Data Segmentation with 2 measures but only 1 possible selection

[davib99]

My problem is kind simple, but I don't know if there's really a way to solve it. I have a data segmentation with 2 measures and it's only one selection allowed. If I selected one value on the "second level" it's ok and working fine as expected. But, the thing is when I click on "the first level filter" it automatically selects all the values on the second level, so it kind of breaks the only one selection rule.

Is there a way to get around with it? 

For example, imagine I insert column1 and column2 into a data segmentation with 1 selection allowed. If I click on the "A" value, it will break the 1 selection allowed rule because it will filter 3 rows of the table, not only 1.

Column1	Column2
A	1
A	2
A	3
B	4
B	5

[Anonymous · Accepted solution]

Hi @davib99 ,

The behavior you mention is one that is enabled by design. When you click on the first slicer, all the options for the second slicer are checked by default, so when you click on “A” three rows will appear and you can manually deselect the other options for the second slicer.

Alternatively, as rohit1991 said, you can create a Composite Key column as a slicer.  The new column is then used as a slicer, allowing for individual filtering.

Best Regards,

Neeko Tang

[rohit1991]

To enforce a single selection in Power BI and prevent the "first level" from selecting multiple rows in the "second level," follow these simple steps:

A) Create a Composite Key Column: Add a calculated column in Power BI: CompositeKey = Column1 & " - " & Column2.

B) Use the Composite Key in the Slicer: Replace your slicer with the CompositeKey column. This ensures each selection is unique and tied to a specific row, avoiding the issue of selecting all values under a "first level" filter.

C) Optional: Use Visuals to Show Separate Columns: If you need to display Column1 and Column2 separately in a table or chart, include them in the visual, but rely on the CompositeKey for filtering.

This approach is simple and solves the problem effectively.