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Helper I

## dax

I have a table

col a     colb

x           0

x           1

y           2

y           0

y           1

z           0

I want to create a new table

New Table

col A    col B(sum of values for each col a)

x            1

y            3

z            0

can someone help me with the dax

1 ACCEPTED SOLUTION
Solution Sage

Hi @Surya1,

In your scenario, you can use SUMMARIZE() function to create a new calculated table. Please refer to following sample:

```Table-2 =
SUMMARIZE ( 'Table-1', 'Table-1'[col a], "col B", SUM ( 'Table-1'[col b] ) )```

Thanks,
Xi Jin.

6 REPLIES 6
Solution Sage

Hi @Surya1,

In your scenario, you can use SUMMARIZE() function to create a new calculated table. Please refer to following sample:

```Table-2 =
SUMMARIZE ( 'Table-1', 'Table-1'[col a], "col B", SUM ( 'Table-1'[col b] ) )```

Thanks,
Xi Jin.

Resolver III

Do you absolutely need to do this through DAX? The query editor is very easy to use. You can get the same results by selecting

Edit Queries -> right click on your table and select Duplicate -> Go to the newly duplicated table and left click on col a and select Group By -> Group by col a and select Sum under Operation

This will give you the results that you want. If you have to do it through DAX via a calculated table, I can provide instructions for that as well, just let me know.

Helper I
Resolver III

You will select New Table -> The formula will be the following:

`col a = DISTINCT(Table1[col a])`

You will then go to the Relationships tab and form a relationship between the col a of both tables.

Then return to the new table and select New Column -> The formula will be:

`Col b := CALCULATE(SUM(Table1[col b]))`

Where Table1 is the original table name.

Helper I
This dax gives the sum of colb.I want the sum of col b tat comes under each cola.
Resolver III

Did you form a relationship between the two tables? If you did not then you are going to get the sum total for all value in col a. After forming the relationship, you get the following:

The solution that @v-xjiin-msft posted should also work as well.

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