Distinct Count of a column based on another column with data filter capability

[vvibhakar]

Hi,

I need help in creating a calculated column to calculate distinct count of 1 column based on another. Below is the sample data I have:

User	Product	Category
104	XYZ	A1
104	ABC	A1
104	PQR	A2
104	XYZ	A2
104	ABC	A3
237	XYZ	B1
237	ABC	B2
237	DEF	B3
237	DEF	B4
237	DEF	B4
569	PQR	C1
569	XYZ	C2
569	ABC	C2
569	DEF	C2
569	PQR	C1

In this case, User 104 & 237 has 3 distinct products and user 569 has 4 distinct products.
So I'd like to have a calculated column were 3 & 4 are mentioned and I can use it in Chiclet slicer, so when I click on 3 I get Users 104 & 237 as output. 

So I used the formula :

CALCULATE(DISTINCTCOUNT(Table[Product]),ALLEXCEPT(Table,Table[User]))

This gives me perfect result.
But now on the report I want to apply filter on Category, say A1, then for User 104, distinct product should only show 2. But the above formula still shows 3. 

Any help is appreciated.

Cheers.
Vivek

[Anonymous · Accepted solution]

Hi @vvibhakar 

Calculated column can not be dynamic impacted by slicer, but measure can.

For reference: What is the difference between Power BI calculated columns and measures?

Here I suggest you to create a table with whole number you want to select in slicer ,then create a count measure and filter your user result by another filter measure based on this count measure.

1. Create a table for slicer.

Count Slicer = GENERATESERIES(1,5) 

2.Create measures.

DISTINCTCOUNT = CALCULATE(DISTINCTCOUNT('Table'[Product]),ALLEXCEPT('Table','Table'[User],'Table'[Category]))

Filter = IF(ISFILTERED('Table'[Category]),1,IF([DISTINCTCOUNT] in VALUES('Count Slicer'[Value]),1,0))

Create a table visual by User column and [DISTINCTCOUNT] measure, then add [Filter] measure into filter field in this visual and set it to show items when value =1. Result is as below.

By default :

Select 3 in Value:

Select A1 in Category:

Best Regards,
Rico Zhou

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

[vvibhakar]

This is awesome. Thank you so much.