How to scale down a range of numbers with a known min and max value

Question

So I am trying to figure out how to take a range of numbers and scale the values down to fit a range   The reason for wanting to do this is that I am trying to draw ellipses in a java swing jpanel   I want the height and width of each ellipse to be in a range of say 1-30   I have methods that find the minimum and maximum values from my data set  but I won t have the min and max until runtime   Is there an easy way to do this

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Based on Charles Clayton s response  I included some JSDoc  ES6 tweaks  and incorporated suggestions from the comments in the original response    x000D   x000D      x000D     Returns a scaled number within its source bounds to the desired target bounds  x000D      param  number  n - Unscaled number x000D      param  number  tMin - Minimum  target  bound to scale to x000D      param  number  tMax - Maximum  target  bound to scale to x000D      param  number  sMin - Minimum  source  bound to scale from x000D      param  number  sMax - Maximum  source  bound to scale from x000D      returns  number  The scaled number within the target bounds  x000D      x000D  const scaleBetween    n  tMin  tMax  sMin  sMax    gt    x000D    return  tMax - tMin     n - sMin     sMax - sMin    tMin  x000D    x000D   x000D  if  Array prototype scaleBetween     undefined    x000D        x000D       Returns a scaled array of numbers fit to the desired target bounds  x000D        param  number  tMin - Minimum  target  bound to scale to x000D        param  number  tMax - Maximum  target  bound to scale to x000D        returns  number  The scaled array  x000D        x000D    Array prototype scaleBetween   function tMin  tMax    x000D      if  arguments length     1    tMax     undefined    x000D        tMax   tMin  tMin   0  x000D        x000D      let sMax   Math max    this   sMin   Math min    this   x000D      if  sMax - sMin    0  return this map num   gt   tMin   tMax    2   x000D      return this map num   gt   tMax - tMin     num - sMin     sMax - sMin    tMin   x000D      x000D    x000D   x000D                                                                      x000D     Usage x000D                                                                      x000D   x000D  let nums    10  13  25  28  43  50   tMin   0  tMax   100  x000D      sMin   Math min    nums   sMax   Math max    nums   x000D   x000D     Result    0 0  7 50  37 50  45 00  82 50  100 00   x000D  console log nums map n   gt  scaleBetween n  tMin  tMax  sMin  sMax  toFixed 2   join         x000D   x000D     Result    0  30 769  69 231  76 923  100   x000D  console log  -4  0  5  6  9  scaleBetween 0  100  join         x000D   x000D     Result    50  50  50   x000D  console log  1  1  1  scaleBetween 0  100  join         x000D   as-console-wrapper   top  0  max-height  100   important    x000D   x000D   x000D

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I ve taken Irritate s answer and refactored it so as to minimize the computational steps for subsequent computations by factoring it into the fewest constants   The motivation is to allow a scaler to be trained on one set of data  and then be run on new data  for an ML algo    In effect  it s much like SciKit s preprocessing MinMaxScaler for Python in usage   Thus  x     b-a  x-min   max-min    a  where b  a  becomes x    x b-a   max-min    min -b a   max-min    a which can be reduced to two constants in the form x    x Part1   Part2   Here s a C  implementation with two constructors  one to train  and one to reload a trained instance  e g   to support persistence    public class MinMaxColumnSpec            lt summary gt          To reduce repetitive computations  the min-max formula has been refactored so that the portions that remain constant are just computed once          This transforms the forumula from         x     b-a  x-min   max-min    a         into x    x b-a   max-min    min -b a   max-min    a         which can be further factored into         x    x Part1   Part2          lt  summary gt      public readonly double Part1  Part2            lt summary gt          Use this ctor to train a new scaler           lt  summary gt      public MinMaxColumnSpec double   columnValues  int newMin   0  int newMax   1                if  newMax  lt   newMin              throw new ArgumentOutOfRangeException  newMax    newMax must be greater than newMin             var oldMax   columnValues Max            var oldMin   columnValues Min             Part1    newMax - newMin     oldMax - oldMin           Part2   newMin    oldMin    newMin - newMax     oldMax - oldMin                    lt summary gt          Use this ctor for previously-trained scalers with known constants           lt  summary gt      public MinMaxColumnSpec double part1  double part2                Part1   part1          Part2   part2             public double Scale double x    gt   x   Part1    Part2

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I sometimes find a variation of this useful   Wrapping the scale function in a class so that I do not need to pass around the min max values if scaling the same ranges in several places Adding two small checks that ensures that the result value stays within the expected range   Example in JavaScript  class Scaler     constructor inMin  inMax  outMin  outMax        this inMin   inMin      this inMax   inMax      this outMin   outMin      this outMax   outMax         scale value        const result    value - this inMin     this outMax - this outMin     this inMax - this inMin    this outMin       if  result  lt  this outMin          return this outMin        else if  result  gt  this outMax          return this outMax             return result         This example along with a function based version comes from the page https   writingjavascript com scaling-values-between-two-ranges

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Here s some JavaScript for copy-paste ease  this is irritate s answer      function scaleBetween unscaledNum  minAllowed  maxAllowed  min  max      return  maxAllowed - minAllowed     unscaledNum - min     max - min    minAllowed      Applied like so  scaling the range 10-50 to a range between 0-100   var unscaledNums    10  13  25  28  43  50    var maxRange   Math max apply Math  unscaledNums   var minRange   Math min apply Math  unscaledNums    for  var i   0  i  lt  unscaledNums length  i        var unscaled   unscaledNums i     var scaled   scaleBetween unscaled  0  100  minRange  maxRange     console log scaled toFixed 2           0 00  18 37  48 98  55 10  85 71  100 00   Edit    I know I answered this a long time ago  but here s a cleaner function that I use now   Array prototype scaleBetween   function scaledMin  scaledMax      var max   Math max apply Math  this     var min   Math min apply Math  this     return this map num   gt   scaledMax-scaledMin   num-min   max-min  scaledMin       Applied like so     -4  0  5  6  9  scaleBetween 0  100         0  30 76923076923077  69 23076923076923  76 92307692307692  100

