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nvizard
Regular Visitor

Count distinct values across two separate rows in two separate tables

Hi all,

I am trying to count the number of distinct values that occur in two separate columns in two separate tables (count the number of distinct values as though the two columns were appended) 

For example, if I had the two columns below :

1                       1

2                       3

3                       5

4                       7

5                       9

 

it should return a count of 7, 3 values were not counted (1,3,5) as they appeared twice.

 

Many thanks for your help

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

@nvizard 

 

You may try the measure below.

Measure =
COUNTROWS (
    DISTINCT ( UNION ( VALUES ( Table1[Column1] ), VALUES ( Table2[Column1] ) ) )
)

 

Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

View solution in original post

3 REPLIES 3
v-chuncz-msft
Community Support
Community Support

@nvizard 

 

You may try the measure below.

Measure =
COUNTROWS (
    DISTINCT ( UNION ( VALUES ( Table1[Column1] ), VALUES ( Table2[Column1] ) ) )
)

 

Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable

just do union of these two columns which will remove duplicates and create new table with unique values.

 

New table=Union(table1,table2)

 

Thanks & regards,
Pravin Wattamwar
www.linkedin.com/in/pravin-p-wattamwar

If I resolve your problem Mark it as a solution and give kudos.

Try one of the two

 

EXCEPT(union(all(table[column1]),all(table[column1])),INTERSECT(all(table[column1]), all(table[column2])))
or
EXCEPT(union(distinct(table[column1]),distinct(table[column1])),INTERSECT(distinct(table[column1]), distinct(table[column2])))

 

Appreciate your Kudos. In case, this is the solution you are looking for, mark it as the Solution. In case it does not help, please provide additional information and mark me with @
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