Forum Discussion
Mentzer
3 years agoFrequent Visitor
Counting Old Dates for a Specific Condition
Hello people, How can I deal with the following situation? I have Table1 that has all the Groups that certain ID has gone through. The Table2 has the quantity of distinct Groups that the ID has g...
- Anonymous3 years ago
Hi Mentzer ,
Here I suggest you to try this code to create a calculated column.
Count Group in Oldest Date = VAR _Oldest_Date = CALCULATE ( MIN ( Table1[Date] ), FILTER ( Table1, Table1[ID] = EARLIER ( Table2[ID] ) ) ) VAR _GROUPinOldestDate = CALCULATE ( SELECTEDVALUE ( Table1[Group] ), FILTER ( Table1, Table1[ID] = EARLIER ( Table2[ID] ) && Table1[Date] = _Oldest_Date ) ) VAR _Count_Group = CALCULATE ( COUNT ( Table1[Group] ), FILTER ( Table1, Table1[ID] = EARLIER ( Table2[ID] ) && Table1[Group] = _GROUPinOldestDate ) ) RETURN IF ( Table2[Change Group] > 1, _Count_Group, BLANK () )Result is as below.
Best Regards,
Rico ZhouIf this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
tamerj1
Community Champion
3 years agoHi Mentzer
Please try
Count Changed =
SUMX ( VALUES ( Table1[ID] ), IF ( [Changed Group] > 1, 1 ) )Mentzer
3 years agoFrequent Visitor
In the followin part, the Changed Group is not recognized:
IF ( [Changed Group] > 1, 1 )
- Anonymous3 years agoNot applicable
Hi Mentzer ,
Here I suggest you to try this code to create a calculated column.
Count Group in Oldest Date = VAR _Oldest_Date = CALCULATE ( MIN ( Table1[Date] ), FILTER ( Table1, Table1[ID] = EARLIER ( Table2[ID] ) ) ) VAR _GROUPinOldestDate = CALCULATE ( SELECTEDVALUE ( Table1[Group] ), FILTER ( Table1, Table1[ID] = EARLIER ( Table2[ID] ) && Table1[Date] = _Oldest_Date ) ) VAR _Count_Group = CALCULATE ( COUNT ( Table1[Group] ), FILTER ( Table1, Table1[ID] = EARLIER ( Table2[ID] ) && Table1[Group] = _GROUPinOldestDate ) ) RETURN IF ( Table2[Change Group] > 1, _Count_Group, BLANK () )Result is as below.
Best Regards,
Rico ZhouIf this post helps, then please consider Accept it as the solution to help the other members find it more quickly.