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PowerBIUserX's avatar
PowerBIUserX
Frequent Visitor
2 years ago
Solved

Trend analysis using linear regression gradient per category

Hi,

 

I'm building a report and need to calculate a trend over time per category, to identify top increases/decreases over time.

The end goal is to have a trend value (gradient of linear regression) per category, to then filter on the top 10 by trend value.


I've seen the following article on linear regression, but don't know how to go about applying it per categoryhttps://community.fabric.microsoft.com/t5/Desktop/Trend-line-values/td-p/1641410

Additionally (nice to have), I would like to exclude any potential errors, for example by removing the top5%/bottom 5% of data (by value per category).

 

I appreciate any help!

 

Few comments:

  • My Value from below is actually the result of a measure
  • There are between 100-200 categories.
  • The trend value in itself is rather meaningless and just used to sort/filter on categories.

Essentially, I would like to get from:


To this:

 

  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi PowerBIUserX 

    I add a month column in the table

    Then modifiy the Lines measure

    Lines =
    VAR a =
        LINESTX (
            FILTER ( ALLSELECTED ( 'Table' ), [Category] IN VALUES ( 'Table'[Category] ) ),
            [Value_measure],
            [Month]
        )
    RETURN
        MINX ( a, [Slope1] )
    

    Output

    Best Regards!

    Yolo Zhu

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

     

     

     

4 Replies

  • Anonymous's avatar
    Anonymous
    Not applicable

    Hi PowerBIUserX 

    You can refer to the following solution.

    Sample data 

    I create a sample measure

    Value_measure = SUM('Table'[Value]) 

    Then create a measure

    Lines =
    VAR _minxy =
        MINX (
            FILTER ( ALLSELECTED ( 'Table' ), [Category] IN VALUES ( 'Table'[Category] ) ),
            [Value_measure]
        )
    VAR _maxy =
        MAXX (
            FILTER ( ALLSELECTED ( 'Table' ), [Category] IN VALUES ( 'Table'[Category] ) ),
            [Value_measure]
        )
    VAR x_minxy =
        MINX (
            FILTER (
                ALLSELECTED ( 'Table' ),
                [Category]
                    IN VALUES ( 'Table'[Category] )
                        && [Value_measure] = _minxy
            ),
            MONTH ( [Date] )
        )
    VAR x_maxy =
        MINX (
            FILTER (
                ALLSELECTED ( 'Table' ),
                [Category]
                    IN VALUES ( 'Table'[Category] )
                        && [Value_measure] = _maxy
            ),
            MONTH ( [Date] )
        )
    RETURN
        DIVIDE ( _maxy - _minxy, x_maxy - x_minxy )
    

    Best Regards!

    Yolo Zhu

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

     

  • PowerBIUserX's avatar
    PowerBIUserX
    Frequent Visitor

    Hi Anonymous ,

    Thank you for your reply. While it works on the limited dataset I initially provided, I do not think the solution works in all cases.
    If I understand correctly, the measures you suggested calculate the slope [ (y2-y1)/(x2-x1) ] of the maximum point and minimum point in the dataset, but not the slope of the regression. In the dataset it works because they are already linear functions.

    On the following dataset (non-linear function), the slope I am expected is -3.5%, but your measure returns -45%.

    Do you have an idea on how to achieve the same but with the slope of the linear regression?


    • Anonymous's avatar
      Anonymous
      Not applicable

      Hi PowerBIUserX 

      I add a month column in the table

      Then modifiy the Lines measure

      Lines =
      VAR a =
          LINESTX (
              FILTER ( ALLSELECTED ( 'Table' ), [Category] IN VALUES ( 'Table'[Category] ) ),
              [Value_measure],
              [Month]
          )
      RETURN
          MINX ( a, [Slope1] )
      

      Output

      Best Regards!

      Yolo Zhu

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