Forum Discussion
Distinct Count of Two Columns
- 9 years ago
Hi Simon_Nuss,
Currently I don’t think only using measure could achieve this.
In addition to concatenate those two columns, we could create a special column to mark the same value in those two columns.
Then use the two distinct value to minus the sum of the same value column, take use of your example here:
Create the calculated column with the following formula:
Samevalue = if(
LOOKUPVALUE( Table1[Gold], Table1[Gold], Table1[Sliver] ) <> BLANK(),
1,
0)
Modify the count measure with the following:
Measure := DISTINCTCOUNT(Table1[Gold])+DISTINCTCOUNT(Table1[Sliver])-calculate(DISTINCTCOUNT(Table1[Sliver]), Table1[Samevalue]=1)
See the result:
Hope this should be helpful.
Regards
ankitpatira Thanks for the quick response Ankitpatira. That's close but not quite what I'm looking for. Also, my data set is 1.1 billion rows - I cannot afford a calculated column.
Take the below example:
| Gold Medals | Silver Medals |
| USA | Australia |
| USA | Canada |
| Germany | Spain |
| Greece | USA |
The distinct count of Gold Medals is 3, Silver medals is 4. However, I am looking for the distinct count of the population which in this case would be 6. This can be done in SQL by performing a distinct count of the union of [Gold Medals] and [Silver Medals]. I'm chasing a DAX solution.
Thanks,
Simon
Hi Simon_Nuss,
Currently I don’t think only using measure could achieve this.
In addition to concatenate those two columns, we could create a special column to mark the same value in those two columns.
Then use the two distinct value to minus the sum of the same value column, take use of your example here:
Create the calculated column with the following formula:
Samevalue = if(
LOOKUPVALUE( Table1[Gold], Table1[Gold], Table1[Sliver] ) <> BLANK(),
1,
0)
Modify the count measure with the following:
Measure := DISTINCTCOUNT(Table1[Gold])+DISTINCTCOUNT(Table1[Sliver])-calculate(DISTINCTCOUNT(Table1[Sliver]), Table1[Samevalue]=1)
See the result:
Hope this should be helpful.
Regards