One issue.. It if you have duplicate values it counts it as one. For example, if my table column is what people chose as a favorite animal, it would have a lot of people choosing dogs. There would be for example 50 dogs, but this formula would count the 50 instances of "Dog" as one. (Just a simple example). If there's 50, you want the count to reflect that.
MeasureHappy = calculate(count(MyTable[MyColumn]), MyTable[MyColumn] <> BLANK())
Changing from distinctcount, to just plainly count will resolve this, for anyone not looking to count dups as a singular. (If you want a zero instead of null
MeasureHappy = calculate(count(MyTable[MyColumn]), MyTable[MyColumn] <> BLANK()) + 0
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