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msuser48
Helper I
Helper I

Calculate COUNT of DISTINCT values with GROUP BY

I have two tables, where I am trying to make the right CalculatedColumn in table 2:

 

Table 1:

IdDifference
11
11
21
21
32
32

Table 2 (DESIRED OUTPUT TABLE)

NumbersCalculatedColumn
12
21
30

 

I need the right code for my "CalculatedColumn" in table 2.

 

The goal is to count the DISTINCT "Difference" value from table 1 for EACH Id, and add these together.

- So, you can see in table 1, both ID 1 & ID 2 have the distinct value of 1 EACH.

- 1+1 = 2, therefore the calculated column should put 2 as value where "Numbers" column = 1.

 

 

My attempt:

Counter =
SWITCH (
TRUE(),
Table2[Numbers] = 1, CALCULATE(SUM(Table1[Difference]),Table1[Difference] = 1),

Table2[Numbers] = 2, CALCULATE(SUM(Table1[Difference]),Table1[Difference] = 2
)

.. and so on.

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

Hi @msuser48 ,

 

Please follow these steps:

(1) Create a new column

Counter =
SWITCH (
TRUE(),
'Table 2'[Number] = 1, CALCULATE(DISTINCTCOUNT('Table 1' [ID]),'Table 1' [Difference] = 1),
'Table 2'[Number] = 2, CALCULATE(DISTINCTCOUNT('Table 1' [ID]),'Table 1' [Difference] = 2),
'Table 2'[Number] = 3, CALCULATE(DISTINCTCOUNT('Table 1' [ID]),'Table 1' [Difference] = 3)
)

 

(2)Final output

vjialluomsft_0-1669773363439.png

 

 

Best Regards,

Gallen Luo

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

1 REPLY 1
v-jialluo-msft
Community Support
Community Support

Hi @msuser48 ,

 

Please follow these steps:

(1) Create a new column

Counter =
SWITCH (
TRUE(),
'Table 2'[Number] = 1, CALCULATE(DISTINCTCOUNT('Table 1' [ID]),'Table 1' [Difference] = 1),
'Table 2'[Number] = 2, CALCULATE(DISTINCTCOUNT('Table 1' [ID]),'Table 1' [Difference] = 2),
'Table 2'[Number] = 3, CALCULATE(DISTINCTCOUNT('Table 1' [ID]),'Table 1' [Difference] = 3)
)

 

(2)Final output

vjialluomsft_0-1669773363439.png

 

 

Best Regards,

Gallen Luo

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

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