
)

ebeery_0-1609804242586.png

 

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