grouping ranges

[jcastr02]

I have a table with the monthly call volume per store.  I'd like to do a visual like such below but I'm having a hard time even determining what the grouping ranges should be.  Is there a tool that can help me determine the grouping ranges (0-X, X-Y, Y-Z).  **I'm not limited to 3 groupings only, it's just for example purposes.   Thanks

Monthly Call Volume                    % of Total

0-X / Month                                     10%

X-Y / Month                                      40%
Y-Z / Month                                      50%

[Anonymous · Accepted solution]

Hi @jcastr02 ,

According to your description, here are my steps you can follow as a solution.

(1) This is my test data. 

(2) We can create calculated columns.

Rank = RANKX(FILTER('Table','Table'[store]=EARLIER('Table'[store])),'Table'[Call Volume],,ASC,Dense) 

Monthly Call Volume = 
var _a=CALCULATE(MIN('Table'[Call Volume]),FILTER('Table','Table'[store]=EARLIER('Table'[store]) && 'Table'[Rank]=1))
var _b=IF('Table'[Rank]=1,0 & "-" & _a)
var _c=CALCULATE(MIN('Table'[Call Volume]),FILTER('Table','Table'[store]=EARLIER('Table'[store]) && 'Table'[Rank]=EARLIER('Table'[Rank])))
var _d=CALCULATE(MIN('Table'[Call Volume]),FILTER('Table','Table'[store]=EARLIER('Table'[store]) && 'Table'[Rank]=EARLIER('Table'[Rank])+1))
RETURN SWITCH(TRUE(),
'Table'[Rank]=1,0 & "-" & _a & " /Month", 
'Table'[Rank] <> 1,_c & "-" & _d & " /Month")

(3) We can create a measure.

% of Total = 
var _a=CALCULATE(SUM('Table'[Call Volume]),FILTER(ALLSELECTED('Table'),'Table'[store]=MAX('Table'[store])))
var _b=CALCULATE(SUM('Table'[Call Volume]),FILTER(ALLSELECTED('Table'),'Table'[Rank]=MAX('Table'[Rank]) &&'Table'[store]=MAX('Table'[store])))
RETURN DIVIDE(_b,_a,0)

(4) Then the result is as follows.

If the above one can't help you get the desired result, please provide some sample data in your tables (exclude sensitive data) with Text format and your expected result with backend logic and special examples. It is better if you can share a simplified pbix file. Thank you.

Best Regards,

Neeko Tang

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

[Ashish_Mathur · Accepted solution]

Hi,

The grouping ranges will have to be created manually.  You may refer to my solution to a similar problem in the attached PBI file.

[Ashish_Mathur]

Hi,

The grouping ranges will have to be created manually.  You may refer to my solution to a similar problem in the attached PBI file.

[Anonymous · Accepted solution]

Hi @jcastr02 ,

According to your description, here are my steps you can follow as a solution.

(1) This is my test data. 

(2) We can create calculated columns.

Rank = RANKX(FILTER('Table','Table'[store]=EARLIER('Table'[store])),'Table'[Call Volume],,ASC,Dense) 

Monthly Call Volume = 
var _a=CALCULATE(MIN('Table'[Call Volume]),FILTER('Table','Table'[store]=EARLIER('Table'[store]) && 'Table'[Rank]=1))
var _b=IF('Table'[Rank]=1,0 & "-" & _a)
var _c=CALCULATE(MIN('Table'[Call Volume]),FILTER('Table','Table'[store]=EARLIER('Table'[store]) && 'Table'[Rank]=EARLIER('Table'[Rank])))
var _d=CALCULATE(MIN('Table'[Call Volume]),FILTER('Table','Table'[store]=EARLIER('Table'[store]) && 'Table'[Rank]=EARLIER('Table'[Rank])+1))
RETURN SWITCH(TRUE(),
'Table'[Rank]=1,0 & "-" & _a & " /Month", 
'Table'[Rank] <> 1,_c & "-" & _d & " /Month")

(3) We can create a measure.

% of Total = 
var _a=CALCULATE(SUM('Table'[Call Volume]),FILTER(ALLSELECTED('Table'),'Table'[store]=MAX('Table'[store])))
var _b=CALCULATE(SUM('Table'[Call Volume]),FILTER(ALLSELECTED('Table'),'Table'[Rank]=MAX('Table'[Rank]) &&'Table'[store]=MAX('Table'[store])))
RETURN DIVIDE(_b,_a,0)

(4) Then the result is as follows.

If the above one can't help you get the desired result, please provide some sample data in your tables (exclude sensitive data) with Text format and your expected result with backend logic and special examples. It is better if you can share a simplified pbix file. Thank you.

Best Regards,

Neeko Tang

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