Grouping 1 value into 2 groups

[Krzysiek]

Hi!

I have a table like 1., I know how to do 2. and I would like to do something like 3.

Could you help me? I have no clue how to do it.

For example,

I want to count rows with A - the result should be 6,

I want to count rows with B - the result should be 6,

but if I check A and B - the result shoulb be 11 (not equals 6 + 6).

[Krzysiek]

Thank you! It will help a little.

I have list of schools and list of big cities (9).
Every school is related to big city.

But if school is between 2 cities then one school is related to both (and the name looks like for example "Liverpool/Manchester".

When I click on filter Liverpool I would like to see data connected to Liverpool (and Liverpool/Manchester) on the graph.

If I click Manchester I would like to see data for Manchester (and Liverpool/Manchester).

But if I check both I would like see data for Machester and Liverpool (and Liverpool/Manchester).

[v-jiascu-msft]

Hi @Krzysiek,

How about expanding them like the snapshot? You can choose both of them with "CTRL".

Best Regards,

Dale

[Anonymous]

@Krzysiek

On the lines of what @v-jiascu-msft suggested, you don't need to modify your original table to split the multi valued column into 2 rows.

Instead, you can create a new separate table which will have just 2 columns - City and Big City.

Then make a relation between this new table and your own table. You can use the slicer with the field from this new table.

Hope it helps

Regards

[v-jiascu-msft]

Hi @Krzysiek,

I'm afraid we can't do like that. But we can create some measures. What's your real scenario?

countA =
CALCULATE (
    COUNT ( Table1[Letter] ),
    SEARCH ( "A", 'Table1'[Letter], 1, 0 ) > 0
)

countB =
CALCULATE (
    COUNT ( Table1[Letter] ),
    SEARCH ( "B", 'Table1'[Letter], 1, 0 ) > 0
)

countAB =
CALCULATE (
    COUNT ( Table1[Letter] ),
    FILTER (
        'Table1',
        SEARCH ( "A", 'Table1'[Letter], 1, 0 ) > 0
            || SEARCH ( "B", 'Table1'[Letter], 1, 0 ) > 0
    )
)

Best Regards,

Dale