Skip to main content
cancel
Showing results for 
Search instead for 
Did you mean: 

Register now to learn Fabric in free live sessions led by the best Microsoft experts. From Apr 16 to May 9, in English and Spanish.

Reply
Anonymous
Not applicable

Getting the count of common values from two filtered columns

Capture.PNG

Hi,

I have a problem filtering two columns and finding the common among those two columns. I have two columns Process and Task as shown above. I want to take the distinct count of ID's which falls into(Application && accepted) and (Review && over).

I tried this way p1=Calculate(Distinctcount(Table(ID)),Filter'Table',Table[Process]=''Application''&& Table[Task]="accepted"))

                        P2 =Calculate(Distinctcount(Table(ID)), Filter'Table',Table[Process]="Review" && Table[Task]="over"))

                        Permits=if(Table[p1]=Table[p2], Blank(),"True")

But I don't get what I expected. From the above table , I want to get the count as 1. Is there any other way in DAX to show the count. Please help!

             
1 ACCEPTED SOLUTION
v-juanli-msft
Community Support
Community Support

Hi @Anonymous

From this information:

"I want to see the count of ID'S which have(Application, accepted ) and (Review,Over). In my sample data, the count of ID'S which have all these statues are ID 1. So the count is 1. That's how I want to see it."

 

I make a test as below

Create measures

Applicationaccepted_count = COUNTROWS(FILTER(ALLEXCEPT(Sheet3,Sheet3[ID]),Sheet3[merge]="Applicationaccepted"))

Reviewover_count = COUNTROWS(FILTER(ALLEXCEPT(Sheet3,Sheet3[ID]),Sheet3[merge]="Reviewover"))

distinct_count = CALCULATE(DISTINCTCOUNT(Sheet3[ID]),FILTER(ALL(Sheet3),[Applicationaccepted_count]>0&&[Reviewover_count]>0))

10.png

Best Reagrds

Maggie

 

View solution in original post

6 REPLIES 6
v-juanli-msft
Community Support
Community Support

Hi @Anonymous

From this information:

"I want to see the count of ID'S which have(Application, accepted ) and (Review,Over). In my sample data, the count of ID'S which have all these statues are ID 1. So the count is 1. That's how I want to see it."

 

I make a test as below

Create measures

Applicationaccepted_count = COUNTROWS(FILTER(ALLEXCEPT(Sheet3,Sheet3[ID]),Sheet3[merge]="Applicationaccepted"))

Reviewover_count = COUNTROWS(FILTER(ALLEXCEPT(Sheet3,Sheet3[ID]),Sheet3[merge]="Reviewover"))

distinct_count = CALCULATE(DISTINCTCOUNT(Sheet3[ID]),FILTER(ALL(Sheet3),[Applicationaccepted_count]>0&&[Reviewover_count]>0))

10.png

Best Reagrds

Maggie

 

Anonymous
Not applicable

From the sample data it looks like it should  be a count of 2 for each, not sure about the 1?

 

That's what Im getting using these two measures:

Application Accepted = 
    CALCULATE(
        DISTINCTCOUNT(Table1[ID ] ),
        FILTER(
            Table1,
                AND( Table1[Process] ="Application", Table1[Task] ="Accepted")
        )
    )


Review Over = 
    CALCULATE(
        DISTINCTCOUNT(Table1[ID ] ),
        FILTER(
            Table1,
                AND( Table1[Process] ="Review", Table1[Task] ="Over")
            )
        )
    
Anonymous
Not applicable

I want to see the count of ID'S which have(Application, accepted ) and (Review,Over). In my sample data, the count of ID'S which have all these statues are ID 1. So the count is 1. That's how I want to see it.

Anonymous
Not applicable

This should be what you are looking for.  Will give you a new table, then can do whatever counts off of that

New Table = 
Var ApplicationAccepted = SELECTCOLUMNS( FILTER( Table1, AND( Table1[Process] ="Application", Table1[Task] ="Accepted") ), "ID", Table1[ID] ) RETURN Var ReviewOver = SELECTCOLUMNS( FILTER( Table1, AND( Table1[Process] ="Review", Table1[Task] ="Over") ), "ID", Table1[ID] ) RETURN INTERSECT( ApplicationAccepted, ReviewOver)
Measure = DISTINCTCOUNT(New Table[ID] )
AlB
Super User
Super User

Hi @Anonymous

 

Are p1, p2, Permits measure? columns?

 

Anonymous
Not applicable

P1 and P2 are the the count of ID's, they are measure to get the count. I want to get the count of ID's which are common among those P1 and P2

Helpful resources

Announcements
Microsoft Fabric Learn Together

Microsoft Fabric Learn Together

Covering the world! 9:00-10:30 AM Sydney, 4:00-5:30 PM CET (Paris/Berlin), 7:00-8:30 PM Mexico City

PBI_APRIL_CAROUSEL1

Power BI Monthly Update - April 2024

Check out the April 2024 Power BI update to learn about new features.

April Fabric Community Update

Fabric Community Update - April 2024

Find out what's new and trending in the Fabric Community.