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Problem Solved
Individually counting the categories
Ouro = CALCULATE(COUNT('Table'[Categoria]), 'Table'[Categoria] = "Ouro")
Problem Solved
Individually counting the categories
Ouro = CALCULATE(COUNT('Table'[Categoria]), 'Table'[Categoria] = "Ouro")
Hi, @TFernandes_2022
What is the format of your source data? I briefly simulated some data, hoping to match your situation.
Measure = CALCULATE(COUNT('Table'[Categoria]),ALLEXCEPT('Table','Table'[Categoria]))
Result:
Is this similar to what you expect?
Best Regards,
Community Support Team _Charlotte
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi @v-zhangti , thanks for your help. Your mock table is exactly the same as my source data, but the result I needed is the combination of the two tables.
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