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

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?

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 < 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 => (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]

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
{
    /// <summary>
    /// 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
    /// </summary>
    public readonly double Part1, Part2;

    /// <summary>
    /// Use this ctor to train a new scaler.
    /// </summary>
    public MinMaxColumnSpec(double[] columnValues, int newMin = 0, int newMax = 1)
    {
        if (newMax <= 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));
    }

    /// <summary>
    /// Use this ctor for previously-trained scalers with known constants.
    /// </summary>
    public MinMaxColumnSpec(double part1, double part2)
    {
        Part1 = part1;
        Part2 = part2;
    }

    public double Scale(double x) => (x * Part1) + Part2;
}

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);
    }
}

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

I sometimes find a variation of this useful.

1. 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
2. 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 < this.outMin) {
      return this.outMin;
    } else if (result > 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

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

Based on Charles Clayton's response, I included some JSDoc, ES6 tweaks, and incorporated suggestions from the comments in the original response.

/**_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) => {_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 => (tMin + tMax) / 2);_x000D_
    return this.map(num => (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 => 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(', '));

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