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aksl
Frequent Visitor

Creating groups based on numbers

I'd like to create groups based on numbers (i.e. if column X is 0-9, returns "0-9 members", if column S is 10-24, returns "10-24 members", etc.).  If I use the grouping or list function, it won't let me use <=> and lists the category as every number within that group, makes any visualization outrageously messy, and won't accommodate updated data within the number spans unless it is one of the exact numbers that has already been used.  I can only seem to use one IF function in a column - is that correct?  Can anyone suggest a work-around?

 

Thank you!

2 ACCEPTED SOLUTIONS

Select the table/query in question. Hit "Edit Queries".

 

In there, you have option to add blank query. Do so and go to Advanced Editor.

Paste in "M" as below (you can adjust range as needed or add more arugments).

 

(vals) =>
let ValBucket =
    {
        {(x)=>x<10, "0-9 members"},
        {(x)=>x<25, "10-24 members"},
        {(x)=>x<50, "25-49 members"},
        {(x)=>true, "50+ members"}
    },
    Result = List.First(List.Select(ValBucket, each _{0}(vals))){1}
in
    Result

Once confirmed, rename the query to something that makes sense (Ex. BucketFunc).

 

Now, still in Query Editor, go back to your table where you want to add custom buckets/bands.

 

In "Add column" ribbon tool, select "Invoke Custom Function".

 

Choose a name for the column and select "BucketFunc" you just created. "vals" should be column where the values you want to band belongs.

1.JPG

 

Result should look like below.

2_Result.JPG

View solution in original post

v-chuncz-msft
Community Support
Community Support

@aksl,

 

You could also use DAX to add a calculated column.

Group Name =
SWITCH (
    TRUE (),
    Table1[column X] >= 0
        && Table1[column X] <= 9, "0-9 members",
    Table1[column X] >= 10
        && Table1[column X] <= 24, "10-24 members",
    "other"
)
Community Support Team _ Sam Zha
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

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