Measure aggregation

[nickbeltran7]

I have the following measure in PowerBI: SPACES = (SUM('Properties'[Spaces]'). This measure aims to return the total number of spaces each Id has. Additionally, I have two other columns in two different tables. One column is called "Id" and the other column is "Month" in a time dimension table. When I pull a matrix with Id in rows, and Month in columns, and SPACES as values, it performs an aggregation rowwise and columnwise, which is what I want. However, when the aggregation is done rowwise at the asset level, the aggregation returns the SPACES in one month instead of the sum of SPACES each month. Note: Even though each month has the same number of spaces, I still want it to sum up the spaces of the twelve months. I am attaching an image of the matrix so you can understand me better. For example, for ID 302, I want the total to be 1,500, not 125. If anyone could provide any sort of guidance on this, I would appreciate it.

[Anonymous · Accepted solution]

Hi, @nickbeltran7

Based on the information you gave, I have created two tables

Create a New Measure named SPACE.

I'm trying to use your measure and follow your way, and then here's my idea: create another measure, create a virtual table in it, create a virtual table with the id and Month columns using the Cartesian product, and add a new column to this virtual table with [SPACE] as the value. SUMX is then used to calculate the sum of each row. The DAX expression is as follows:

Measure =

VAR a = ADDCOLUMNS(CROSSJOIN(VALUES('Table'[ID]),VALUES('Properties'[Month])),"Me",[SPACE])

RETURN

SUMX(FILTER(a,[ID] in VALUES('Table'[ID])&&[Month] IN VALUES('Properties'[Month])),[Me])

Here's my rendering:

I hope my ideas can help you.

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Best Regards

Yongkang Hua

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[nickbeltran7]

Indeed It did work. Thank you so much!

[Anonymous]

Hi, @nickbeltran7 
If the solution works, can you accept it as a solution?

Best Regards

Yongkang Hua

[Fowmy]

@nickbeltran7 

Not sure how you have set up your model, please share an image of your model.