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Phil88's avatar
Phil88
New Member
1 year ago
Solved

Normalize data ranges for Line Chart

Hi all,   i have four monetary datasets A B C D on a timeline that range between some thousands to millions and want to bring them together in a single line chart just to see the trends. By default...
Need Help
  • Rupak_bi's avatar
    Rupak_bi
    1 year ago

    Hi Phil88 ,

    Thanks for sharing the sample file. This helped me to understand my mistake. please modify the measure as below

    Linie_A = 
var min_val = calculate(min(Tabelle[A]),allselected(Tabelle[Datum]))
var max_val = calculate(max(Tabelle[A]),allselected(Tabelle[Datum]))
var current_val = max(Tabelle[A])
var factor = 100/(max_val - min_val)
return
(current_val-min_val)*factor

    Please refer the last graph .

     

    If this solves,  Plz accept as solution

     

     

Rupak_bi's avatar
Rupak_bi
Super User
1 year ago

hi Phil88 ,

You can scale them from 0 to 100 and plot in a single chart. 

You need to create separate measures for each lines as per below syntax
measure = 

var min_val = calculate(min(table , value),allselected(table, X-axix column))

var max_val = calculate(max(table , value),allselected(table, X-axix column))

var current_val = max(table,value)

var range_val = max_val - min_val

return

divide(current_val,range_val)

 

this should work. plz share sample data for exact representation. 

Now to see the actual value, in the data label section, you can refer the respective columns.

    • Rupak_bi's avatar
      Rupak_bi
      Super User
      1 year ago

      Hi Phil88 ,

      Thanks for sharing the sample file. This helped me to understand my mistake. please modify the measure as below

      Linie_A = 
var min_val = calculate(min(Tabelle[A]),allselected(Tabelle[Datum]))
var max_val = calculate(max(Tabelle[A]),allselected(Tabelle[Datum]))
var current_val = max(Tabelle[A])
var factor = 100/(max_val - min_val)
return
(current_val-min_val)*factor

      Please refer the last graph .

       

      If this solves,  Plz accept as solution

       

       

      • Phil88's avatar
        Phil88
        New Member
        1 year ago

        Thanks, works perfect