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For convenience  here is Irritate s algorithm in a Java form  Add error checking  exception handling and tweak as necessary   public class Algorithms        public static double scale final double valueIn  final double baseMin  final double baseMax  final double limitMin  final double limitMax            return   limitMax - limitMin     valueIn - baseMin     baseMax - baseMin     limitMin            Tester   final double baseMin   0 0  final double baseMax   360 0  final double limitMin   90 0  final double limitMax   270 0  double valueIn   0  System out println Algorithms scale valueIn  baseMin  baseMax  limitMin  limitMax    valueIn   360  System out println Algorithms scale valueIn  baseMin  baseMax  limitMin  limitMax    valueIn   180  System out println Algorithms scale valueIn  baseMin  baseMax  limitMin  limitMax     90 0 270 0 180 0

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Let s say you want to scale a range  min max  to  a b    You re looking for a  continuous  function that satisfies  f min    a f max    b   In your case  a would be 1 and b would be 30  but let s start with something simpler and try to map  min max  into the range  0 1    Putting min into a function and getting out 0 could be accomplished with  f x    x - min       gt    f min    min - min   0   So that s almost what we want   But putting in max would give us max - min when we actually want 1   So we ll have to scale it           x - min                                  max - min f x    ---------       gt    f min    0   f max     ---------   1        max - min                                 max - min   which is what we want   So we need to do a translation and a scaling   Now if instead we want to get arbitrary values of a and b  we need something a little more complicated           b-a  x - min  f x    --------------    a           max - min   You can verify that putting in min for x now gives a  and putting in max gives b   You might also notice that  b-a   max-min  is a scaling factor between the size of the new range and the size of the original range   So really we are first translating x by -min  scaling it to the correct factor  and then translating it back up to the new minimum value of a   Hope this helps

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I came across this solution but this does not really fit my need  So I  digged a bit in the d3 source code  I personally would recommend to do it like d3 scale does    So here you scale the domain to the range  The advantage is that you can flip signs to your target range  This is useful since the y axis on a computer screen goes top down so large values have a small y     public class Rescale       private final double range0 range1 domain0 domain1       public Rescale double domain0  double domain1  double range0  double range1            this range0   range0          this range1   range1          this domain0   domain0          this domain1   domain1             private double interpolate double x            return range0    1 - x    range1   x             private double uninterpolate double x            double b    domain1 - domain0     0   domain1 - domain0   1   domain1          return  x - domain0    b             public double rescale double x            return interpolate uninterpolate x              And here is the test where you can see what I mean  public class RescaleTest         Test     public void testRescale             Rescale r          r   new Rescale 5 7 0 1           Assert assertTrue r rescale 5     0           Assert assertTrue r rescale 6     0 5           Assert assertTrue r rescale 7     1            r   new Rescale 5 7 1 0           Assert assertTrue r rescale 5     1           Assert assertTrue r rescale 6     0 5           Assert assertTrue r rescale 7     0            r   new Rescale -3 3 0 1           Assert assertTrue r rescale -3     0           Assert assertTrue r rescale 0     0 5           Assert assertTrue r rescale 3     1            r   new Rescale -3 3 -1 1           Assert assertTrue r rescale -3     -1           Assert assertTrue r rescale 0     0           Assert assertTrue r rescale 3     1

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Here s how I understand it     What percent does x lie in a range  Let s assume you have a range from 0 to 100  Given an arbitrary number from that range  what  percent  from that range does it lie in  This should be pretty simple  0 would be 0   50 would be 50  and 100 would be 100    Now  what if your range was 20 to 100  We cannot apply the same logic as above  divide by 100  because   20   100   doesn t give us 0  20 should be 0  now   This should be simple to fix  we just need to make the numerator 0 for the case of 20  We can do that by subtracting    20 - 20    100   However  this doesn t work for 100 anymore because    100 - 20    100   doesn t give us 100   Again  we can fix this by subtracting from the denominator as well    100 - 20     100 - 20    A more generalized equation for finding out what   x lies in a range would be    x - MIN     MAX - MIN      Scale range to another range  Now that we know what percent a number lies in a range  we can apply it to map the number to another range  Let s go through an example   old range    200  1000  new range    10  20    If we have a number in the old range  what would the number be in the new range  Let s say the number is 400  First  figure out what percent 400 is within the old range  We can apply our equation above    400 - 200     1000 - 200    0 25   So  400 lies in 25  of the old range  We just need to figure out what number is 25  of the new range  Think about what 50  of  0  20  is  It would be 10 right  How did you arrive at that answer  Well  we can just do   20   0 5   10   But  what about from  10  20   We need to shift everything by 10 now  eg     20 - 10    0 5    10   a more generalized formula would be     MAX - MIN    PERCENT    MIN   To the original example of what 25  of  10  20  is     20 - 10    0 25    10   12 5   So  400 in the range  200  1000  would map to 12 5 in the range  10  20     TLDR  To map x from old range to new range   OLD PERCENT    x - OLD MIN     OLD MAX - OLD MIN  NEW X     NEW MAX - NEW MIN    OLD PERCENT    NEW MIN

[math] How to scale down a range of numbers with a known min and max value

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